TITLE: Two different sequences of random variables each converge in distribution; does their sum?
QUESTION [4 upvotes]: My question is about basic probability.
We have two sequences of random variables, $ \{ X_n \}$ and $\{ Y_n \}$, such that each converge in distribution - i.e. there exist random variables $X$ and $Y$ such that:
$X_n \rightarrow X$ as $n \rightarrow \infty$ in law, and $Y_n \rightarrow Y$ as $n \rightarrow \infty$ in law.
What can we say about the sequence $X_n + Y_n$? I would like to say that it converges in law to $X +Y$ - but is this true?
Many apologies if this question has already been asked and answered. Thank you so much!!
John
p.s. Feel free to add some assumptions like independence of the $X_n$ and $Y_n$ in the process of answering this question.

REPLY [2 votes]: In general, this is not true. If one of the two sequences converges in probability (which is the case of $X$ or $Y$ are constant), the result holds by Slutsky's theorem.


TITLE: Problem involving triangle. Find $x$ in the figure
QUESTION [2 upvotes]: I need to find $x$ in the triangle above.
I tried to do basic things, like sum of a triangle's internal angles $= 180^\circ$ but I only found $2$ equations for $3$ variables
Any help is appreciated

REPLY [0 votes]: Using the same labeling as the author of the trig proof:
Cut off $BE$ on $BC$, such that $BE=BA$. Then, using the fact that $BD$ is the bisector of  $\angle B$, we can easily show that  $\triangle ADE$ is equilateral, so that $EA=ED$. Similarly, we can show that $\triangle AEC$ is isosceles (with base angles of $40^\circ$), so that $EA=EC$. Therefore $ED=EC$, so that, in $\triangle EDC, \angle EDC = \angle ECD = x$.
But the external angle $BED (=\angle BAD) = 20^\circ$ , therefore $2x = 20^\circ$ ,
yielding that $x=10^\circ$.


TITLE: how to prove that a matrix is definitely invertible
QUESTION [1 upvotes]: T $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} a & -b \\ b & a \end{smallmatrix} \bigr)$
prove that any given matrix on image of T excluding the zero matrix is definitely an invertible matrix.
oh and a,b,c,d $\in\mathbb{R}$

REPLY [3 votes]: The determinant is $a^2 + b^2$, which is non-zero for all $(a,b) \ne (0,0)$.


TITLE: Difference between adherent point, boundary point and limit point ?
QUESTION [4 upvotes]: What are their differences I cannot distinguish them from their definitions.

REPLY [4 votes]: Let $S$ be a set in some topological space $(X,\tau)$. Denote by $S^\circ,S^-$ the interior and closure of $S$, respectively.

If $x \in S^-$, $x$ is an adherent point of $S$.
If $x \in S^- \setminus S^\circ$, $x$ is an boundary point of $S$.
If there exists a sequence $(x_j)_{j=0}^\infty$ in $S$ so that $x_j \neq x$ and $x_j \to x$, $x$ is a limit point of $S$.

Assuming that the topology on $X$ is induced by a metric $d$, then

If $\forall r > 0, B(x;r) \cap S \neq \emptyset$, $x$ is an adherent point.
If $\forall r > 0, B(x;r) \cap S \neq \emptyset, B(x;r) \cap S^c \neq \emptyset$, $x$ is a boundary point.
If $\forall r > 0, B(x;r) \setminus \{x\} \cap S \neq \emptyset$, $x$ is a limit point.


TITLE: A Turing machine algorithm which determines other algorithms
QUESTION [1 upvotes]: Let $X$ be a Turing machine algorithm, which run as following:

For a Turing machine algorithm A,
X(A)=0 if A(A)=0
X(A)$\neq$0 if A(A)$\neq$0

We can code X easily. However, what's the result of X(X)?
It runs forever? or We don't know the result?

