TITLE: Flux of $v=(3y_1-2y_2^2+y_3^2,3y_1^2-2y_2+y_3^2,3y_1^2-2y_2^2+y_3)$ through teardrop shape
QUESTION [2 upvotes]: $v=(3y_1-2y_2^2+y_3^2,3y_1^2-2y_2+y_3^2,3y_1^2-2y_2^2+y_3)$
Let $\varphi:K\to R^3$, where $K=[0,2\pi]\times[0,1]$ and $\varphi$ is given by $\varphi(x) = \{ (\cos x_1)(x_2-x_2^3), (\sin x_1)(x_2-x_2^3), x_2^2+ \frac13 x_2^6 \}$. Let $\tilde Y =\varphi(K)$. Calculate flux of $v$ throught $\tilde Y$.

So phi is this funny shape right here from what I understand:

I tried using the determinant method but I end up with these huge equations that I don't think I'll be able to solve. I also tried calculating the pullback, but again it's super long. Last idea I had was to maybe do a change of variables and set $u:=x_2-x_2^3$, but it doesn't really solve the problem.
How would you approach this? Detailed calculations of integrals in the explanation would be appreciated.

REPLY [3 votes]: Since the surface is closed, you may use the divergence theorem. In particular, if $V$ is the region bounded by $Y=\phi(K)$, then the flux of $v$ through $Y$ is given by
$$
\text{flux}= \int_Y v\cdot \text{ d} S= \int_{V}\text{div}(v)\text{ d}V= 2|V|,
$$
where $|V|$ denotes the volume of $V$.
We used the fact that the divergence of $v$ is
$$
\text{div}(v)= \frac{\partial}{\partial y_1}(3y_1-2y_2^2+ y_3^2)+ \frac{\partial}{\partial y_2}(3y_1^2-2y_2+y_3^2)+\frac{\partial}{\partial y_3}(3y_1^2-2y_2^2+y_3)=
2.
$$So we only need to compute the volume of the region enclosed by $Y$. Again, by the divergence theorem,
$$
|V|= \int_V\text{ d}V= \int_{Y}(0, 0,y_3)\cdot\text{ d}S= \int_{K}(0, 0, x_2^2+\frac13 x_2^6)\cdot (\frac{\partial\phi}{\partial x_1}\times \frac{\partial\phi}{\partial x_2})\text{ d}x_1\text{ d}x_2,
$$
because $\text{div}(0,0,y_3)=1$. We compute,
$$
\frac{\partial\phi}{\partial x_1}\times \frac{\partial\phi}{\partial x_2}= \det\begin{pmatrix} i & j & k \\ -\sin(x_1)(x_2-x_2^3) & \cos(x_1)(x_2- x_2^3) & 0 \\ (1-3x_2^2)\cos(x_1) & (1-3x_2^2)\sin(x_1) & 2x_2+ 2x_2^5\end{pmatrix}= (\cos(x_1) 2x_2^2(1-x_2^2)(1+x_2^4), \sin(x_1)2x_2^2(1-x_2^2)(1+x_2^4), -x_2(1-x_2^2)(1-3x_2^2)),
$$
thus,
$$
|V|= \int_K -x_2(1-x_2^2)(1-3x_2^2)(x_2^2+\frac13 x_2^6)\text{ d}x_1\text{d}x_2 \= \int_{x_1=0}^{2\pi}\int_{x_2=0}^1 -x_2^3+ 4x_2^5-\frac{4}{3}x_2^7+\frac43x_2^9- x_2^{11} \text{ d}x_1\text{d}x_2= \frac{3\pi}{5}.
$$
As a result, given that my calculations are correct, the flux of $v$ through the surface $Y$ is $\frac{6\pi}{5}$.


TITLE: Length of intersecting chords inside circle, given angle, vertex and circle radius
QUESTION [1 upvotes]: Let's assume a circle with the known radius of R and origin O. We know the position of the intersection of two lines within that circle: vector A. Those lines both terminate at the perimeter of the circle and we know the value of the angle a between those lines, for simplicity let's assume AB is at a right angle (Y-up). How does one determine the length of line CA or just the position of C (Note that it is not necessarily at the same angle as AO).
I think this one is hard to explain without a picture so either bare with the crappy ASCII art.
          B
   C..|''''..
  .'\ |      '.              B
 :   \|        :      C      |
:     A         :      \     |
:       O       :       \  a |
:               :        \.''|
 :             :          \  |
  '.         .'   CA = ?   \ |
    ''.....''               \|
                             A

Or see this image.

I will implement it with programming / scripting but need to figure out which method to use, I could "project" point A onto the circle if there's no better option.
Thanks in advance!

REPLY [0 votes]: Following formula can be also used to determine the length of $AC$ (see $\mathrm{Fig.\space 1}$)
$$AC=\sqrt{r^2-\left(x_A^2+y_A^2\right)\sin^2\left(\omega+\tan^{-1}\left(\dfrac{x_A}{y_A}\right)\right)}-\sqrt{x_A^2+y_A^2}\cos \left(\omega+\tan^{-1}\left(\dfrac{x_A}{y_A}\right)\right)$$
This formula gives the length of $AC$ for a point $A$ defined anywhere in the interior of a circle and two points $B$ and $C$ lying on its circumference making $\measuredangle BAC=\omega\enspace$ when measured counterclockwise from $B$ to $C$. $\mathrm{Fig.\space 1}$ to $\mathrm{Fig.\space 4\enspace}$ illustrate how $\angle BAC$ is defined. Furthermore, $x_A$ and $y_A$ are not signless lengths, but signed $x$- and $y$-Cartesian coordinates of $A$


TITLE: How can $\{1,2\}$ and $\{n \in \mathbb{N}: \text{there exists} \ x,y,z \in \mathbb{N^{+}}\text{such that}\ x^{n}+y^{n}=z^{n}\}$ be equal?
QUESTION [0 upvotes]: On my textbook, it says that $X_{1}=\{1,2\}$ and
$$X_{2}=\lbrace n \in \mathbb{N}: \text{there exists} \  x,y,z \in \mathbb{N^{+}}\text{such that}\ x^{n}+y^{n}=z^{n}\rbrace,$$
and $X_{1}=X_{2}.$
It also pointed out that this is true is because of Andrew Wiles's proof of Fermat's last theorem.
I don't know anything about Fermat's last theorem, would you give me some hints how to relate to this?

REPLY [3 votes]: Fermat's Last Theorem states that (where $n$ is a natural number):
$$\text{no three positive integers} \quad x,y,z\in\mathbb{N}\quad\text{satisfy the equation}\quad x^n+y^n=z^n \quad\text{for}\quad n>2$$
Therefore since
$$X_{2}=\lbrace n \in \mathbb{N}: \text{there exists} \  x,y,z \in \mathbb{N^{+}}\text{such that}\ x^{n}+y^{n}=z^{n}\rbrace,$$
we then have $X_2\subseteq X_1 $.
Note that $3^2+4^2=5^2$ and $3^1+2^1=5^1$. Therefore $X_1\subseteq X_2$. Thus $X_1=X_2$.


text:
\section*{References}
\ben

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Press} (1974).

\item[[Ce]] Cerf, J.: La Stratification Naturelle des Espace de Fonctions
Diff\'erentiables r\'eeles et la Th\'eor\`eme de la Pseudoisotopie.
{\em Publ. Math I.H.\'E.S.} {\bf 39} (1970).

\item[[FR]] Fenn, R., Rourke, C.: On Kirby's Calculus of Links. {\em Topology}, Vol.
{\bf 18} (1979) 1-15.

\item[[HW]] Hatcher, A., Wagoner, J.: Pseudo-Isotopies of Compact Manifolds.
{\em Ast\'erisque} {\bf 6}. (1973).

\item[[HKRL]] Hennings,M.A.: Invariants of Links and 3-Manifolds Obtained
from Hopf Algebras, Cambridge Preprint, 1990.

Kauffman, L., Radford, D.:
Invariants of 3-Manifolds derived from Finite Dimensional Hopf Algerbras,
Chicago Preprint, 1994.

Lyubashenko, V.: Invariants of 3-Manifolds and
Projective Representations of Mapping Class Groups via Quantum Groups at
Roots of Unity,  {\em Commun. Math. Phys} {\bf 172} (1995) 467-516.

\item[[Ig]] Igusa, K.: The Space of Framed Functions. {\em Trans. \hspace*{-.3cm}
Amer. \hspace*{-.3cm} Math. \hspace*{-.3cm} Soc.} Vol. {\bf 301-2}  (1987) 431-477.

\item[[J]] J\"anich, K.: Charakterisierung der Signatur von Mannigfaltigkeiten
durch eine Additivit\"atseigenschaft. {\em Invent. Math.} {\bf 6}(1968), 35-40.

\item[[Ke1]]Kerler,T.: Mapping Class Group Actions on Quantum Doubles.
 {\em Commun. Math. Phys.} {\bf 168} (1995) 353-388.

\item[[Ke2]]Kerler, T.: A Topological Hopf Algebra.(Talk presented at UNC,
April 1994, paper in preparation)

\item[[Ke3]]Kerler, T.: Equivalence of a Bridged Link
Calculus and Kirby's Calculus of Links on Non-Simply Connected 3-Manifolds.
{\em Topology Appl.} to appear.\newline
Available at {\tt http://www.math.ohio-state.edu/$\sim$kerler/papers/KC/}

\item[[Ki]] Kirby, R.: A Calculus for framed Links in $S^3\,$, {\em Invent. Math.} {\bf
45} (1978), 35-56.

\item[[KL]] Kerler, T., Lyubashenko,
V.: Non-semisimple Topological Quantum Field Theories for Connected
Surfaces. Preprint.

\item[[Li]] Lickorish, W.B.R.: A Representation of Orientable 3-Manifolds.
 {\em Ann. Math.} {\bf 76} (1962) 531-540.

\item[[Mi]] Milnor, J.: Morse Theory. Annals of Mathematical Studies.
{\em Princeton University Press} {\bf 51} (1969).

\item[[MP]] Matveev, S.,  Polyak, M.: A Geometrical Presentation of the Surface
Mapping Class Group and Surgery. {\em Commun. Math. Phys.} {\bf 160} (1994) 537.

\item[[RM]] Roberts, J.D., Masbaum, G.: On Central Extensions of Mapping
Class Groups. Mathematica Gottingensis, {\bf 42} (1993).

\item[[Ro]] Rohlin, V.A.: A Three Dimensional Manifold is the Boundary of a
Four Dimensional One, {\em Dokl.Akad. Nauk.} SSSR (N.S.) {\bf 114} (1951)
355-357.

\item[[Rf]] Rolfsen, D.: Knots and Links. Mathematics Lecture Series {\bf
7}. {\em Publish or Perish} (1976).

\item[[RT0]] Reshetikhin,N., Turaev,V.: Ribbon Graphs and Their Invariants
Derived from Quantum Groups. {\em Commun. Math. Phys.} {\bf 127}(1990), 1-26.

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Polynomials and Quantum Groups. {\em Invent. Math.} {\bf 103} (1991) 547-598.

\item [[Sw]] Sawin, S.: Extending the Kirby Calculus to Manifolds with
Boundary, MIT-preprint, 1994.

\item[[Tu]] Turaev, V.: Quantum Invariants of 3-Manifolds. {\em Walter de Gryter},
 Berlin (1994) (to appear).

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Orientable Surface. {\em Israel J. Math.} {\bf 45} (1982) 157-174.

\item[[Wa]] Wall, C.T.C.: Non-Additivity of the Signature, {\em Invent. Math.} {\bf 7}
(1969), 269-274.

\item[[Wa2]] Wall, C.T.C.: Determination of the Cobordism Ring, {\em Ann. Math.}
{\bf 72} (1960), 292-311.

\item[[Wc]] Wallace, A.H.: Modifications and Cobounding Manifolds, {\em Cam. J. Math.}
{\bf 12 } (1960), 503-528.

\item[[Wk]] Walker, K.: On Witten's 3-Manifold Invariants. UCSD-Preprint (1991).

\een

\bigskip

{\sc Harvard University, Cambridge, MA, USA}

present address:

{\sc The Ohio State University, Columbus, OH, USA}

{\tt kerler@math.ohio-state.edu}

config: arxiv


TITLE: Example for pairwise triangularizable but not all three.
QUESTION [2 upvotes]: I am not able give an example for the following problem on simultaneous triangularization. So, I thought I will post it here.
Give an example of three linear transformations $A,B$ and $C,$ such that the pairs $\lbrace A,B\rbrace$, $\lbrace B,C\rbrace$ and $\lbrace A,C\rbrace$ are simultaneously triangularizable, but the triplet $\lbrace A,B,C\rbrace$ is not simultaneously triangularizable.
Thank you.
ADDED LATER: For my need, I am looking for an example of linear transformations acting on vector spaces over $\mathbb{C}.$

REPLY [2 votes]: Here is a simple example:
$$ A = \begin{pmatrix}
        0 & 1 & \\
          & 0 & \\
          &   & 0
       \end{pmatrix},
   \quad
   B = \begin{pmatrix}
        0 & & \\
          & 0 & 1 \\
          &   & 0
       \end{pmatrix},
   \quad
   C = \begin{pmatrix}
         0 & & \\
           & 0 & \\
         1 &   & 0
       \end{pmatrix} .
$$
Every pair can be  triangularized (by a permutation matrix, by the way), but $A$, $B$ and $C$ have no common eigenvector, and so these three matrices can't be simultaneously triangularized.


