TITLE: Let $T$ be a Hermitian operator and $\|v\|=1$, prove $\langle T^2(v),v\rangle \ge\langle T(v),v\rangle^2$
QUESTION [4 upvotes]: Let $T$ be a Hermitian operator on an inner product space $V$ over the field of complex numbers, $C$ (of a finite dimension), also assume $\|v\|=1$. Prove:
$$\langle T^2(v),v\rangle \ge\langle T(v),v\rangle^2$$

REPLY [2 votes]: Edit: As pointed out by Robert Israel, this follows immediately from Cauchy-Schwarz. I so wanted to use your (wrong and now deleted hint), that I actually basically reproved Cauchy-Schwarz...
Develop the following
$$
((T-\alpha I)^2v,v)=\alpha^2\|v\|^2-2\alpha(Tv,v)+(T^2v,v).
$$
Note that for every $\alpha\in \mathbb{R}$, the operator $(T-\alpha I)^2=(T-\alpha I)^*(T-\alpha I) $ is hermitian positive so
$$
((T-\alpha I)^2v,v)\geq 0.
$$
We have a quadratic in $\alpha$ which is always nonnegative.
This means that its discriminant is nonpositive:
$$
\Delta=4((Tv,v)^2-(T^2v,v)\|v\|^2)\leq 0.
$$
Hence
$$
(Tv,v)^2\leq (T^2v,v)\|v\|^2.
$$
So your inequality holds for all $\|v\|\leq 1$.


TITLE: full bins with limited capacity after throwing balls
QUESTION [0 upvotes]: Expected  number of full  bins after throwing balls uniformly randomly to bins that have limited capacity
Let us have $N$ bins with the same limited capacity ($N_{max}=C$), in the sense that if a ball is threw into a bin that already has $C$ balls in it, the ball is discarded.
After throwing (uniform random allocation) $L$ balls, what is the expected number of bins that are full?
considerations:

since uniform random allocations, after $L$ launches of balls we should have an uniform distribution of $L/N$ balls in each bin
if $L<C$, no bin can be full (few balls to have the chance to have one bin full), so the expected number of "full" bins should be 0.
if $L \in [C,C\cdot N ) $ it should be expected to have  $\dfrac{L}{C}$ full bins, and $N-L/C$ bins remain free to host a new ball.
if $L>=C\cdot N$ no bin can be filled any more,all the bins reached their capacity. Any attempts of a new launch of a ball is rejected.

This is an idea, is there an analytical closed form way to express that?

REPLY [0 votes]: This answer assumes the model you have described, where balls always choose their target uniformly (and independently of each other) among all bins, even if some are full.
Let $B_i$ denote the number of balls thrown at the $i$th bin after $L$ balls have been thrown.  Since each ball has a $\frac{1}{N}$ chance of being thrown at the $i$th bin, $B_i \sim \textrm{Bin}\left(L, \frac{1}{N} \right)$ is binomially distributed.
The $i$th bin is full if $B_i \ge C$.  Using the probability mass function of the binomial distribution, the probability of this happening is $\sum_{n = C}^L \binom{L}{n} \frac{1}{N^n} \left(1 - \frac{1}{N} \right)^{L - n}$.
By linearity of expectation, the expected number of full bins is the sum of the probabilities that each bin is full.  The bins have the same distribution, and so the expected number of full bins after $L$ balls have been thrown is
$$ N \sum_{n = C}^L \binom{L}{n} \frac{1}{N^n} \left(1 - \frac{1}{N} \right)^{L - n}. $$
Note that this will converge to $N$ as $L \rightarrow \infty$, but will never be equal to $N$, since, as mentioned in the comments, you might get so unlucky as to have all the balls thrown at the same bin.

REPLY [0 votes]: I don't think the third bullet is right, either. If L=C, I would certainly not expect a bin to be full, since the only way that happens is if all L=C balls go into one bin.
Based on the problem description, it seems that the appropriate sample space $S$ is the set of ways to distribute L indistinguishable balls into N distinguishable boxes, independent of the fullness constraint, since any box being full does not prevent attempts to add more balls.
In this sample space, let $A_i$ denote the set of trials with $i \in {1..N}$ such that bin $i$ is full. This occurs whenever bin $i$ has at least C balls, which can happen in $\displaystyle{{L-C+N-1}\choose{N-1}}$ ways.
Now, $A_i$ may contain trials where boxes other than box $i$ are full, so to calculate probabilities we would need to use inclusion/exclusion. But since we only need the expected number of full boxes, I think we can calculate this as $$\frac{1}{|S|}\sum_{i=1}^N |A_i|$$
Every trial with more than one full box gets counted once for each full box, which is exactly the multiple counting needed for the value of the random variable in the expectation calculation. Noting that $|S|=\displaystyle{{L+N-1}\choose{N-1}}$, we get the expectation $$\frac{N\times L!\times(L-C+N-1)!}{(L-C)!\times(L+N-1)!}$$
Note: I'm traveling, without references, but believe this is correct.


TITLE: Bandwidth of laser pulses
QUESTION [1 upvotes]: I am working on a practice test for my optics class.  The professor does not provide the solutions for the test.  I was just wondering if someone could tell me if I am doing this correctly because it seems kind of simple but I am unsure.

A laser emits short pulses of duration 14.0 ps separated in time by
  1.0 ns at a wavelength of 1 um.  From one pulse to the next, the timing of field oscillations is kept perfectly  coherent and phase
  stable using sophisticated feedback control.
What is the bandwidth of the source (in Hertz)?