REPLY [1 votes]: The most naive machine $X$ that meets your requirements simply does the following:

$X$: If the input represents a Turing machine $A$, then simulate $A$'s behavior on the input (that is, simulate it on $\langle A \rangle$) and halt with the simulated output.

(The machine's behavior if the input doesn't represent a Turing machine is not relevant.)  If $X$ is implemented this way, then $X(\langle X \rangle)$ runs forever: the machine will only terminate after the simulated machine terminates, which will happen only after the simulation's simulated machine terminates, etc… turtles all the way down.  However, we can also devise a machine $X'$ such that $X'(\langle X'\rangle)$ halts with output $0$ (or any other output you like), but still behaves the same for all other inputs, and so still meets all your requirements (which, after all, are trivially satisfied when the input is the machine's own description).  First we need something like this:

$Q$: If the first and second inputs are equal, then halt with output $0$.  Otherwise, simulate $X$'s behavior on the second input and halt with the simulated output.

Then we want to build $X'$ out of this, by passing $\langle X'\rangle$ as the first input to $Q$, and the input to $X'$ as the second input to $Q$.  But how do we get $X'$ to have access to its own description?
It turns out that there is a general procedure for constructing Turing machines like this.  Specifically, for any Turing machine $Y$ that takes two inputs, there is a related Turing machine $G_Y$ that takes one input, such that $G_Y(a) = Y(\langle G_Y \rangle, a)$ for any input $a$.  I'll leave the procedure as an exercise (as it's fairly detailed, and anyway you can look it up).  But it should be clear that $X'=G_Q$ behaves just like the original $X$ except for a single input: its own description, on which it halts immediately with output $0$.


TITLE: Binary Operation with Cayley Table
QUESTION [1 upvotes]: I am asked to write out a Cayley table of a binary operation $\ast$ on the set $$ S = \{1, 2, 3\}$$ for which there is no solution for $\ast$ in $S$ to the equation $1 \ast x = 2$.
Here is my attempt.  Is this correct?  If $1 \ast x = 2$ has no solution in $S$, should that entry in the Cayley table be empty or can it be some other integer outside of $S$, i.e. 4,5, or 6?
\begin{array}{cc|c|c|c|}
* && 1 & 2 & 3\\
&&&&\\
\hline
1 &&  &  & \\
\hline
2 && 1 & 3 & 2\\
\hline
3 && 2 & 1 & 3\\
\hline
\end{array}

REPLY [2 votes]: "No solution to $1*x=2$ means that $2$ can not appear in the first row of your Cayley table; if $2$ appeared, that would mean that $1$ star something is equal to $2$.


TITLE: Showing isomorphism of two $C^*$ algebras
QUESTION [3 upvotes]: It seems that quite a standard trick of showing two $C^*$ algebras are as follows:

Let $A$ be a $C^*$ algebra $B$ another $C^*$ algebra. $A' \subseteq A$ be a subalgebra that is closed under $*$. (Not necessarily closed) and is dense in $A$. Let $L:A' \rightarrow B$ be a $*$-isometric homomoprhism which has dense image.  Then $L$ extends to a $*$-isometric isomorphism .

So my aim is showing that $L$ actually extends to an injective map - since  it is not necessarily case that $L$ is injective in its closure.

REPLY [1 votes]: The answer by Martin is of course correct, I just want to emphasise that this statement and the way to see it is essentially elementary functional analysis. In my opinion is very helpful to be able to differentiate when a statement is about the linear structure and when a statement needs the C*.
Central is the following statement: Let $A:V\to W$ be a continuous linear map between normed vector spaces, then there is a unique continuous linear extension $\overline A :\overline V\to\overline W$ between the completions. This can be done as an elementary exercise and is basically the first half of Martin's answer.
(As a remark: The above statement is itself an expression of a more general statement, namely that Lipschitz maps between metric spaces $A,B$ so that $B$ is complete and $A$ is dense in some other metric space $A'$ always have a unique continuous extension $A'\to B$.)
The rest of what you need follows from:

If a continous linear map is isometric on a dense sub-set, then it is isometric on its entire domain.
If a continuous linear map is multiplicative wrt continuous multiplications on a dense sub-set, then it is multiplicative on its entire domain.
If a continuous linear map $A$ satisfies $A(v^*)= A(v)^*$ on a dense sub-set and the $*$ operation is continuous on domain and codomain then the above equality holds on the entire domain.