TITLE: Why any open set in a topology can be expressed as a union of basis elements
QUESTION [2 upvotes]: I am reading Topology by Munkres. The following lemma is proven:

A topology $\mathcal{T}$ equals all unions of elements of the basis $\mathcal{B}$.

The reasonign is like this: let $U \in \mathcal{T}$, choose for each $x \in U$ a basis element $B_x \in \mathcal{B}$. Then:
$$U = \bigcup_{x \in U} B_x$$
I do not understand the final step. Shouldn't it be:
$$U \subseteq \bigcup_{x \in U} B_x$$
Consider a topology on $X$ with a single basis element $B = X$. Then $U \subset B$, if $U \neq X$. Alternatively, pick an open subset of a basis: $U \subset B_i$. Then $U \subset \bigcup_{x \in U} B_x = B_i$.
Munkres defines a basis on $X$ as a collection $\mathcal{B}$ of subsets $B_i$ of $X$ satisfying:
$$\forall x \in X \Rightarrow \exists B_i: x \in B_i$$
$$x \in B_1 \cap B_2 \Rightarrow \exists B_3: x \in B_3 \subset B_1 \cap B_2$$

REPLY [3 votes]: Munkres sort of has 2 notions of a base :

We start with a topology $(X,\mathcal{T})$. A base for this particular space is a collection of open subsets $\mathcal{B}$ such that for each open set $U$ of $X$, and each $x \in U$, there exists some $B_x \in \mathcal{B}$ such that $x \in B_x \subset U$. One can then check that every $U \in \mathcal{T}$ is a union of base elements: for every $x \in U$ pick $B_x$ from the definition. Then $U = \cup \{B_x: x \in U\}$ because, as all $B_x \subset U$, we have the right to left inclusion and as each $x \in U$ is (At least) in its "own" $B_x$, we have the left to right inclusion. So a base is a more compact way to describe a topology. It also satisfies the 2 (necessary) conditions, because $X$ is open and the intersection of two base elements is open,so for each $x$ in it we have such a $B_3$.
We can start with a set $X$. A collection of subsets $\mathcal{B}$ that satisfy those two conditions, is a base for some unique topology $\mathcal{T}$.
It's clear how it should be defined: it's the collection of all unions of subcollections of $\mathcal{B}$. By definition $\mathcal{T}$ then contains $\emptyset$ (as the empty union), and is closed under all unions by construction, and the first condition gives that $X$ is in $\mathcal{T}$, while the second one implies that the intersection of two members of $\mathcal{T}$ is again in $\mathcal{T}$(some minor reasoning required), so this definition indeed gives a topology that $\mathcal{B}$ is a base for: any $O = \cup \mathcal{B'}$ for some $\mathcal{B'} \subset \mathcal{B}$. so any $X \in O$ is in some member of $\mathcal{B'}$ by definition. And it's clear by 1. that this is the only way we could have defined $\mathcal{T}$. Here we use a base to generate a topology ,not to describe a pre-given one more compactly.

Your question probably referred to the first notion: a base for a given space.


text:
\section{Searching for inductive biases}
\label{ss:searching}
To formalize the way we look for inductive biases of a learner $\mathcal{M}$, we consider a training dataset of input/output pairs, $T = \{x_i, y_i\}_{i=1}^n$, and a hold-out set of inputs, $H=\{x_i\}_{i=n+1}^k$. W.l.o.g, we assume that there are two candidate ``rules'' that explain the training data, but do not coincide on the hold-out data: $\mathcal{C}_1(x_i) = \mathcal{C}_2(x_i) = y_i, ~~ 1 \le i \le n$ and $\exists i: \mathcal{C}_1(x_i) \neq \mathcal{C}_2(x_i), n + 1 \le i \le k$.

To compare preferences of a learner  $\mathcal{M}$ toward those two rules, we fit the learner on the training data $T$ and then compare its predictions on the hold-out data $H$ to the outputs of the rules. We refer to this approach as ``intuitive''. Usually, the measures of similarity between the outputs are task-specific: \cite{Mccoy2020} used accuracy of the first term, \cite{Zhang2019} used correlation and MSE, and \cite{Lake2017} used accuracy calculated on the entire output sequence.

We too start with an accuracy-based measure. We define the fraction of perfect agreement (FPA) between a learner $\mathcal{M}$ and a candidate generalization rule $\mathcal{C}$ as the fraction of seeds that generalize \emph{perfectly} in agreement with that rule on the hold-out set $H$.
Larger FPA of $\mathcal{M}$ is w.r.t.\ $\mathcal{C}$,  more biased $\mathcal{M}$ is toward $\mathcal{C}$. However, FPA does not account for imperfect generalization nor allows direct comparison between two candidate rules when both are dominated by a third candidate rule. Hence, below we propose a principled approach based on the description length.

\textbf{Description Length and Inductive Biases}
At the core of the theory of induction~\citep{Solomonoff1964} is the question of continuation of a finite string that is very similar to our setup. Indeed, we can easily re-formulate our motivating example
as a string continuation problem: ``$3\rightarrow6;4\rightarrow$''. The solution proposed by~\cite{Solomonoff1964} is to select the continuation that admits ``the simplest explanation'' of the entire string, i.e.\ that is produced by programs of the shortest length (description length).

Our intuition is that when a continuation is ``simple'' for a learner, then this learner is biased toward it.
{We  consider a learner $\mathcal{M}$ to be biased toward
$\mathcal{C}_1$ over $\mathcal{C}_2$ if the training set and its extension according to $\mathcal{C}_1$ has a shorter description length (for $\mathcal{M}$) compared to that of $\mathcal{C}_2$.} Denoting description length of a dataset $D$ under the learner $\mathcal{M}$ as $L_{\mathcal{M}}(D)$, we hypothesise that if $L_{\mathcal{M}}(\{ \mathcal{C}_1(x_i)\}_{i=1}^k) < L_{\mathcal{M}}( \{ \mathcal{C}_2(x_i)\}_{i=1}^k)$, then  $\mathcal{M}$ is biased toward $\mathcal{C}_1$.

\textbf{Calculating Description Length}
 To find the description length of data under a fixed learner, we use the online (prequential) code~\citep{Rissanen1978,Grunwald2004,Blier2018}.

The problem of calculating $L_{\mathcal{M}}(D)$, $D = \{x_i, y_i\}_{i=1}^k$ is considered as a problem of transferring outputs $y_i$ one-by-one, in a compressed form, between two parties, Alice (sender) and Bob (receiver). Alice has the entire dataset $\{x_i, y_i\}$, while Bob only has inputs $\{x_i\}$.
Before the transmission starts, both parties agreed on the initialization of the model $\mathcal{M}$, order of the inputs $\{x\}$, random seeds, and the details of the learning procedure. Outputs $\{y_i\}$ are sequences of tokens from a vocabulary $V$.
W.l.o.g.\ we fix some order over $\{x\}$. We assume that, given $x$, the learner $\mathcal{M}$ produces a probability distribution over the space of the outputs $y$, $p_{\mathcal{M}}(y | x)$.

The very first output $y_1$ can be sent by using not more than $c = |y_1| \cdot \log |V|$ nats, using a na\"ive encoding.\footnote{As we are interested in comparing candidate generalization rules, the value of the additive constant $c$ is not important, as it is learner- and candidate-independent. In experiments, we subtract it from all measurements.} After that, both Alice and Bob update their learners using the example $(x_1, y_1)$, available to both of them, and get identical instances of $\mathcal{M}_1$.

Further transfer is done iteratively under the invariant that both Alice and Bob start every step $t$ with exactly the same learners $\mathcal{M}_{t-1}$ and finish with identical $\mathcal{M}_t$. At step $t$ Alice would use $\mathcal{M}_{t-1}$ to encode the next output $y_t$. This can be done using $\left(-\log p_{\mathcal{M}_{t-1}}(y_{t} | x_{t})\right)$ nats~\citep{Mackay2003}. Since Bob has exactly the same model, he can decode the message to obtain $y_{t}$ and use the new pair $(x_{t}, y_{t})$ to update his model and get $M_{t}$. Alice also updates her model, and proceeds to sending the next $y_{t+1}$ (if any), encoding it with the help of $\mathcal{M}_{t}$.

The cumulative number of nats transmitted is:
\begin{equation}
\label{eq:online}
    L_{\mathcal{M}}(D) = - \sum_{t=2}^k \log p_{\mathcal{M}_{t-1}}(y_t | x_t) + c.
\end{equation}

The obtained code length of Eq.~\ref{eq:online} depends on the order in which $y$ are transmitted and the procedure we use to update $\mathcal{M}.$  To account for that, we average out the data order by training with multiple random seeds. Further, for larger datasets, full re-training after adding a new example is impractical and, in such cases, examples can be transmitted in blocks.

If we measure the description length of the training data $T$ shuffled with the hold-out data $H$,
both datasets  would have symmetric roles. However, there is a certain asymmetry in the extrapolation problem: we are looking for an extrapolation from $T$, not vice-versa. To break this symmetry, we always transmit outputs for the entire training data as the first block.

While $L_{\mathcal{M}}(D)$ is seemingly different from the ``intuitive'' measures introduced before, we can illustrate their connection as follows. Consider a case where we first transmit the training outputs as the first block and \emph{all} of the hold-out data outputs under $\mathcal{C}$, $\mathcal{C}(H)$, as the second block. Then the description length is equal to cross-entropy of the trained learner on the hold-out data,
recovering a process akin to the ``intuitive'' measuring of inductive biases.
With smaller blocks, the description length also catches whether a learner is capable of finding regularity in the data fast, with few data points; hence it also encodes the speed-of-learning ideas for measuring inductive biases~\citep{Chaabouni2019a}.

Finally, the description length has three more attractive properties when measuring inductive biases: (a) it is domain-independent, i.e.\ can be applied for instance in the image domain, (b) it allows comparisons across models that account for model complexity, and (c) it enables direct comparison between two candidate rules (as we will show in the Section~\ref{s:experiments}).


TITLE: Charging sphere by induction
QUESTION [0 upvotes]: In the above picture, I near positively rod to a neutral sphere.
Then I ground the right side (negative side).
Why electrons enter to sphere from ground?

REPLY [1 votes]: Initially these conducting spheres are neutral bodies but when a positively charged conductor is brought near to them, the charges inside the sphere are redistributed on the surface of the sphere in such a way that the charges have minimum repulsions as you have shown.
But due to the positively charged rod, the spheres aquaire some potential. For general idea of what is happening during earthing let us consider a point charge $+Q$, then initial potential of sphere due to charge $+Q$ is $$V_{i}=kQ/r$$ where r is distance from positive charge $+Q$ to the sphere.
When two conductors of different potentials are connected by a conducting wire, charge transfer takes place between the conductors until they attain same potential.
We know that Earth is also a conductor with huge number of charges and almost negligible potential due to it's large radius (for reference we consider it to be zero). So when the sphere is earthed, charge transfer takes place between Earth and the sphere until the potential of the sphere becomes zero.
In the case you have shown, negative charges flow from Earth to the conductor so that this negative charge cancels the potential on the sphere due to positive charge of the rod and bring the sphere to zero potential.
Let $q_{e}$ be the negative charge that flows from Earth to sphere and $V_{f}$ be the final potential of the sphere which equals to zero. R be radius of the sphere.
We express it as:
$$V_{f}=kQ/r + kq_{e}/R = 0$$
Here $kQ/r$ is potential on the Sphere due to positively charge and  $kq_{e}/R$ is potential on the sphere due to negative charges that flow from earth to conductor. They cancel out each other which brings the sphere to zero potential.
So to bring the sphere to zero potential (reference to Earth), the electrons enter from ground to sphere.


text:
\begin{document}
\begin{abstract}{It is established that non-isotropic vector field with Jacobi operator
of maximal rank is an obstacle for the existence of non-trivial second-order symmetric
parallel tensor field. It follows that such manifold
as pseudo-Riemannian manifold is locally non-reducible.  In particular result is applied
 to widely studied classes of almost para-contact metric manifolds -
 para-contact metric, para-cosymplectic or para-Kenmotsu manifolds
  satisfying nullity and generalized nullity conditions. As corollary we have the
  following theorem: almost para-contact metric manifold with
  maximal rank Jacobi operator of characteristic vector field is locally non-isometric
  to Riemann product. }
\end{abstract}

\maketitle
\section{Introduction}

Many authors has recently studied  the problem of the existence of non-trivial  parallel
quadratic form  on almost para-contact metric manifold which  satisfies generalized nullity conditions.
Let us recall it is said that almost para-contact metric manifold
satisfies generalized nullity condition if
\begin{multline}
\label{nullcond}
R_{XY}\xi = \kappa (\eta(Y)X-\eta(X)Y)+
\mu (\eta(Y)hX-\eta(X)Y)+ \\ \nu(\eta(Y)h'X-\eta(X)h'Y),
\end{multline}
$R$ is the operator of the Riemann curvature, and $\kappa$, $\mu$, $\nu$ are functions,
$\xi$ is the characteristic vector field. The more elaborate terminology is that characteristic vector
field belongs to generalized nullity distribution.
For the studied classes of manifolds the results are that in generic case
  there is no parallel quadratic form, different from metric tensor  up to non-zero  multiplier.
The proofs of these results are based
on   properties of Jacobi operator  $X\mapsto J_\xi=R_{X\xi}\xi$,
particularly its algebraic form.
However more careful analysis shows that,  for above mentioned generic  cases, $J_\xi$
has maximal rank. Going into this direction we have
\begin{proposition}
There is no non-trivial parallel quadratic form if Jacobi operator
of characteristic vector field has maximal rank. Manifold as Riemannian
manifold is locally irreducible.
\end{proposition}
The latter sentence comes from the fact that Riemann product always admit
non-trivial parallel differential quadratic form.

 Let's denote by $\chi(\xi,x)$ the  characteristic polynomial of the Jacobi operator
 \begin{equation}
\chi(\xi,x)=x^n-\omega_1(\xi)x^{n-1}+\ldots (\pm 1)^{n-1}\omega_{n-1}(\xi)x,
\end{equation}
we will show the following
\begin{proposition}
Assuming $\xi$ is non-isotropic, $J_\xi$ has maximal rank if and only if the coefficient
$\omega_{n-1}$ at the lowest power term is non-zero $\omega_{n-1}\neq 0$.
\end{proposition}
Of course this assumption is always satisfied for characteristic vector field on almost
para-contact metric manifold. Yet our results are remain valid in wider framework
of pseudo-Riemannian manifolds. Then fact that $\xi$ is non-isotropic is essential.

Author would like to express his gratitude  to Professor Quanxiang Pan for our fruitfull discussion.

\section{Preliminaries}
Let $(\mathcal M, g)$ be pseudo-Riemannian $n$-dimensional manifold, $\nabla$ denote the Levi-Civita connection, and  $R_{XY}Z$ its curvature
\begin{equation}
R_{XY}Z = \nabla_X\nabla_Y Z -\nabla_Y\nabla_X Z - \nabla_{[X,Y]} Z,
\end{equation}
it is assumed that letters $U$, $V$, $X$,... are used  to denote
vector fields, if it is not stated otherwise. Our convention for the Riemann covariant curvature  tensor  is
\begin{equation}
R(X,Y,Z,W) = g(R_{XY}Z,W).
\end{equation}
For vector field $\xi$ on $\mathcal M$,  $(1,1)$-tensor field
$X \mapsto J_\xi X=R_{X \xi}\xi$, is called  Jacobi operator.
 From the properties of the curvature $J_\xi$ is $g$-self-adjoint and uppper limit of its rank is
$$
   g(J_\xi X,Y)=g(J_\xi Y,X),\quad r < {\rm dim}\,\mathcal M.
$$
If $g$ is definite, $J_\xi$ is semi-simple: eigenvalues are real and tangent space splits
into orthogonal direct sum of corresponding eigen-spaces.

Given Riemannian manifold $\mathcal M$, A. Gray \cite{Gray}, considered  so-called k-nullity distribution $\mathcal N_k$, $k=const.$,
distribution where the curvature has algebraic form  as curvature of constant sectional curvature manifold
\begin{equation}
R_{XY}Z = k ( g(Y,Z)X  -  g(X,Z)Y),
\end{equation}
where $Z$ is section of $\mathcal N_k$.

\subsection{Almost para-contact metric manifolds}(\cite{Blair}, \cite{Zam})

Let $\mathcal M$ be $(2n+1)$-dimensional manifold.
Almost para-contact metric structure $(\varphi, \xi,\eta,g)$  is  quadruple of tensor fields: $(1,1)$-tensor field (affinor) $\varphi$,
characteristic (or Reeb) vector field $\xi$, characteristic 1-form $\eta$,
and pseudo-Riemannian metric $g$, such that, $\varepsilon=\pm 1$,
\begin{eqnarray}
\label{def1} & \varphi^2 = \varepsilon(Id -\eta\otimes \xi), \quad \eta(\xi) =1, & \\[+4pt]
\label {def2} & g(\varphi X, \varphi Y) = \varepsilon(g(X,Y) -\eta(X)\eta(Y)). &
\end{eqnarray}

For $\varepsilon=-1$, structure is customary called almost contact metric.
Eigenvalues of $\varphi$ are imaginary, spectrum contains $\{0,-i,i\}$. It is assumed that the metric
is strictly positive. The latter condition implies that eigenvalues $-i$, $i$ have the same multiplicity
$n$.
The triple $(\varphi, \xi, \eta)$  is called almost contact structure.

For $\varepsilon=+1$, structure is called almost para-contact metric.
Eigenvalues of $\varphi$ are real, the spectrum is  $\{-1,0,+1\}$.  Tangent space decomposes into direct sum of one-dimensional kernel,
and  $n$-dimensional eigen-spaces $\mathcal V(\pm 1)$.
From definition it follows that restriction of the metric  to any of eigen-space is null tensor,
and $\mathcal V(\pm 1)$ are maximal in dimension isotropic subspaces.  Signature of $g$ is
\begin{equation}
 \underbrace{-1,\ldots,-1}_{n}, \underbrace{+1,\ldots,+1}_{n+1}
\end{equation}
Operators
$P_\pm=\varphi \pm Id$, are orthogonal projectors $P_\pm^2=P_\pm $ , $P_{+}P_{-}=P_{-}P_{+}=0$,
onto eigen-spaces $\mathcal V(\pm 1)$.
We have ${\rm Im}\,P_\pm = \mathcal V(\mp 1)$.
In particular
$g(P_\pm X, P_\pm Y)=0$, $g(P_{\pm}X,P_\mp Y)=g(X,Y)$, for $\eta(X)=\eta(Y)=0$.
 Per analogy the triple $(\varphi, \xi, \eta)$,  is called almost para-contact structure.

From now on we adopt in this paper the convention where both almost contact
metric and almost para-contact metric structures are all together called
almost para-contact metric manifolds. We just follow the line where some authors
use term pseudo-Riemannian manifold in wider sense: manifold equipped with non-degenerate
quadratic differential form. So the reader should be aware of this.

For almost para-contact metric structure tensor field $\varPhi(X,Y) = g(X,\varphi Y)$, is skew-symmetric form called fundamental form.
It satisfies
\begin{equation}
\eta\wedge \varPhi ^n \neq 0,
\end{equation}
at every point, so $\varPhi$ has maximal rank everywhere, its kernel is spanned by characteristic vector field $\xi$.

Manifold equipped with almost para-contact metric structure is called almost para-contact metric manifold.
Such manifold is always orientable.

An important notion is normality. Almost para-contact metric manifold is called normal if
\begin{equation}
[\varphi,\varphi](X,Y)-2\varepsilon\, d\eta\otimes\xi =0,
\end{equation}
where $[\varphi,\varphi]$ denotes Nijenhuis torsion of $\varphi$
\begin{equation*}
[\varphi X,\varphi Y]=\varphi^2 [X,Y]+[\varphi X, \varphi Y]-\varphi ([\varphi X,Y]+[X,\varphi Y]).
\end{equation*}

 Non-degenerate hypersurface of almost para-Hermitian manifold can be equipped with
 almost para-contact metric structure.  Thus such hypersurfaces are one of the fundamental  examples of almost para-contact
 metric manifolds.

We just mention some classes of almost para-contact metric manifolds.

\begin{definition}
\label{almcont}
{\rm (\cite{Blair}, \cite{MonNicYud}, \cite{DilPas})}.
{\rm Almost para-contact metric
manifold $(\mathcal M, \varphi,\xi,\eta, g)$ is called }
\begin{enumerate}
\item para-contact metric
\begin{equation*}
d\eta = \varPhi,
\end{equation*}
\item almost para-cosymplectic (or almost para-coKaehler)
\begin{equation*}
d\eta =0, \quad d\Phi =0,
\end{equation*}
\item almost para-Kenmotsu
\begin{equation*}
d\eta =0, \quad d\Phi = 2\eta\wedge\Phi.
\end{equation*}
\end{enumerate}
\end{definition}

Assuming additionally  normality
we obtain following classes of manifolds: para-Sasakian\footnote{Contact metric and normal, etc},
 para-cosymplectic (or para-coKaehler) and para-Kenmotsu.

Let define
$$
                    h=\frac{1}{2}\mathcal L_\xi \varphi, \quad h'=h\circ\varphi ,
$$
$\mathcal L_\xi$ denotes the Lie derivative along $\xi$. Applying $\mathcal L_\xi$ to identity $\varphi \xi=0$,
we obtain $h\xi=0$ ($h'\xi=0$ is evident).
 For given appropriate functions $\kappa$, $\mu$, $\nu$
- the choice depends on the structure in question -   almost para-contact metric manifold, such that
\begin{multline}
R_{XY}\xi = \kappa(\eta(Y)X-\eta(X)Y)+\mu(\eta(Y)hX -\eta(X)hY) + \\
                                          \nu(\eta(Y)h'X-\eta(X)h'Y),
\end{multline}
is called $(\kappa,\mu,\nu)$-space or $(\kappa,\mu,\nu)$-almost para-contact metric manifold,
etc. here authors
adopt different naming conventions.
The Jacobi
operator of characteristic vector field on almost contact metric $(\kappa,\mu,\nu)$-space is
\begin{equation}
J_\xi: X \mapsto R_{X\xi}\xi = -\kappa\eta(X)\xi + (\kappa Id+\mu h +\nu h')X  ,
\end{equation}
cf.  \cite{BlKoufPap}, \cite{Boeckx}, \cite{DacOl}, \cite{DilPas}. Note that $J_\xi$ has
maximal rank if and only if
$$
J_\xi |_{ \eta = 0} = \kappa Id  +\mu h +\nu h',
$$
is invertible on $ \eta = 0$.

\section{Symmetric parallel tensors of pseudo-Riemannian manifolds}
The goal of this section is to prove the following result.
\begin{proposition}
\label{symm}
Let $(\mathcal M,g)$ be $(n+1)$-dimensional pseudo-Riemannian manifold. Let assume there is non-isotropic vector field
$\xi$, $g(\xi,\xi)\neq 0$, with Jacobi operator   of maximal rank. If $\alpha \neq 0$ is parallel differential quadratic form,
then it is proportional to metric tensor $\alpha=c g$.
\end{proposition}
\begin{proof}
We may assume $g(\xi,\xi)=\epsilon$, $\epsilon =\pm 1$. So  there is $(0,2)$-tensor $\alpha(X,Y)=\alpha(Y,X)$ symmetric and parallel $\nabla \alpha =0$, such that our quadratic differential form is just $X \mapsto \alpha(X,X)$.
Moreover by assumption $r={\rm dim (Im}\,J_\xi)=n$.
We set
\begin{equation}
i_\xi\alpha(X)=\alpha(\xi,X), \quad i_\xi g(X)=g(\xi,X).
\end{equation}
Let consider the case  $i_\xi\alpha\neq 0$. For
$\alpha$ and $g$  are both parallel
\begin{equation}
i_\xi\alpha(J_\xi \,\cdot) =0, \quad i_\xi g(J_\xi\, \cdot)=0,
\end{equation}
hence
$$
{\rm Im }\,J_\xi \subset ( \iota_\xi\alpha =0 ) \cap ( \iota_\xi g=0 ),
$$
and assumption that $J_\xi$ has maximal rank, implies
$$
{\rm Im }\,J_\xi = ( \iota_\xi\alpha =0 ) = ( \iota_\xi g=0 ),
$$
in particular forms $i_\xi\alpha$ and $i_\xi g$ are
collinear
$i_\xi\alpha = \epsilon\alpha(\xi,\xi)i_\xi g$.
We apply now the second covariant derivatives $\nabla_{Y,\xi}^2$, and $\nabla_{\xi,Y}^2$ resp. to the both sides of the identity
\begin{equation}
\label{coll}
\alpha(\xi,X)=\epsilon \alpha(\xi,\xi)g(\xi,X),
\end{equation}
we have to take into account that by assumption $\alpha$ is covariant constant.
Subtracting resulting equations, having in
mind that $R_{Y\xi}\xi = \nabla_{Y,\xi}^2-\nabla_{\xi,Y}^2$, we obtain
\begin{equation}
\alpha(J_\xi Y, X) =\epsilon\alpha(\xi,\xi)g( J_\xi Y, X),
\end{equation}
which by maximality  follows
\begin{equation}
\alpha = c g, \quad c=\epsilon\alpha(\xi, \xi) \neq 0,
\end{equation}
and by $0=\nabla \alpha = dc\otimes g $, we have $c=const$.