Note that this is basically the same statement $3$ times: Continuous linear maps that preserve a continuous structure on a dense sub-set automatically do so on the entire domain.


TITLE: Prove that $\mathscr B$ is satisfiable if and only if $(\exists y_1)...(\exists y_n)\mathscr B$ is satisfiable.
QUESTION [1 upvotes]: Prove that $\mathscr B$ is satisfiable if and only if $(\exists y_1)...(\exists y_n)\mathscr B$ is satisfiable.
This question is from "Introduction to Mathmatical Logic" by Elliot Mendelson , forth edition , exercise 2.20.
The question is as follows:

Prove that if the free variables of $\mathscr B$ are $y_1,...,y_n$ , then $\mathscr B$ is satisfiable if and only if $(\exists y_1)...(\exists y_n)\mathscr B$ is satisfiable.

My attempt:
I first start by proving a lemma

lemma: $\mathscr B$ is satisfiable if and only if $(\exists x_i)\mathscr B$ is satisfiable.

proof: [$\to$]

If $\mathscr B$ is satisfiable , then there is a sequence $s$ in an interpretation $M$ such that $s$ satisfies $\mathscr B$.
$s$ doesn't satisfy $\lnot \mathscr B$
$s$ doesn't satisfy $(\forall x_i) \lnot \mathscr B$
$s$ satisfies $\lnot (\forall x_i) \lnot \mathscr B$
$s$ satisfies $(\exists x_i)\mathscr B$
$(\exists x_i)\mathscr B$ is satisfiable.

[$\leftarrow$]

$(\exists x_i)\mathscr B$ is satisfiable.
$\lnot (\forall x_i) \lnot \mathscr B$ is satisfiable
if $\lnot (\forall x_i) \lnot \mathscr B$ is satisfiable , then there is a $s$ in an interpretation $M$ such that $s$ satisfies $\lnot (\forall x_i) \lnot \mathscr B$.
$s$ doesn't satisfy $(\forall x_i) \lnot \mathscr B$.
There exists a sequence $s'$ that differs from $s$ atmost the $i_{th}$ element such that $s'$ doesn't satisfy $\lnot \mathscr B$
$s'$ satisfies $\mathscr B$

Thus the lemma is proven.
Now , to prove the main theorem , I will use induction.

$\mathscr B$ is satisfiable if and only if $(\exists y_n)\mathscr B$ is satisfiable.
$\mathscr B$ is satisfiable if and only if $(\exists y_m)\mathscr B$ is satisfiable. [where $1 \leq m<n$]

With this , the main theorem is proven. [end]
Now , I have some skepticism about my proof , for the reason that I didn't use a single time in my proof the information that:

The free variables of $\mathscr B$ are $y_1,...,y_n$.

So I think there is a big flaw with my proof which I am simply ignoring and not able to find.Can someone tell me what is that flaw?

REPLY [1 votes]: There is no flaw in your proof. The requirement in the problem that the free variables of $\mathscr{B}$ are $y_1, \ldots, y_n$ is not necessary as the statement is true for any $y_1, \ldots, y_n$, which may include variables that do not occur free in $\mathscr{B}$ and need not includes all the free variables of $\mathscr{B}$: if $y$ is not free in $\mathscr{B}$, then $\mathscr{B}$ and $(\exists y)\mathscr{B}$ are logically equivalent and hence equisatisfiable; if $y$ and $z$ both occur free in $B$, then $(\exists y)(\exists z)\mathscr{B}$ and $(\exists z)\mathscr{B}$ are equisatisfiable. Your proof deals correctly with both of these cases.