In the case  $i_\xi\alpha =0$,  in the similar way  as above we find
$\alpha(X,Y) = 0$, $g(\xi,Y)=0$. For $\xi$ is non-isotropic and
$i_\xi\alpha =0$, $\alpha$ must vanish which contradicts our assumption $\alpha \neq 0$.
Hence the case $i_\xi\alpha =0$ is not  possible.
\end{proof}

\subsection{Trace form}
Let $\mathcal M$ be $(n+1)$-dimensional pseudo-Riemannian manifold. Let recall formula
for characteristic polynomial
\begin{equation}
\chi(\xi,x)=det(x Id -J_{\xi})=x^{n+1} -\omega_1(\xi) x^{n}+\ldots (-1)^{n}\omega_{n}(\xi)x.
\end{equation}
Note that $\omega_i$'s as a functions
$$
  : \xi \mapsto \omega_{i}(\xi)
$$
are all smooth differential forms. Indeed, let
$J^{i}_{\xi}$ denotes the $i$-th exterior power of $J_{\xi}$.
It is $(i,i)$-tensor field - interpreted as endomorphism acting on $i$-th degree polivectors
on $\mathcal M$. By definition on simple polivector $\mathcal W = V_1\wedge\ldots V_i$
$$
J^{i}_{\xi}(V_1\wedge\ldots V_i) = J_{\xi}(V_1)\wedge\ldots J_{\xi}(V_i),
$$
then $\omega_{i}=(-1)^{n} tr(J^{i}_{\xi})$. From the
definition of Ricci tensor we have $Ric(\xi,\xi)=\omega_{1}(\xi)$. So
we may think of $\omega_{i}$, $i > 1$, as a kind of Ricci tensors of higher
degrees.

Let us recall the metric tensor gives rise to canonical symmetric bilinear form on
polivectors: we denote it by $g^{\wedge k}$
\begin{equation}
g^{\wedge k}(X_1\wedge\ldots X_k) = det [ g(X_i,X_j)].
\end{equation}

The metric $g^{\wedge k}$ and $J^{k}_{\xi}$ are compatible in the sense
that the latter is $g^{\wedge k}$-self-adjoint.
\begin{proposition}
Let assume $J^{n}_{\xi}\neq 0$. Then $J^n_{\xi}$ has rank one. In particular there is $n$-form $\tau_{\xi}$ and
$n$-multivector $\mathcal W_{\xi}$, such that
\begin{equation}
J^{n}_{\xi}	= \tau_{\xi}\otimes \mathcal W_{\xi},\quad \omega_{n}(\xi)=\tau_{\xi}(\mathcal W_{\xi}),
\end{equation}
and $\omega_{n}(\xi) = 0$ if and only if $\xi$ is isotropic.
\end{proposition}
 \begin{proof}
 Let extend $\xi$ to local frame $(\xi, X_1,\ldots,X_{n})$, then
 \begin{equation}
 \label{base}
\mathcal W_i = \xi\wedge X_{1}\ldots  \wedge \hat{ X}_{i}\wedge\ldots X_n, \quad 1 \leq i \leq n,
\end{equation}
span the  kernel of $J^{n}_\xi$,
we see that dimension of kernel is $n$, so
if $J^{n}_\xi \neq 0$  its image is one-dimensional subspace, so
there is polivector $\mathcal W$ and $n$-form $\tau$, such that
\begin{equation}
J^{n}_\xi = \tau\otimes \mathcal W,\quad
\omega_{n}=tr(J^{n}_\xi) = \tau(\mathcal W),
\end{equation}
$\tau$ and $\mathcal W$ are determined up to re-scaling
\begin{equation}
	\tau \mapsto f\tau, \quad \mathcal W \mapsto f^{-1}\mathcal W,
\end{equation}
where $f$ is arbitrary non-vanishing function.

Condition $tr(J^{n}_{\xi})=\tau(\mathcal W)=0$ means that  $\mathcal W$ itself belongs to
the kernel of $J^{n}_{\xi}$. Every element of the kernel is linear
combination of polivectors as in (\ref{base}), therefore $\mathcal W$
 has decomposition
 $$
 \mathcal W = \xi\wedge \mathcal W_{0},
$$
For some simple polivector $X_{1}\wedge\ldots X_{n}$ we have
$$
 J^{n}_\xi (X_1\wedge\ldots X_{n}) = \mathcal W,
$$
Then
\begin{equation}
\xi\wedge \mathcal W_0 = \mathcal W = J^{n}_\xi (X_1\wedge\ldots X_{n})
= (R_{X_1\xi}\xi)\wedge\ldots (R_{X_{n}\xi}\xi),
\end{equation}
which follows that up to reoder
\begin{equation}
R_{X_1\xi}\xi = c_{0} \xi + c_{1}X_{1}+\ldots c_{n}X_{n}, \quad c_0 \neq 0,
\end{equation}
from other hand
$$
  0=g(R_{X_{1}\xi}\xi,\xi)=c_{0}g(\xi,\xi)+\sum\limits_{i=1}^{n}c_{i}g(X_{i},\xi),
$$
If $g(\xi,\xi)\neq 0$, we
would take all $X_{1},\ldots X_{n}$, such that $g(X_{i},\xi)=0$
then from the above equation we will have $c_{0}=0$ - contradiction.
Hence $g(\xi,\xi)=0$.
\end{proof}
Note Jacobi operator has maximal rank if and only if its exterior power $J^n_\xi \neq 0$.
\begin{corollary}
Jacobi operator of non-isotropic vector field has maximal rank if and only if coefficient at lowest term
of its characteristic polynomial is non-zero.
\end{corollary}

Note there are non-trivial
parallel quadratic differential forms on Riemannian products.
\begin{corollary}
Let $(\mathcal M, g)$ be pseudo-Riemannian manifold. If every point admits
locally defined vector field with maximal rank Jacobi operator, then
$\mathcal M$ is locally irreducible.
\end{corollary}

\section{Applications to almost para-contact metric manifolds}

Applications are direct. Assume $(\mathcal M, \varphi,\eta,g)$ is almost
para-contact metric manifold.
\begin{proposition}
If Jacobi operator of characteristic vector field has maximal rank, then
 non-zero parallel second order differential form is proportional to
pseudo-length form. In particular manifold is locally irreducible.
\end{proposition}

In case of $\kappa$-nullity spaces, $\kappa \neq 0$, we just rephrase above result.

\begin{corollary} Non-zero parallel second order differential form on
$\kappa$-nullity almost para-contact metric manifold is proportional to pseudo-length form
provided $\kappa \neq 0$.
\end{corollary}

The case of $(\kappa,\mu)$-spaces requires study of singular values of
operator $\kappa Id +\mu h$.  Let denote by
$\omega(x)=\sum\limits_{i=0}c_ix^{n-i}$, $c_0=1$, the characteristic polynomial of $h$.
Those values are solutions $(\kappa, \mu)$ of polynomial equation
\begin{equation}
det(\kappa Id+\mu h) = \sum\limits_{i=0}c_i \kappa^{n-i}\mu^i=0,
\end{equation}
are singular values.

The case of almost contact
metric manifold is simpler due to fact that $h$ is
diagonalizable. Denoting by $\lbrace \lambda_1,\ldots\lambda_k \rbrace$ spectrum of $h$,
we see that
$(\kappa,\mu)$ are singular if they satisfy one of the equations
\begin{equation}
  \kappa+\lambda_i \mu =0,\quad i=1,\ldots k.
\end{equation}

Let have a look at some particular classes of manifolds where we posses more
detailed information concerning operator $h$.

\noindent{\bf Example 1.}
Almost Kenmotsu $(\kappa,\mu)$-nullity manifolds. There is strong result which
asserts that $\kappa = -1$, and $\mu=0$.
So for such class of manifolds Jacobi operator of $\xi$ is of maximal rank.
If we take instead
$h'=h\circ \varphi$ (generalized $(\kappa,\mu)'$-nullity spaces),
then $\kappa \leq -1$ , for $\kappa =-1$, $h'=0$ and for
$\kappa < -1$, $\mu =- 2$, eigenvalues of $h'$ are $0$, $\pm\sqrt{-k-1}$,
from these conditions there is one-point singularity $(-2,-2)$.
Beyond that pair of values vector field $\xi$ has maximal Jacobi operator.

\noindent{\bf Example 2.} Contact metric $(\kappa,\mu)$-nullity spaces.
For such manifold $\kappa \leq 1$, if $\kappa =1$, then $h=0$. For $\kappa <1 $,
 eigenvalues of $h$ are $0$, $\pm\sqrt{1-k}$ and singular values are
 pairs $(\kappa,\mu)$ which satisfy $\kappa^2 -(1-k)\mu^2=0$.
 So, beyond  these points Jacobi operator has maximal rank.
\footnote{Minimal polynomial of $h$ is $x(x^2-(1-\kappa))$}

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$(k,\mu)$-spaces},
IIllinois J. of Math. {\bf 44 (1)}, (2000), 213--219.
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Diff. Geom. Appl. {\bf 30} (2012), 665--693.

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{\bf 69}, 211--220.

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\end{document}


TITLE: Countable unions and the axiom of countable choice
QUESTION [19 upvotes]: Let us denote by ACC the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by UCC the assertion that a countable union of countable sets is countable.
UCC is a simple theorem of ZF+ACC.
Proof Suppose for every $i\in\omega$ we have $X_i$ a countable set, and $X_i\cap X_j=\varnothing$ for $j\neq i$.
Since $X_i$ is countable $O_i=\{f\colon X_i\to\omega\mid f\ \text{ injective}\}$ is non-empty. We can choose $f_i\in O_i$ by the axiom of countable choice, and define: $$F\colon\bigcup X_i\to \omega\times\omega\colon\qquad x\mapsto\langle n,f_n(x)\rangle$$
Where $n$ is the unique $n\in\omega$ such that $x\in X_n$.
The Cantor pairing function shows that $\omega\times\omega$ is countable and we are done.

Is the opposite assertion is true, namely ZF+UCC implies ACC? If the answer is negative, does that imply at least some other weaker form of choice?

As noted by Emil Jeřábek below, UCC implies the axiom of countable choice for countable sets (the latter abbreviated as CCF).
Digging through the paper mentioned by godelian in the comments, I reached [1] in which Howard constructs a model of ZFA in which CCF holds and UCC does not, and by the transfer theorem of Pincus constructs this over ZF. Therefore we have: $$\text{ACC}\Rightarrow\text{UCC}\Rightarrow\text{CCF}$$ The first implication is irreversible in ZFA, by the comment of godelian, and the second irreversible in ZF by [1]. Both papers are two decades old, is there any known progress?

Bibliography:

Howard, P. The axiom of choice for countable collections of countable sets does not imply the countable union theorem. Notre Dame J. Formal Logic Volume 33, Number 2 (1992), 236-243.

REPLY [7 votes]: The implication $ACC \implies UCC$ is irreversible in $ZF$. This follows again from the transfer theorem of Pincus:

$UCC$ is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely $UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the $\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993

It is known that $¬ACC$ is boundable, and hence injectively boundable.

Then we can apply the transfer theorem of Pincus to the conjunction $UCC \wedge ¬ACC$ and we are done.
SECOND PROOF: Browsing through the "Consequences..." book I've just found another less direct proof of the same fact. I'm adding it here to avoid the interested person from looking it up for himself (especially since the AC website is not working these days). It involves form 9, known as $W_{\aleph_0}$: a set is finite if and only if it is Dedekind finite. Now, $ACC \implies W_{\aleph_0}$ is provable in $ZF$ (in fact this was already proved by Dedekind), while $UCC$ does not imply $W_{\aleph_0}$. The latter follows from the fact that in the basic Fraenkel model, $\mathcal{N}1$, $UCC$ is valid while $W_{\aleph_0}$ is not, and such a result is transferable by considerations of Pincus that can be found at Pincus, D.: "Zermelo-Fraenkel consistency results by Fraenkel Mostowski methods", J. of Symbolic Logic vol 37, pp. 721-743 - 1972 (doi:10.2307/2272420)


TITLE: Comparing different approaches: What is the probability question that a student chosen at random studies both chemistry and physics?
QUESTION [2 upvotes]: The question was the following:

There are $28$ students in a class, $15$ study chemistry, $18$ study physics and $2$ study neither chemistry nor physics.

Calculate the probability that a student chosen at random studies both chemistry and physics.
My approach was as follows:
We pick a random person, the chance that this person studies physics is $\frac{18}{28}$ (because $18$ people study physics out a total of $28$). Now given that he studies physics, what is the chance that he also studies chemistry? I would say $P(\text{chem}|\text{phy}) = \frac{15}{26}$, because we already now that the person studies physics we can exclude the two persons who study neither given $26$ in total, of which $15$ study physics. Concluding that:
$P(\text{phy and chem}) = P(\text{phy})P(\text{chem}|\text{phy}) = \frac{18}{28}\frac{15}{26} = 0.37$.
However, the solutions provides elsewjere stated the following approach:
$P(\text{phy and chem}) = P(\text{phy}) + P(\text{chem}) - P(\text{phy}\cup \text{chem})$.
Where $P(\text{phy}) = \frac{18}{28}$, $P(\text{chem}) = \frac{15}{28}$, and $P(\text{phy}\cup \text{chem}) = 1 - P(\overline{\text{phy}\cup \text{chem}}) = 1 - P(\overline{\text{phy}} \cap \overline{\text{chem}}) = 1 - \frac{2}{28} = \frac{26}{28}$.
Concluding that: $P(\text{phy and chem}) = \frac{18}{28}+\frac{15}{28} - \frac{26}{28} = \frac{1}{4} = 0.25$.
The two answers differ. I think I made a mistake in calculating the conditional probability but I can't identify where. Please comment on this.