TITLE: Subset of knight's move in chess.
QUESTION [10 upvotes]: A particle is allowed to move in the $\mathbb{Z}\times \mathbb{Z}$ grid by choosing any of the two jumps:
1) Move two units to right and one unit up
2) Move two units up and one unit to right.
Let $P=(30,63)$ and $Q=(100,100)$, if the particle starts at origin then?
a) $P$ is reachable but not $Q$.
b) $Q$ is reachable but not $P$.
c) Both $P$ and $Q$ are reachable.
d) Neither $P$ nor $Q$ is reachable.

I could make out that the moves given are a subset of that of a knight's in chess.
I think that it'd never be able to reach $(100,100)$ but I'm not sure of the reason. It has got to do something with the move of the knight but I cannot figure out what.
I don't have a very good idea about chess, so I'd be glad if someone could answer elaborately.

REPLY [0 votes]: Let a be the count of the first move and b be the count of the second move.
$$(x,y) = (2,1)a + (1,2)b$$
Where x an y are nonnegative integers representing our position and a position is reachable if a and b are both nonnegative integers.
$$x = 2a + b$$
$$y = 2b + a$$
We have two linear equations and two unknowns, so we expect exactly one pair of a and b for each pair of x and y, the question then is are a and b nonnegative integers?
$$2y = 4b + 2a$$
$$2y - x = 3b$$
$$b=\frac{2y-x}{3}=y-\frac{x+y}{3}$$
$$x = 2a + \frac{2y-x}{3}$$
$$3x = 6a + 2y-x$$
$$4x -2y = 6a$$
$$a = \frac{2x-y}{3}=x-\frac{x+y}{3}$$
Requiring that a and b are nonnegative integers now gives us several conditions.

$2y \geqslant x$
$x \geqslant 2y$
$x+y$ is divisible by 3

Your first position fails critera 1, your second position fails critera 2.


TITLE: Axiom of choice - Proving an equivalence
QUESTION [1 upvotes]: I am reading a book on set theory and I got stuck on a simple step of a proof involving an equivalence of the axiom of choice.

Axiom of Choice (1) For every set $a\neq \emptyset$ there exists a function $f:\mathcal{P}\left (a\right )\setminus \left \{\emptyset\right \}\to a$ such that $f\left (b\right )\in b$ for every $b\subseteq a$, $b\neq \emptyset$
Axiom of Choice (2) Let $\left \{a_i\right \}_{i\in I}$ be a non-empty family of pairwise disjoint non-empty sets (i.e. $I\neq \emptyset$ and $a_i\cap a_j=\emptyset$ if $i\neq j$). Then there exists a set $d$ such that $d\cap a_i$ is a unit set for every $i\in I$.

I could understand the proof $\left (2\right )\Rightarrow \left (1\right )$, but I cannot understand the converse.
The proof starts by supposing that $\left (1\right )$ is true. We let $\left \{a_i\right \}_{i\in I}$ be a non-empty family of pairwise disjoint non-empty sets and define $a=\bigcup_{i\in I}a_i\neq \emptyset$. Therefore there exists a function $f:\mathcal{P}\left (a\right )\setminus \left \{\emptyset\right \}\to a$ be a function such that $f\left (b\right )\in b$ for every $b\in \mathcal{P}\left (a\right )\setminus \left \{\emptyset\right \}$. If $d$ is the image of $f$, then (the book says) we have $d\cap a_i=\left \{f\left (a_i\right )\right \}$. My question is: Why? It is obvious that one of the inclusions holds, but what about $\subseteq$?