REPLY [1 votes]: I think that N. F. Taussig's answer is spot on, including his demonstration that the OP's analysis is inaccurate.  My answer explores the OP's mis-step more deeply.
Let A denote the event that the student studies chemistry.
Let B denote the event that the student studies physics.
Let C denote the event that the student studies chemistry or physics or both.
Then 15/26 = p(A|C), not p(A|B).  Meta-cheating, the OP assumed (correctly) that the problem provides sufficient info for an answer, and assumed (incorrectly) that the solution must involve bayes theorem :
p(A|B) x p(B) = p(AB).
Since there didn't seem to be any (other) way of directly computing p(A|B), the OP assumed (incorrectly) that 15/26 = p(A|B).
For someone new to conditional probability (i.e. bayes theorem), this is an unfair problem.  Conditional probability problems are tricky and require a step by step development of the student's intuition.
BTW... I was impressed with the OP's metacheating, so I upvoted.


TITLE: If the set of limit points of $A \subseteq \mathbb{R}^n$ is countable, then $A$ is countable
QUESTION [2 upvotes]: Let $A \subseteq \mathbb{R}^n$. Prove that if $A'$ is countable, then $A$ is countable.

$A'$ is the collection of all limit points of $A$.
This is the problem on Real Analysis. I know how to prove injection on $A$ to $\mathbb N=\{0,1,2,\ldots\}$, but I have no idea how to do the above problem.

REPLY [0 votes]: One thing to realize about limits (that goes a bit against the everyday use of the word) is that the extremely boring sequence $3, 3, 3, 3, 3, 3, \ldots$ is also a sequence with a well defined limit; obviously this limit is $3$.
In other words: the popular romantic image of a limit as 'something you can get arbitrarily close to but never really reach' only makes up a subset of all limits in the mathematical sense. Many (somewhat boring) sequences have limits that are reached many times as part of the sequence itself.
I think that if you are sufficiently aware of this, the question (especially when taking the contra-positive approach as suggested by Brian) becomes kind of easy, but please let me know if you need more hints.


TITLE: Maximum area of a right-angled triangle with a fixed perimeter - can it be done without calculus?
QUESTION [0 upvotes]: The area, $A$, of a right-angled triangle with fixed perimeter $P$ will be maximised when the triangle is isosceles (i.e. is a multiple of the $1, 1, \sqrt{2}$ right-angled triangle). I know how to prove this using calculus, but I am interested to know whether there is an alternative proof that doesn't need advanced techniques. In particular, Simple proof that equilateral triangles have maximum area shows that a proof without calculus is possible in the general case (to show that the area of any triangle with fixed perimeter will be maximum when the triangle is equilateral), so I think a simple proof should also be possible in the right-angled case.

REPLY [1 votes]: Let a , b , c be the sides of the right angle triangle right angle at C.
$$ \therefore a^2+b^2=c^2 \,,where\,a+b+c=p\, ,\,perimeter\, of \,triangle$$
Using $AM\ge GM$ between $a^2,b^2$ we get
$$\frac{a^2+b^2}{2} \ge (ab)$$
$$\therefore ab\le \frac{c^2}{2}$$
Hence maximum area is $A=\frac{1}{2}ab=\frac{c^2}{4}$
$$ $$ AM = GM holds when $a=b$ ttherfore $c=\sqrt{2}a$
Hence sides are $a,a,\sqrt{2}a$ hence $a=(\frac{p}{2}(2-\sqrt{2}))$ where p is constant perimeter.


TITLE: Quantum field theory
QUESTION [0 upvotes]: Electrons are wave packets in quantum field theory.The process in which there is an exchange of electrons between materials in quantum field theory can be described as the exchange of wave packets?

REPLY [1 votes]: No. A field theory has little to do with the wave behaviour of elementary particles. It is in fact the quantum mechanics that postulates elementary particles as entities with wave-like properties. A field is a concept that is useful in many fields of physics and you don't necessarily have just "particle" fields. Fields are in fact just tools to describe energy transfer and forces acting on some distance.
Imagine you have a space - a sheet of paper with coordinate lines drawn. Now fill this paper with points and assign some number to each point you draw. This is how you visualise a field. You can think of a field as a function that gives you some number for every point in space.
An example is a temperature field - you can assign every point in your room a number in degrees of celsius or farenheit(whichever you prefer).
A field that assigns a number to each point in space is called a scalar field - because numbers in mathematics are called scalars. You can also have a field that gives you a vector - an arrow - for every point in space - an example of this kind of field is wind: for wind in atmosphere, you can assign a direction where the wind is blowing and a speed or force with which it is blowing.
Electromagnetic field given by Maxwell's equations in classical electrodynamics are fields. Electromagnetism are just numbers that arise from these equations for every point in space.
Field theory looks at different particles as different fields. Meaning that singular particles make little sense in field theory because it is all just some numbers in every point of a space. The particle field has some ground value if you do not pump energy into it, which we call a vacuum. If you do pump some energy into the field, you can excite it and create a particle. So particles in field theory in general are really just bunches of energy pumped into some kind of field. You can have "photons" as vector fields of electric and magnetic vectors in classical field theory, that is really just a flow of energy from one place to another. You can also have nonzero classical field even if you don't pump any energy into it: for example pernament magnets.
You can have a magnet that stays magnetic forever and does not require any energy to stay magnetic.
By interaction in any field theory(classical or quantum) we usually mean just transfer of energy between different kinds of fields.
Now enter the "new" physics - relativity and quantum. Relativity tells us, that matter is just a form of energy and vice versa. This fact changes our perception of physical fields. Now physical fields, which are just a transfer of energy, are as real as the energy they transfer. Knowing this fact and given sufficient amount of energy, we can ask ourselves, can we turn this abstract field into some tangible matter? Quantum field theory says yes, we can.
Quantum field theory views particles as an universal fields. Particles are always there and are everywhere - vacuum is a sea of particles in some ground states. Pump it enough energy and you can excite this vacuum into tangible particles that you can measure.
But just because you have a vacuum it does not mean you have no field. A vacuum in field theory can have some nonzero field, in quantum field theory it has two kinds of nonzero fields: you can have a nonzero field vacuum like the example with pernament magnet - it just has some number in every point in space. Usually we have the freedom to set this value to something, but there is one elementary field that we can not do this - that is the case of the weird Higgs field. It has something to do with the function that governs behaviour of this field - it has multiple ground states, multiple "vacuums".
With quantum physics, field theory also acquires some weird nuances. Because in quantum physics, there is the famous heisenberg's uncertainty relation, which can be "transformed" to not speak about momentum and position, but about time and energy, you can create particles from literally nothing given sufficiently short times. This is another kind of particles you can have even in the absolute vacuum.
Both of these facts give rise to the so-called "vacuum energy" and make the quantum world so weird and complicated. It is not just the waves, it is much deeper and weirder.


TITLE: A newbie text for learning the independence of CH
QUESTION [3 upvotes]: As a newbie, I'm trying to learn a (rigorous) proof of the independence of CH, but I'm struggling a bit with my current text.  I really like the approach and style of
An Introduction To Set Theory, W.A.R. Weiss, 2014, but I can't see it clearly indicated that this PDF covers the theorems of the independence of CH.
If I work my way through this PDF, will I see the forcing proof (originally by Cohen) that the negation of CH is consistent?

REPLY [8 votes]: What an odd question. I'm not familiar with the book, but I was able to ascertain quickly that the answer is no.
Searching the pdf for the words "force" or "forcing" gives no results. Searching for "Cohen" gives two results, which allude to Cohen's proof but do not indicate it will be covered in the current book. The table of contents indicates nothing about the independence of $\mathsf{CH}$ (and this topic would certainly require its own chapter!). So no, it's safe to say you won't learn Cohen's forcing proof from this book.
What you will learn, in Chapter 13, is Gödel's theorem that $\mathsf{GCH}$ is consistent with $\mathsf{ZFC}$ - that is, $\mathsf{ZFC}$ does not prove the negation of the (generalized) continuum hypothesis. This is one half of the theorem that $\mathsf{(G)CH}$ is independent of $\mathsf{ZFC}$. The other half is Cohen's proof that the negation of $\mathsf{(G)CH}$ is consistent with $\mathsf{ZFC}$.


TITLE: Partial integration of dot products and the $\nabla$ operator
QUESTION [0 upvotes]: I am working my way through a lecture notes draft (section 5.1.) on kinetic theory of plasmas, and I'm having trouble with the derivation of the law of energy conservation in a Vlasov-Poisson system. The equations I'm working with here are:

Next, I wish to calculate the rate of change of electric energy as follows:

I understand that the first part comes from Poynting's theorem (in Planck units), Leibniz' rule and using the electrostatic approximation to substitute in $E = -\nabla \varphi$. Now what I'm having trouble with is both of the parts where partial integration is used.
So, first, how do I get the result that
$E = \int d^{3} \vec r \frac {\nabla \varphi}{4\pi} \cdot \frac{\partial(\nabla \varphi)}{\partial t} = -\int d^{3} \vec r \frac {\varphi}{4\pi} \frac{\partial(\nabla^{2} \varphi)}{\partial t}$?
My attempt at integrating is this:
$E = \int d^{3} \vec r \frac {\nabla \varphi}{4\pi} \cdot \frac{\partial(\nabla \varphi)}{\partial t} = [{\nabla \varphi \frac {d}{dt} (\nabla \varphi \frac{1}{4\pi}) }]\Biggr|_{-\infty}^{\infty} -\int d^{3} \vec r \frac {\varphi}{4\pi} \frac{\partial}{\partial r}\frac {\partial}{\partial t} \nabla \varphi = -\int d^{3} \vec r \frac {\varphi}{4\pi} \frac{\partial}{\partial t}\frac {\partial}{\partial r} \nabla \varphi = -\int d^{3} \vec r \frac {\varphi}{4\pi} \nabla \frac {\partial}{\partial t} \nabla \varphi = -\int d^{3} \vec r \frac {\varphi}{4\pi} \frac{\partial(\nabla^{2} \varphi)}{\partial t}$,
but I'm unsure how this works out since I'm integrating a dot product, and the derivative is partial. Am I on the right track here? Should I be using the divergence theorem? From what I understand $\nabla$ is taken over the space, so it corresponds to a derivative with respect to $\vec r$.
The second thing I'm not sure of is integrating the first term on the second line of (5.3) by parts, since there's two integrals and a dot product. Do I integrate by parts in $\vec v $ or $\vec r$? If I multiply by $\varphi$, and integrate by parts, I get terms with $\varphi \vec v$ and $\varphi f_{\alpha}$, so I'm not grasping where I would get the ${f_{\alpha} \vec v}$ term in the result.
Any help is greatly appreciated, I'm only familiar with very simple calculations with partial integration and the double integrals, dot products and $\nabla$ are confusing me.
EDIT: 1st part answered, still need help on the second part!

REPLY [2 votes]: So, first, how do I get the result that

$E = \int d^{3} \vec r \frac {\nabla \varphi}{4\pi} \cdot \frac{\partial(\nabla \varphi)}{\partial t} = -\int d^{3} \vec r \frac {\varphi}{4\pi} \frac{\partial(\nabla^{2} \varphi)}{\partial t}$?

You use the usual rules for integration by parts (from vector calculus). The relevant result, written here with a bunch of parentheses to try and clarify what is operating on what, is:
$$
\int \vec \nabla\left(f(\vec x,t)\right) \cdot \vec \nabla \left( g(\vec x, t)\right)
=\int \vec \nabla \cdot \left(f\vec \nabla \left(g\right)\right) - \int f\nabla^2g
$$
Just use
$$
f = \phi
$$
and
$$
g = \frac{\partial \phi}{\partial t}
$$
And apparently "use 5.2" means to use something like:
$$
\int \vec \nabla \cdot \left(\phi\frac{\partial }{\partial t}\vec \nabla \phi\right) = \oint_{at\;\infty} \vec dA \cdot (\phi\frac{\partial }{\partial t}\vec \nabla \phi)= 0
$$
Which jives if $\phi$ falls off like $1/r$ or faster (since the surface at $\infty$ grows in area like $r^2$).
You also use the fact that partial derivative wrt $t$ and wrt $\vec x$ commute.


TITLE: Finding the sketch of the original function from a sketch of its derivative
QUESTION [0 upvotes]: Let's just say I had a graph like the one above but I didn't know its equation. Let's also say it's the graph of a derivative function and I had to find the graph of the original function between x = -2 and x = 4. How would finding the area underneath the graph, between these two points help me to sketch the original function?
NOTE: I have a homework question very similar to this but with a different graph. I just wanted to figure out or understand the method behind how finding the area underneath the derivative function between these two points can help with the sketch of the original function.