REPLY [3 votes]: Recall that $a=\bigcup_{i\in I}a_i$ is trivially a superset of each of the $a_i$'s. Therefore each $a_i$ is an element of $\mathcal P(a)$.
By assumption, no $a_i$ is empty, so $a_i$ is in the domain of the choice function. Now take $\{f(a_i)\mid i\in I\}$ to be $d$, and by the fact each $a_i$ is disjoint we get that $d\cap a_i$ is a singleton for all $i\in I$.


text:
\section{Conclusion}
\label{sec:conclusion}

Parameter selection for meta-heuristics is a topic that is widely
regarded in the literature. In this paper we focus on the parameter
selection for PSO and its influence on the convergence of the particle
swarm. We introduce a new convergence indicator $\Phi_t$ that can be
used to prove for a selection of parameters whether the swarm will finally
converge or diverge. If in expectation the difference of two
consecutive convergence indicators is negative the swarm will finally
converge. We introduce a series of equations for calculating this
difference of convergence indicators and explain how to
numerically solve these equations using cubic spline interpolation
where the results have only minor errors.
Finally we provide experiments that confirm the correctness of
presented equations and their numerical evaluation.

config: arxiv


TITLE: Tangent space of coordinate axes
QUESTION [3 upvotes]: Let $X$ be the union of the $3$ coordinate axes in $\mathbb{A}^{3}$. I want to compute the tangent space of $X$ at every $p \in \mathbb{A}^{3}$.
One can check that the only singular point of $X$ is the origin so if $p$ is the origin then the tangent space of $X$ at the origin is equal to $\mathbb{A}^{3}$.
Question: how do we express the answer if $p \in \mathbb{A}^{3} \setminus \{(0,0,0)\}$? Would it be correct to say:
Note that $I(X)=(xy,xz,yz)$ so if $p=(a,b,c) \in \mathbb{A}^{3}$ then:
$T_{p}(X):=\{(x_{1},x_{2},x_{3}) \in \mathbb{A}^{3}: \sum_{i=1}^{3} \frac{\partial f}{\partial x_{i}}(a,b,c)(x_{i})=0\ \text{for every f in I(X)}\}$
In case $p$ is not the origin, is there a "nice" way to express $T_{p}(X)$ other than the above?

REPLY [2 votes]: Sure there is an easy way to compute $T_p(X)$ for $(0,0,0)\neq p\in X$ !
Just use that $T_p(X)=T_p(Y)$for any subscheme $Y\subset \mathbb A^3$ that coincides with $X$ in an open  neighbourhood of $p$: the tangent space $T_p(X)$ only depends on the germ of $X$ at $p$.
So here  take for $Y$ the unique coordinate axis axis on which $p$ lies.
For example if $p=(0,b,0)$ with $b\neq 0$, then $T_p(X)\subset T_p(\mathbb A^3)$ is the line $k\cdot (0,1,0)\subset k^3$


TITLE: Surgery on trivial knots
QUESTION [3 upvotes]: I know a theorem that any closed orientable 3 manifold can be obtained from the sphere $S^3$ by surgery along a framed knot.
I think I read or heard somewhere that as a surgery link, we can take unknots with integer framing.
I am confused whether if this means that we can find a surgery link whose connected components are unknots with integer framing AND each of the components are "separated" (there exists disjoint balls containing each of the components. Is this the same as linking number is zero?)
Or does it mean that each component is unknot, but it might be they are linked?

REPLY [6 votes]: It means each component is an unknot, but the entire link is not necessarily trivial.
If you take the Lickorish-Wallace theorem for granted (every 3-manifold is obtained by integral surgery on a link in $S^3$), it is not too hard to see how to make every component of such a link an unknot using Kirby moves. You can blow up and then do handle slides to change the crossings of any component of your link without changing the resulting $3$-manifold. It is a classical fact that every knot can be turned into the unknot by changing some of its crossings. Hence we can always alter a framed link in $S^3$ in such a way that we get a new framed link with all components unknots.
Here's a picture of the "Kirby move" that changes crossings (from Gompf and Stipsicz's $4$-Manifolds and Kirby Calculus):

If $M$ is a $3$-manifold obtained by surgery on a framed unknot, then $M$ is $S^3$, $S^2 \times S^1$, or a lens space. If $M$ is obtained by surgery on a framed unlink, then $M$ is a connected sum of copies of $S^3$, $S^2 \times S^1$, and lens spaces. There are many $3$-manifolds that are not of this type. One example is the Poincaré homology sphere $\Sigma(2,3,5)$. Hence in general a $3$-manifold can only be obtained by integral surgery on a nontrivial link with all components unknots, not necessarily an unlink.


TITLE: Generating function for number of ways n people can pick a total of r1 chairs of type 1, r2 chairs of type 2 etc
QUESTION [2 upvotes]: This is a homework question for my combinatorics class that I just need to be pointed in the right direction to start.
Find a generating function $x_1, x_2, . . . , x_m$ whose coefficient of $x_1^{r_1} x_2^{r_2} . . . x_m^{r_m}$ is
the number of ways $n$ people can pick a total of $r_1$ chairs of type $1$, $r_2$ chairs of type $2$, . . . $r_m$ chairs of type $m$ if (a, b, and c are all unique and separate scenarios)
(a) Each person picks one chair
(b) Each person picks either two chairs of one type or no chairs at all
(c) Person $i$ picks up to $i$ chairs of exactly one type

REPLY [0 votes]: Basic idea: [I am solving only part a]
a)

Let $P(r_1,r_2,...,r_m)$ denote the number of ways n people can pick a total of $r_1$ chairs of type 1, $r_2$ chairs of type 2, . . . $r_m$ chairs of type m such that each person picks one chair. Clearly, $$n=r_1+r_2+...+r_m$$
Claim
$$P(r_1,r_2,...,r_m) = {n\choose r_1}*{(n-r_1)\choose r_2}...{(n-(r_1+r_2+...+r_{m-2})\choose r_{m-1}}*{r_m\choose r_m}  $$
$$= n! / (r_1!r_2!...r_m!) = {n \choose r_1,r_2,..,r_m}$$[Known as multinomial coefficient]
Proof :You select $r_1$ people from n who will receive chair of type 1 in ${n\choose r_1}$ ways, next you select $r_2$ people from remaining ones who will receive chair of type 2 in ${(n-r_1)\choose r_2}$ ways and so on.

Thus your generating function F is, $$ F(x_1,x_2,..,x_m) = \sum_{r_1+r_2+...+r_m=n}P(r_1,r_2,...,r_m)*x_1^{r_1}*x_2^{r_2}...*x_2^{r_m}$$
 $$=\sum_{r_1+r_2+...+r_m=n}{n \choose r_1,r_2,..,r_m}*x_1^{r_1}*x_2^{r_2}...*x_m^{r_m} = (x_1+x_2+...+x_m)^n$$
This is also known as multinomial distribution. You can find out the generating function for b) and c) in similar way.


TITLE: Quotient map $\pi : X \rightarrow X / \mathrm{ker}(A)$ is open for a bounded linear operator $A$
QUESTION [2 upvotes]: I'd like to show: if $A : X \rightarrow Y$ is a bounded linear operator between Banach-spaces, then $\pi : X \rightarrow X / \mathrm{ker}(A)$ is a open map.
I found a proof, which I do not really like, namely by first embedding $X$ into $X \times (X / \ker (A))$ by $x \mapsto (x, \pi(x))$ - which is not too hard to be proven an open map - and then using the general fact from topology that any projection from a product-space is open. Then $\pi$ is the composition of the two maps that have been shown to be open.
However, I was hoping that there would be a more elegant way to prove this.
Thanks for any suggestions!

REPLY [1 votes]: Notice that $\pi$ is surjective linear mapping. Since $\mathrm{ker}(A)$ is closed subspace of $X$ the quotient space $X/\mathrm{ker}(A)$ endowed with the quotient topology is Banach space. Now you can apply the Open Mapping theorem.