REPLY [1 votes]: I wanted to make a comment but my points weren't enough. Anyway, can we approximate the answer? If yes then you can look here for some techniques of approximating integration: Approximating Integral


TITLE: Testing for trapped surfaces
QUESTION [0 upvotes]: If $M$ is a $n-1$ dimensional Riemannian submanifold in a $1+n$ dimensional space-time manifold $(V,g)$ of pseudo-Riemannian signature $(1,n)$ and $\nabla$ be the Riemann-Christoffel connection on it.
One chooses two future directed null geodesics orthogonal to M say $X^+$ and $X^-$ and defines two rank $2$ tensors $K^+$ and $K^-$ such that their components are,
$K^+ _{ab} = \nabla _a X^+_b$
$K^- _{ab} = \nabla _a X^- _b$
(where $a$ and $b$ run over the indices on $M$)
Let $h_{ab}$ be the components of the induced metric on $M$ from $g$ of $V$. Then one defines the following "null mean curvatures", $\chi ^+$ and $\chi ^-$ along $X^+$ and $X^-$ as,
$\chi ^ + = h^{ab}K^+_{ab}$
$\chi ^ - = h^{ab}K^-_{ab}$
Now apparently $h$ and $g$ can be related as,
$h = g + \frac{1}{2}(X^+ \otimes X^- + X^- \otimes X^+)$

This is something I am not very clear about. I guess there is some abuse of notation about what is called $h$ since in the two definitions the dimensions of $h$ don't seem to match.

With the above definition one can show that,
$h^{\mu \nu} \nabla _\mu X^+ _\nu = g^{\mu \nu} \nabla _\mu X^+ _\nu$
(where $\mu$ and $\nu$ run over indices of the space-time $V$)
A similar expression holds with $X^-$ and the proof of this crucially needs both the properties that $X$'s are null as well as geodesic. This relates to the notion of ``expansion" of a geodesic congruence and hence shows that null mean curvature is exactly the expansion for the geodesic congruence to which $X$'s would be tangent vectors to.
Now one calls such a $M$ a ``Trapped Surface" if both the null mean curvatures $\chi ^+$ and $\chi ^-$ are negative.
I run into some confusions when I try doing this test on $r=constant$ and $t=constant$ spheres in the Schwarzschild metric,
$ds^2 = -(1-\frac{2M}{r})dt^2 + \frac{dr^2}{(1-\frac{2M}{r})} + r^2(d\theta ^2 + sin^2 \theta d\phi ^2)$
I should get both the null mean curvatures of all spheres inside $r=2M$ to be negative and hence all of them are trapped surfaces. To get this I have to choose the future directed null geodesics orthogonal to the 2-spheres as,
$X^{+/-} = (\frac{+/-1}{(1-\frac{2M}{r})}, -1 , 0, 0)$

Now I am not very clear as to how to justify that these are both "future directed" inside the event horizon (i.e the surface $r=2M$). (I have some arguments of my own but not very clear) That these are null and orthogonal geodesics to the two spheres is clear.

For the above one gets that $\chi ^{+/-} = -\frac{2}{r}$ and hence justifying that all spheres inside the event horizon are trapped surfaces.

I would also like to know as to what is the coordinate independent definition for a geodesic to be "future directed" and in how general a space-time can a notion of being "future directed" be imposed globally and continuously.
Given a general pseudo-Riemannian metric how does one detect the presence or absence of a trapped surface? How does one find such a surface? (The above definition seems to give a test of being trapped if one is given a surface)

REPLY [1 votes]: Quite a lot of questions you have there. First a bit of nitpick:

You didn't give enough information to define the null second fundamental forms. Just knowing a geodesic is not enough. You need a whole family of them. To be precise, you need $X^\pm$ to be null vector fields along $M$ orthogonal to $TM$, and then you extend them by parallel transport along themselves in the null direction.

For a bit more about the geometry, look at Galloway's Beijing Lecture Notes or B. O'Neill's Semi-Riemannian Geometry.
About the relationship between $h$ and $g$: the induced metric on a Riemannian submanifold of a semi-Riemannian manifold can be treated as a projection operator on the total tangent space restricted to over the submanifold. In other words, let $p\in M\subset V$, then because $M$ is Riemannian, you can split $T_pV = T_pM \oplus  (T_pM )^\perp$, then $h$ uniquely lifts from $Sym_2(T_pM)$ to $Sym_2(T_pV)$ by extending trivially in the perpendicular directions. This lift coincides with $g + X^-\otimes X^+ + X^+\otimes X^-$ (if you normalize $g(X^-,X^+) = -1$). (Note, this fixing of normalization is important. You didn't specify it in your question.)
Now, you should get out of the habit of working in Schwarzschild coordinates for the trapped region. That coordinate system is only good for thinking about the exterior of the black hole. Inside the trapped region, the surfaces of constant $r$ are in fact space-like... The picture would be much clearer if you think in terms of Kruskal coordinates.
About "future directed":

Looks like you are missing a bit of basic causal theory? A space-time is called time-orientable if it admits a continuous, non-vanishing time-like vector field $\tau$. For time-orientable space-times, the notion of future and past are well-defined: a causal vector $v$ is said to be future oriented if $g(v,\tau) < 0$, and past-oriented if $g(v,\tau) > 0$. There exists Lorentzian manifolds which are not time-orientable. See any General Relativity textbook for more details and examples (esp. Hawking-Ellis and O'Neill).

Detection of trapped surface:

Outside of spherical symmetry, where you can detect trapped surfaces by using the metric dual of the one form $dr$, where $r$ is the area-radius (if $dr$ is space-like, then the sphere is non-trapped), there's no general mechanism to detect trapped surfaces, other than a priori knowledge that both of its null mean curvature are negative (as in, this is an open problem). If you restrict to a spatial slice, then there are some work in the Riemannian geometry literature concerning this problem by looking at the Jang equation (I think Marcus Khuri worked on it for a bit, don't know if he is still on it). There are also related work on dynamical horizons and marginally outermost trapped surfaces by Ashtekar, Galloway, Lars Andersson, and others. But if you do find a general way to detect presence of trapped surfaces without local computations, do let me know, because you'd be about a third way there toward solving weak cosmic censorship.


TITLE: define appropriate big-M in linearize KKT Complementary slackness conditions
QUESTION [1 upvotes]: When we transform our maximization problem to its equivalent KKT, It becomes  non-linear because of Complementary slackness.
$$lambda\times S=0,$$
We can easily linearize above condition with replace it with to two inequalities:
$$lambda \le M\times Y$$
$$S\le M\times (1-Y)$$
Where $M$  is a very large positive number and $Y$ is a binary variable in which when $Y= 0$, $lambda=0$ and $S$ can take any value and when $Y= 1$, $S=0$ and $lambda$ can take any value.
To improve the quality of solving our problem whit aforementioned method with optimization solvers, it is necessary to define a appropriate $M$.
Does anybody Know how can define a good $M$?
Is there any method to linearize Complementary slackness whitout the presence of big-M?

REPLY [2 votes]: This is to provide more detail on the comment about the use of SOS1 variables to implement the complementarity condition $\lambda \cdot s = 0$ with $\lambda, s\ge 0$.
Let $y_1,y_2\ge 0$ form a SOS1 set (i.e. only one of them can be nonzero). Then
$$
\begin{align}
  \lambda \le y_1\\
  s \le y_2
\end{align}
$$
will do what the poster wants. (Of course, more directly, we can also form a SOS1 set consisting of $s,\lambda$).


TITLE: What is the physics behind the energy transferred in a 'rotating collision'?
QUESTION [1 upvotes]: https://www.youtube.com/watch?v=eLXHLRa37_g
In this video, we can see that rotating the paper somehow makes it less susceptible to being damaged. I can tell that the paper has some angular momentum. But how do we explain this phenomena in terms of energy transfer? It sort of remind me of how using simple equations, ( for momentum and energy) we can show that objects moving faster in a linear collision have less energy transferred to them. Is there an analogous description for 'rotating collisions'?
Moreover, would it be correct to say that the paper becomes harder when spinning very fast?

REPLY [0 votes]: It is not just that the paper becomes hard, it is the friction that does the cutting. Angle grinders spin very fast, the glue and 3 pieces of paper add some rigidity and if you add that to the 'centrifugal' forces it is tough enough to cut through many things. I was surprised it did not burn with the wood, you could see smoke from that cut and the wood was burned on the surface of the cut.
As Paul B said in the comments above the paper could have cut all those materials without the grinder if you could just move it fast enough. The way it cuts is by removing a very small amount of material at a time, it just does it very fast due to the rpms.
Drew K


TITLE: Strength of languages - midpoints+betweenness vs congrunece+betweenness
QUESTION [3 upvotes]: Congrunece relation is a 4-ary relation between points: $\delta(x,y;z,t)$ iff $|xy|=|zt|$.
Betweennes relation is a 3-ary relation between points: $\beta(x,y,z)$ iff $y \in xz$.
Midpoint relation is a 2-ary operation on points: $x \circ y = z$ iff $\beta(x,z,y) \land \delta(x,z;z,y)$.
Is there any general way of defining congruence using only the notions of midpoints and betweenness in any affine space $\mathbb R^n$? Or, perhaps, is there a separate way for any $n$?

REPLY [1 votes]: This cannot be done. If we work in $\mathbb R^2$, then the transformation $(x,y) \mapsto (x, 2y)$ preserves betweenness and midpoints. (In general, any affine transformation has this property.) So any property that you can define in terms of midpoints and betweenness will also be preserved by this transformation.
But the transformation $(x,y) \mapsto (x,2y)$ does not preserve congruence: the segment from $(0,0)$ to $(1,0)$ is congruent to the segment from $(0,0)$ to $(0,1)$, but after the transformation this is no longer the case.
In general, betweenness, midpoints, and other properties such as:

collinearity of points,
concurrence of lines,
parallel lines,
ratios of lengths along parallel lines,
ratios of areas,

are properties of affine spaces: they are preserved by affine transformations. Lengths of line segments, and comparisons between lengths of non-parallel line segments, are not preserved in this way; they are outside the realm of affine geometry. (The same is true for circles, angles, and other more complicated objects.)


TITLE: Is this Diophantine problem solvable without invoking Fermat's Last Theorem?
QUESTION [3 upvotes]: Let $a,b,c,n$ be positive integers with $a<b<c$ and $n\geq 3$ odd. Given that $a^n + b^n < 2c^n$, can one prove that $a^{n+2}+b^{n+2}\neq c^{n+2}$ without invoking Wiles' theorem ? Or is this actually equivalent to Fermat's Last Theorem ?

REPLY [0 votes]: As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $a\leq b$, so your condition just becomes $a\neq b$. In fact, $a\neq b$ is quite extraneous too, because from the Fermat equation we will obtain $\sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.


text:
\section{Overlap concentration: proof of inequality \eqref{over-concen}}\label{appendix-overlap}

The derivations below will apply for any $t\in[0,1]$ so we drop all un-necessary notations and indices. Only the dependence of the free energies in $R(\epsilon)\equiv R_n(t,\epsilon)$ matters, so we denote $F(R(\epsilon))\equiv F_{n,t,\epsilon}(\bW(t),\tilde{\bW}(t,\epsilon))$ and $f(R(\epsilon))\equiv f_{n}(t,\epsilon)$.

Let $\mathcal{L}$ be the $R(\epsilon)$-derivative of the Hamiltonian \eqref{Ht} divided by $n$:
\begin{align}
\mathcal{L}(\bx,\bX,\tilde \bZ) =\mathcal{L} \equiv \frac1n \frac{d\mathcal{H}_{n,t,\epsilon}}{dR(\epsilon)}= \frac{1}{n}\Big(\frac{\|\bx\|^2}{2} - \bx\cdot \bX - \frac{\bx\cdot \tilde \bZ}{2\sqrt{R(\epsilon)}} \Big)\,.\label{def_L}
\end{align}
The overlap fluctuations are upper bounded by those of $\mathcal{L}$, which are easier to control, as
\begin{align}
\mathbb{E}\big\langle (Q - \mathbb{E}\langle Q \rangle_{t})^2\big\rangle_{t} \le 4\,\mathbb{E}\big\langle (\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle_{t})^2\big\rangle_{t}\,.\label{remarkable}
\end{align}
The bracket is again the expectation w.r.t. the posterior of the interpolating model \eqref{t_post}.
A detailed derivation of this inequality can be found in appendix \ref{proof:remarkable_id} and involves only elementary algebra using the Nishimori identity
and integrations by parts w.r.t.\ the gaussian noise $\tilde \bZ$.

We have the following identities: for any given realisation of the quenched disorder
\begin{align}
 \frac{dF}{dR(\epsilon)}  &= \langle \mathcal{L} \rangle_{t} \,,\label{first-derivative}\\

 \frac{1}{n}\frac{d^2F}{dR(\epsilon)^2}  &= -\big\langle (\mathcal{L}  - \langle \mathcal{L} \rangle_t)^2\big\rangle_{t}+
 \frac{1}{4 n^2R(\epsilon)^{3/2}} \langle \bx\rangle_{t} \cdot \tilde \bZ\,.\label{second-derivative}
\end{align}
The gaussian integration by part formula \eqref{GaussIPP} with hamiltonian \eqref{Ht} yields
\begin{align}
\frac{\mathbb{E} \big\langle \tilde \bZ\cdot  \bx \big\rangle_t}{\sqrt{R(\epsilon)}}
	 =   \mathbb{E}\big\langle \|\bx\|^2 \big\rangle_t - \EE\|\langle \bx \rangle_t\|^2 \overset{\rm N}{=}\mathbb{E}\big\langle \|\bx\|^2 \big\rangle_t - \EE \big\langle\bX\cdot \bx \big\rangle_t= \mathbb{E}\big\langle \|\bx\|^2 \big\rangle_t - n\,\EE \langle Q\rangle_t\,.\label{NishiTildeZ}
\end{align}

Therefore averaging \eqref{first-derivative} and \eqref{second-derivative} we find

\begin{align}
 \frac{df}{d R(\epsilon)} &= \mathbb{E}\langle \mathcal{L} \rangle_{t}
 \overset{\rm N}{=}-\frac{1}{2}\mathbb{E}\langle Q\rangle_{t}\,,\label{first-derivative-average}\\

 \frac{1}{n}\frac{d^2f}{dR(\epsilon)^2} &= -\mathbb{E}\big\langle (\mathcal{L} - \langle \mathcal{L} \rangle_{t})^2\big\rangle_{t}
 +\frac{1}{4n^2R(\epsilon)} \mathbb{E}\big\langle \|\bx- \langle \bx \rangle_t\|^2\big\rangle_t\,.\label{second-derivative-average}
\end{align}
We always work under the assumption that the map $\epsilon\in [s_n, 2s_n] \mapsto R(\epsilon)\in [R(s_n), R(2s_n)]$ is regular, and do not repeat this assumption in the statements below. The concentration inequality \eqref{over-concen} is  a direct consequence of the following result (combined with Fubini's theorem):

\begin{proposition}[Total fluctuations of $\mathcal{L}$]\label{L-concentration} Let the sequences $\lambda_n$ and $\rho_n$ verify \eqref{app-scalingRegime}. Then $$\int_{s_n}^{2s_n} d\epsilon\,\mathbb{E}\big\langle (\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle_{t})^2\big\rangle_{t} \le C\Big(\frac{\lambda_n\rho_n}{ns_n}\big(1+\lambda_n\rho_n^2\big)\Big)^{1/3}$$
for a constant $C>0$ that is independent of $n$, as long as the r.h.s. is $\omega(1/n)$.
\end{proposition}

The proof of this proposition is broken in two parts, using the decomposition

\begin{align*}
\mathbb{E}\big\langle (\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle_{t})^2\big\rangle_{t}
& =
\mathbb{E}\big\langle (\mathcal{L} - \langle \mathcal{L}\rangle_t)^2\big\rangle_{t}
+
\mathbb{E}\big[(\langle \mathcal{L}\rangle_t - \mathbb{E}\langle \mathcal{L}\rangle_t)^2\big]\,.
\end{align*}

Thus it suffices to prove the two following lemmas. The first lemma expresses concentration w.r.t.\ the posterior distribution (or ``thermal fluctuations'') and is a direct consequence of concavity properties of the average free energy and the Nishimori identity.

\begin{lemma}[Thermal fluctuations of $\cal L$]\label{thermal-fluctuations}
	We have $$\int_{s_n}^{2s_n} d\epsilon\,
  \mathbb{E} \big\langle (\mathcal{L} - \langle \mathcal{L}\rangle_t)^2 \big\rangle_t  \le \frac{\rho_n}{n}\Big(1+\frac{\ln2}{4}\Big)\,.$$
\end{lemma}
\begin{proof}

We emphasize again that the interpolating free energy \eqref{fnt} is here viewed as a function of $R(\epsilon)$. In the argument that follows we consider derivatives of this function w.r.t. $R(\epsilon)$.
By \eqref{second-derivative-average}
\begin{align}
\mathbb{E}\big\langle (\mathcal{L} - \langle \mathcal{L} \rangle_t)^2\big\rangle_t
& =
-\frac{1}{n}\frac{d^2f}{dR(\epsilon)^2}
+\frac{1}{4n^2R(\epsilon)} \big(\mathbb{E}\big\langle \|\bx\|^2 \big\rangle_t - \EE\|\langle \bx \rangle_t\|^2\big)
\nonumber \\ &
\leq
-\frac{1}{n}\frac{d^2f}{dR(\epsilon)^2} +\frac{\rho_n}{4n\epsilon} \,,
\label{directcomputation}
\end{align}
where we used $R(\epsilon)\geq \epsilon$ and $\frac1n\mathbb{E}\langle \|\bx\|^2\rangle_t \overset{\rm N}{=} \mathbb{E}[X_1^2]=\rho_n$. We integrate this inequality over $\epsilon\in [s_n, 2s_n]$. Recall the map
$\epsilon\mapsto R(\epsilon)$ has a Jacobian $\ge 1$, is $\mathcal{C}^1$
and has a well defined $\mathcal{C}^1$ inverse since we have assumed that it is regular. Thus integrating \eqref{directcomputation} and performing a change of variable (to get the second inequality) we obtain
\begin{align*}
\int_{s_n}^{2s_n} d\epsilon\, \mathbb{E}\big\langle (\mathcal{L} - \langle \mathcal{L} \rangle_t)^2\big\rangle_t
& \leq
- \frac{1}{n}\int_{s_n}^{2s_n} d\epsilon \,\frac{d^2f}{dR(\epsilon)^2} + \frac{\rho_n}{4n}\int_{s_n}^{2s_n} \,\frac{d\epsilon}{\epsilon}
\nonumber \\ &
\leq
- \frac{1}{n}\int_{R(s_n)}^{R(2s_n)} dR(\epsilon) \,\frac{d^2f}{dR(\epsilon)^2}
+ \frac{\rho_n}{4n}\int_{s_n}^{2s_n} \,\frac{d\epsilon}{\epsilon}
\nonumber \\ &
=
\frac{1}{n}\Big(\frac{df}{dR(\epsilon)}(R(s_n))
- \frac{df}{dR(\epsilon)}(R(2s_n))\Big)+ \frac{\rho_n}{4n}\ln 2\,.
\end{align*}

We have $|f'(R(\epsilon))| = |\mathbb{E}\langle Q\rangle_t/2|\le \rho_n/2$ so the first term is certainly smaller in absolute value than $\rho_n/n$. This concludes the proof of Lemma \ref{thermal-fluctuations}.
\end{proof}

The second lemma expresses the concentration w.r.t.\ the quenched disorder variables
and is a consequence of the concentration of the free energy onto its average (w.r.t. the quenched variables).

\begin{lemma}[Quenched fluctuations of $\cal L$]\label{disorder-fluctuations}
	Let the sequences $\lambda_n$ and $\rho_n$ verify \eqref{app-scalingRegime}. Then $$\int_{s_n}^{2s_n} d\epsilon\,
  \mathbb{E}\big[ (\langle \mathcal{L}\rangle_t - \mathbb{E}\langle \mathcal{L}\rangle_t)^2\big] \le C\Big(\frac{\lambda_n\rho_n}{ns_n}\big(1+\lambda_n\rho_n^2\big)\Big)^{1/3}$$
  for a constant $C>0$ that is independent of $n$, as long as the r.h.s. is $\omega(1/n)$.
\end{lemma}

\begin{proof}
Consider the following functions of $R(\epsilon)$:
\begin{align}\label{new-free}
 & \tilde F(R(\epsilon)) \equiv F(R(\epsilon)) +S\frac{\sqrt{R(\epsilon)}}{n} \sum_{i=1}^n\vert \tilde Z_i\vert\,,
 \nonumber \\ &
 \tilde f(R(\epsilon)) \equiv \mathbb{E} \,\tilde F(R(\epsilon))= f(R(\epsilon)) + S\sqrt{R(\epsilon)} \mathbb{E}\,\vert \tilde Z_1\vert\,.
\end{align}
Because of
\eqref{second-derivative} we see that the second derivative of $\tilde F(R(\epsilon))$ w.r.t. $R(\epsilon)$ is negative so that it is concave. Note $F(R(\epsilon))$ itself is not necessarily concave in $R(\epsilon)$, although $f(R(\epsilon))$ is. Concavity of $f(R(\epsilon))$ is not obvious from \eqref{second-derivative-average} (obtained from differentiating $\EE\langle\mathcal{L}\rangle_t$ w.r.t. $R(\epsilon)$) but can be seen from \eqref{secondDer_f_pos} (obtained instead by differentiating $-\frac12\EE\langle{Q}\rangle_t$) which reads $\frac{d}{dR(\epsilon)}\EE\langle Q\rangle_t=-2\frac{d^2}{dR(\epsilon)^2}f\ge 0$.

Evidently $\tilde f(R(\epsilon))$ is concave too.
Concavity then allows to use the following standard lemma:
\begin{lemma}[A bound for concave functions]\label{lemmaConvexity}
Let $G(x)$ and $g(x)$ be concave functions. Let $\delta>0$ and define $C^{-}_\delta(x) \equiv g'(x-\delta) - g'(x) \geq 0$ and $C^{+}_\delta(x) \equiv g'(x) - g'(x+\delta) \geq 0$. Then
\begin{align*}
|G'(x) - g'(x)| \leq \delta^{-1} \sum_{u \in \{x-\delta,\, x,\, x+\delta\}} |G(u)-g(u)| + C^{+}_\delta(x) + C^{-}_\delta(x)\,.
\end{align*}
\end{lemma}

First, from \eqref{new-free} we have
\begin{align}\label{fdiff}
 \tilde F(R(\epsilon)) - \tilde f(R(\epsilon)) = F(R(\epsilon)) - f(R(\epsilon)) + S\sqrt{R(\epsilon)}  A_n
\end{align}
with $A_n \equiv \frac{1}{n}\sum_{i=1}^n \vert \tilde Z_i\vert -\mathbb{E}\,\vert \tilde Z_1\vert$.
Second, from \eqref{first-derivative}, \eqref{first-derivative-average} we obtain for the $R(\epsilon)$-derivatives
\begin{align}\label{derdiff}
 \tilde F'(R(\epsilon)) - \tilde f'(R(\epsilon)) =
\langle \mathcal{L} \rangle_t-\mathbb{E}\langle \mathcal{L} \rangle_{t} + \frac{SA_n}{2\sqrt{R(\epsilon)}} \,.
\end{align}
From \eqref{fdiff} and \eqref{derdiff} it is then easy to show that Lemma \ref{lemmaConvexity} implies
\begin{align}\label{usable-inequ}
\vert \langle \mathcal{L}\rangle_t - \mathbb{E}\langle \mathcal{L}\rangle_t\vert&\leq
\delta^{-1} \sum_{u\in \{R(\epsilon) -\delta,\, R(\epsilon),\, R(\epsilon)+\delta\}}
 \big(\vert F(u) - f(u) \vert + S\vert A_n \vert \sqrt{u} \big)\nonumber\\
 &\qquad\qquad\qquad\qquad
  + C_\delta^+(R(\epsilon)) + C_\delta^-(R(\epsilon)) + \frac{S\vert A_n\vert}{2\sqrt \epsilon}
\end{align}
where $C_\delta^-(R(\epsilon))\equiv \tilde f'(R(\epsilon)-\delta)-\tilde f'(R(\epsilon))\ge 0$
and $C_\delta^+(R(\epsilon))\equiv \tilde f'(R(\epsilon))-\tilde f'(R(\epsilon)+\delta)\ge 0$.
We used $R(\epsilon)\ge \epsilon$ for the term $S\vert A_n\vert/(2\sqrt \epsilon)$. Note that $\delta$ will
be chosen later on strictly smaller than $s_n$ so that $R(\epsilon) -\delta \geq \epsilon - \delta \geq s_n -\delta$ remains
positive. Remark that by independence of the noise variables $\mathbb{E}[A_n^2]  = (1-2/\pi)/n\le 1/n$.
We square the identity \eqref{usable-inequ} and take its expectation. Then using $(\sum_{i=1}^pv_i)^2 \le p\sum_{i=1}^pv_i^2$, and
that $R(\epsilon)\le 2s_n+\lambda_n\rho_n$, as well as the free energy concentration Proposition \ref{prop:fconc} (under the assumption that $\lambda_n$ and $\rho_n$ verify \eqref{app-scalingRegime}),

\begin{align}\label{intermediate}
 \frac{1}{9}\mathbb{E}\big[(\langle \mathcal{L}\rangle_t - \mathbb{E}\langle \mathcal{L}\rangle_t)^2\big]
 &
 \leq \,
 \frac{3}{n\delta^2} \Big(C\lambda_n^2\rho_n^3 +S(2s_n+\lambda_n\rho_n+\delta)\Big)
 \nonumber \\ & \qquad \qquad\qquad\qquad\qquad+ C_\delta^+(R(\epsilon))^2 + C_\delta^-(R(\epsilon))^2
 + \frac{S}{4n\epsilon} \,.
\end{align}
Recall $|C_\delta^\pm(R(\epsilon))|=|\tilde f'(R(\epsilon)\pm\delta)-\tilde f'(R(\epsilon))|$. By \eqref{first-derivative-average}, \eqref{new-free}  and $R(\epsilon)\ge \epsilon$ we have
\begin{align}
|\tilde f'(R(\epsilon))|  \leq \frac12\Big(\rho_n  +\frac{S}{\sqrt{R(\epsilon)}} \Big)\leq \frac12\Big(\rho_n  +\frac{S}{\sqrt \epsilon} \Big)\label{boudfprime}
\end{align}
Thus, as $\epsilon\ge s_n$,
$$|C_\delta^\pm(R(\epsilon))|\le \rho_n  +\frac{S}{\sqrt{\epsilon-\delta}}\le \rho_n  +\frac{S}{\sqrt{s_n-\delta}}\,.$$ We reach

\begin{align*}
 &\int_{s_n}^{2s_n} d\epsilon\, \big\{C_\delta^+(R(\epsilon))^2 + C_\delta^-(R(\epsilon))^2\big\}\\
 &\qquad\leq
 \Big(\rho_n +\frac{S}{\sqrt {s_n-\delta}}\Big)
 \int_{s_n}^{2s_n} d\epsilon\, \big\{C_\delta^+(R(\epsilon)) + C_\delta^-(R(\epsilon))\big\}
 \nonumber \\
  &\qquad\leq  \Big(\rho_n +\frac{S}{\sqrt {s_n-\delta}}\Big)
 \int_{R(s_n)}^{R(2s_n)} dR(\epsilon)\, \big\{C_\delta^+(R(\epsilon)) + C_\delta^-(R(\epsilon))\big\}
 \nonumber \\
&\qquad= \Big(\rho_n +\frac{S}{\sqrt {s_n-\delta}}\Big)\Big[\Big(\tilde f(R(s_n)+\delta) - \tilde f(R(s_n)-\delta)\Big)\nn
&\qquad\qquad\qquad\qquad\qquad+ \Big(\tilde f(R(2s_n)-\delta) - \tilde f(R(2s_n)+\delta)\Big)\Big]
\end{align*}

where we used that the Jacobian of the $\mathcal{C}^1$-diffeomorphism $\epsilon\mapsto R(\epsilon)$ is $\ge 1$ (by regularity) for
the second inequality. The mean value theorem and \eqref{boudfprime}
imply $|\tilde f(R(\epsilon)-\delta) - \tilde f(R(\epsilon)+\delta)|\le \delta(\rho_n  +\frac{S}{\sqrt{s_n-\delta}})$. Therefore

\begin{align*}
 \int_{s_n}^{2s_n} d\epsilon\, \big\{C_\delta^+(R(\epsilon))^2 + C_\delta^-(R(\epsilon))^2\big\}\leq
 2\delta \Big(\rho_n +\frac{S}{\sqrt{s_n-\delta}}\Big)^2\,.
\end{align*}

Set $\delta  = \delta_n = o(s_n)$. Thus, integrating \eqref{intermediate} over $\epsilon\in [s_n, 2s_n]$ yields
\begin{align*}
 &\int_{s_n}^{2s_n} d\epsilon\,
 \mathbb{E}\big[(\langle \mathcal{L}\rangle_t - \mathbb{E}\langle \mathcal{L}\rangle_{t})^2\big]\nonumber\\
 &\qquad\qquad\leq \frac{27s_n}{n\delta_n^2}\Big(C\lambda_n^2\rho_n^3 +S(2s_n+\lambda_n\rho_n+\delta_n)\Big) +18\delta_n \Big(\rho_n +\frac{S}{\sqrt {s_n-\delta_n}}\Big)^2 +  \frac{9 S\ln 2}{4n}  \nonumber\\
 &\qquad\qquad\leq\frac{Cs_n\lambda_n\rho_n}{n\delta_n^2}(1+\lambda_n\rho_n^2)+\frac{C\delta_n}{s_n}+\frac{C}{n}
\end{align*}
where the constant $C$ is generic, and may change from place to place. Finally we optimize the bound choosing $\delta_n^3=  s_n^2\lambda_n\rho_n(1+\lambda_n\rho_n^2)/n$. We verify the condition $\delta_n =o(s_n)$: we have $(\delta_n/s_n)^3=O(\lambda_n\rho_n(1+\lambda_n\rho_n^2)/(ns_n))$ which, by \eqref{app-scalingRegime}, indeed tends to $0_+$ for an appropriately chosen sequence $s_n$. So the dominating term $\delta_n/s_n$ gives the result.
\end{proof}


TITLE: Does $A=\{a|\forall x\in \emptyset\ H(x,a) \}$ make sense?
QUESTION [6 upvotes]: What happens if I define a set $A=\{a|\forall x\in \emptyset\ H(x,a)  \}$, where $H$ is some property ? $\forall x\in \emptyset\ H(x,y)$ should be always true, since it is vacuously true, right? So this set $A$ shouldn't exists, or it would be "the set of all sets".

REPLY [10 votes]: (to complement a bit on Martin's answer with regards to arbitrary collections with respect to ZFC)
The modern approach is that $\{a\mid P(a)\}$, for some property $P$, is a class. This class may or may not be a set. The main difference is that sets can be (read: are) elements of other classes; classes cannot be elements of other classes.
If we assume (for the sake of discussion) that ZFC is consistent then it has a model, let us call it $V$. From the Russell and Cantor paradoxes, we know that $V$ cannot think of $V$ as a set. In fact, $V$ is a cocky piece of mathematical object and thinks of himself larger than any possible set, and that every possible set is already an element.
However if we consider the $V$ externally then it may be a set in some larger universe (which is possibly a set in a larger universe, and so on...). The point is that from an internal point of view, $V$ cannot work with arbitrary collections of elements. If we have a class, which is defined in such a nice way as above then we can write theorems about this class despite the fact that it is not an element of the model. This means that we cannot say that $A\in V$ if $A$ is such class, but we can talk about "elements in $A$" as if $A$ were to be a set.
Consider $V=\{a\mid a=a\}$, this is the entire universe. The entire collection of sets available to us in a certain interpretation of ZFC. $V$ is not a set at all from its own point of view. It is still a definable collection, and we may prove theorems about this class. However handling these collections require a bit more carefulness.
Another important example is the constructible universe denoted by $L$, which is essentially the smallest transitive class-model of ZFC with the same ordinals as the original universe. There exists some formula $\varphi(x)$ such that $\{x\mid \varphi(x)\}=L$. That is to say this formula defines a proper class, but it we can still "use" this class almost as if we were using sets when proving theorems.
For example, we can write the following theorem "If $\forall x\forall y(\varphi(x)\land\varphi(y))$ then there exists $\psi(x,y)$ such that $\{\langle x,y\rangle\mid\psi(x,y)\}$ is a well-ordering".
That is to write "There exists a definable well ordering of $L$", which is a theorem involving two classes, but in a completely valid way.
This point is quite delicate, and you should probably study some more logic and set theory in order to understand it fully. However classes are syntactic objects (as Miha comments below) and we can give them some meaning when we interpret ZFC with a model (i.e. as a semantic object).
So while classes are not objects per se of ZFC, in some cases we can still work with them if we are careful enough.
This notion is formalized fully within the set theory of von Neumann-Bernays-Gödel. This set theory is a conservative extension of ZFC, meaning if we proved something in this setting, and we did so using only sets, then there is a proof from ZFC.
In this set theory, proper classes are objects of the universe and "The class of all sets which do not contain themselves" has a perfectly valid semantical meaning.


TITLE: The definition of absolute continuity without disjointness of intervals is satisfied only by Lipschitz functions
QUESTION [1 upvotes]: A function $f: [a,b] \rightarrow \mathbb{R}$ is Lipschitz continuous if and only if $\forall \epsilon >0, \exists \delta>0$ such that for every finite collection of closed intervals in $[a,b]$, $\{[a_k,b_k]\}_{k=1}^N$ (not necessarily disjoint) with
$\sum_{k=1}^N b_k-a_k < \delta$ then $\sum_{k=1}^N |f(b_k)-f(a_k)|<\epsilon$
The easy part is to prove that a Lipschitz function satisfy the property. But I'm having trouble with the other part.
I tried to mimic the proof that $\sqrt{x}$ does not satisfy the property with not necessarily disjoint intervals but unfortunately I got stuck. I saw this exercise on chapter 5 of Bruckner, real analysis.
I would really appreciate any hints or suggestions.

REPLY [3 votes]: Suppose $f$ is not Lipschitz continuous. Let $\epsilon>0$ be given. I will construct intervals with $\sum |a_k-b_k|<\epsilon$ but $$\sum|f(a_k)-f(b_k)|\ge 1/2\tag1$$
Since $f$ is not Lipschitz, there exist points $s<t$ such that $$|f(s)-f(t)|>\epsilon^{-1}|s-t|\tag2$$
Case 1: $|s-t|<\epsilon$.
Use the interval $[s,t]$ as $[a_k,b_k]$, repeating it $N$  times where $N$ is chosen so that $$\frac{\epsilon}{2} \le  N|s-t| < \epsilon\tag3$$
(Such $N$ always exists, why?) Observe that (2) and (3) imply (1).
Case 2: $|s-t| \ge \epsilon$.
Divide the interval $[s,t]$ into $N$ equal subintervals $[s_k,t_k]$; choose the number of pieces $N$ so that
$$\frac{\epsilon}{2}\le \frac{|s-t|}{N} < \epsilon\tag4$$
(This is always possible, why?) The triangle inequality implies that at least one piece $[s_k,t_k]$ satisfies  $$|f(s_k)-f(t_k)|>\epsilon^{-1}|s_k-t_k|\tag5$$
Use the single interval $[s_k,t_k]$ to achieve (1), using (4) and (5).


TITLE: Convergence, Integration
QUESTION [0 upvotes]: Assume that f is a non-negative real function, and let $a>0$ be a real number.
Define $I_a(f)$ to be
$I_a(f)$=$\frac{1}{a}\int_{0}^{a} f(x) dx$
We now assume that $\lim_{x\rightarrow \infty} f(x)=A$ exists.
Now I want to proof whether $\lim_{a\rightarrow \infty} I_a(f)=A$ is true or not. I have concluded that this is not always true.
My approach has been to construct the following counterexample $f(x)=A+\frac
{1}{x}$, it is easily seen that $\lim_{x\rightarrow \infty} f(x)=A$.
By integrating the chosen function I get that
$\lim_{a\rightarrow \infty}I_a(f)=A+\lim_{a\rightarrow \infty}\frac{1}{a}\int_{0}^a \frac{1}{x}dx = A+\lim_{a\rightarrow \infty} [\frac{\log(a)}{a}-(-\infty)]\rightarrow \infty$.
Therefore I concluded that $\lim_{x\rightarrow \infty} f(x)=A$ does not in general imply that $\lim_{a\rightarrow \infty}I_a(f)=A$.
I am of course unsure whether my calculations are correct, because you could also write the $\log(0)$ as a limit of $\log(\epsilon)\rightarrow \log(0)$, an then by division by $a$, and letting $a\rightarrow \infty$ get that the whole expression with the integral of $\frac{1}{x}$ goes to zero, I this case, it might be true that $\lim_{x\rightarrow \infty} f(x)=A$ implies that $\lim_{a\rightarrow \infty}I_a(f)=A$.
Does anybody have an idea to this?

REPLY [1 votes]: Your counter-example works because f is not integrable on $(0,a)$ for some $a$. If $f$ is integrable on $(0,a)$ for every a, it's true.
WLOG, assume $\lim\limits_{x\to \infty} f(x) = 0$. Otherwise, since $f$ is integrable, we may show that $\frac{1}{a}\int_0^a (f(x)-A) dx \to 0$. For every $\epsilon >0$, there exists $M > 0$ s.d. $\forall x\geq M$, $|f(x)| < \epsilon$. Then for $a \geq M$,
$$\left|\frac{1}{a} \int_0^a f(x) dx \right|
= \left| \frac{1}{a}\int_0^M f(x) dx + \frac{1}{a}\int_M^a f(x) dx \right|
< \frac{1}{a}\left| \int_0^M f(x)dx \right| + \frac{a-M}{a}\epsilon
$$
Take $a\to \infty$ first then $\epsilon \to 0$.


TITLE: Finding the probability function of a random sum of Bernoulli variables (stochastic processes)
QUESTION [0 upvotes]: Edit: Sorry for the Latex, I'm new to it and trying to fix it right now
I am trying to find the probability function of a random sum of Bernoulli variables in this scenario:
Starting from 9AM, clients arrive (independently), before a store opens, and stand in line to enter the store when it opens. An employee of the store randomly checks the line $M$ minutes after 9AM, with $M$ a Poisson variable with parameter $v$.
The number of clients that arrive during any $1$ minute interval is either $1$ or $0$ with probabilities $p$ and $1 - p$.
Let $S_m$ be the total number of clients in line when the line is checked at $M = m$ minutes after 9AM.
I need to find
1) $\Pr(S_m = k)$ for $k = 0,1,2,\ldots$
2) $\operatorname{E}(M \mid S_m = k)$
For 1) I found that
$$\Pr(S_m = k) = \sum_{M=1}^\infty {M \choose k} p^k (1-p)^{M-k}\frac{{e^{-v}v^M }}{M!}$$
For 2), I started by trying to find $\Pr(M = m \mid S_m = k)$ but I'm having a hard time finding it.
I started by doing $$\Pr(M = m \mid S_m = k) = \frac{P(M = m, S_m = k)}{P(S_m = k)}$$ and
plugged in the denominator my answer to $1$, but I don't know how to find the numerator and am really unsure of what I've done so far.
Could anyone help me out?
Thank you!

REPLY [0 votes]: $P(M=m,S_m=k)=P(S_m=k| M=m) \cdot P(M=m)$. You know the second term. For the first term, it's just $\binom{m}{k}p^k(1-p)^{m-k}$.