I think the answer to this would be 1/t which is 1 /14 ps = 71 Ghz
I think it is just that simple but I want to know if I am wrong so I do not make that mistake on the actual test.  Also just to clarify the bandwidth of the pulse is the range of frequencies in that pulse of light right?

REPLY [1 votes]: About 30 GHz,
according to this.
Check "Bandwidth-limited Pulses".
Because of Fourier transform, the product of the temporal duration and spectral width is ≈ 0.44 for Gaussian-shaped pulses. See also this.


TITLE: Need one example solving trigonometry.
QUESTION [0 upvotes]: Calculate $\sin \beta, \tan \beta, \cot \beta, \cos(2\beta)$ if $\cos \beta = {5 \over 13}$ and $\beta \in (0^{\circ},90^{\circ})$.
I'm a student and I forgot how to solve it correctly...I need just one example.
Thanks.

REPLY [0 votes]: There are many ways:-

Try making a triangle and use $5$ as length of base and $13$ as length of Hypotenuse. [Hint: $5^2 +12^2 = 13^2$]
$\cos \beta = 5/13$ apply $\sin^2 \beta + \cos^2 \beta = 1$
For $\cos (2 \beta)$ use $\cos(2 \beta)= 2\cos^2 \beta -1$


TITLE: Assume $\{x\}$ is transitive. What can you say about $x$?
QUESTION [1 upvotes]: Definition. A set $X$ is transitive if every element of $X$ is a subset of $X$.

Assume $\{x\}$ is transitive. What can you say about $x$?
I think, $x$ is the set of natural numbers.

REPLY [3 votes]: HINT: The set $\{x\}$ is transitive if and only if each element of $\{x\}$ is a subset of $\{x\}$. The only element of $\{x\}$ is $x$, so if $\{x\}$ is transitive, then $x\subseteq\{x\}$. This means that if $y\in x$, then $y\in\{x\}$.

If there is a $y\in x$, what must it be equal to? This makes $x$ a very strange set; why?
If there is no $y\in x$, what is $x$? Is $\{x\}$ transitive in this case?


TITLE: Explicit integral representation theory
QUESTION [16 upvotes]: The representation theory of finite groups over the complex numbers is classical und it is usually quite easy to compute the set of isomorphism classes of irreducible representations, at least for small examples. Now, sometimes one is not content with the representation theory over the complex numbers or even over any field, but one wants to consider representations over $\mathbb{Z}$ or $\mathbb{Z}_{(p)}$. At least for the non-expert it is not so easy to obtain a complete list of isomorphism classes of indecomposable representations in these cases even for small examples. So, what I want to ask is the following:
1) Can one formulate a "guide" how to obtain such a list?
2) Is there a place in the literature where a list of indecomposables/irreducibles is given for some small examples as $S_3$ (the symmetric group on three elements)? In the last example I am especially interested.

REPLY [25 votes]: You might be asking about four separate types of modules:

irreducible Z[G] modules,
Z-forms of irreducible Q[G] modules,
indecomposable Z[G] modules, or
indecomposable Z[G] modules that are finitely generated and free as Z-modules.

I'll assume the last is the main concern.
The irreducible modules of ZS3 are all finite and have an elementary abelian p-group as their additive group.  For p=2,3 there are 2 each, and for p>3, there are 3 each.
The irreducible CS3 modules are all realizable over Q.  Every such module may be realized over Z, but the two-dimensional representation has two distinct Z-forms, giving four total "irreducible" Z-free ZS3 modules, that is, four total Z-forms of irreducible QS3 modules.
Indecomposable ZS3 modules up to isomorphism are more complicated than the human mind can possibly comprehend.  Indeed, even those in which S_3 acts as the identity are much too complex.  Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).
The indecomposable ZS3 modules that are free as Z-modules are classified in:
Lee, Myrna Pike.  "Integral representations of dihedral groups of order 2p."
Trans. Amer. Math. Soc. 110 (1964) 213–231.
MR 156896
doi:10.2307/1993702
There are 10 of them, and the Krull-Schmidt theorem fails for them.  Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique.  In other words, integral representations of even very small groups are quite complicated.


TITLE: Compute the expected value of the surplus water.
QUESTION [1 upvotes]: Water flows in and out of a dam such that the daily inflow is uniform on
  $[0, 2]$ (liters) and the daily outflow is uniform on $[0, 1]$,
  independently of the inflow. Each day the surplus water (if there is
  any) is collected for an irrigation project. Compute the expected
  amount of surplus water in a given day.

Let $X=$ inflow and $Y=$outflow. Then $X\sim \text{unif}[0,2]$ and $Y\sim \text{unif}[0,1]$ and we denote the surplus as $g(X,Y)=Z=X-Y.$
By independence, it follows that $f(x,y)=f_X(x)f_Y(y)=1\cdot1/2=1/2.$ The range of $Z$ is $[0,2]$, since on a given day one can have zero outflow and 2 inflow or zero inflow and zero outflow. The formula for the expected value of $g(X,Y)$ is
$$E(g(X,Y))=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x,y)f(x,y) \ dxdy \tag{1}$$
But how do I determine the bounds for this doubleintegral?

REPLY [1 votes]: We should have $g(X,Y)=Z=\max(0,X-Y).$
Note that $0 \le x \le 2$ and $0 \le y \le 1$.
\begin{align}E(g(X,Y))&= \int_0^1\int_0^2 g(x,y) f(x,y) \,\, dxdy \\
&=\int_0^1\int_y^2 (x-y) f(x,y)\,\, dxdy\end{align}


TITLE: Verify the identity $\frac{\tan(a+b)}{\tan(a-b)}$ = $\frac{\sin(a)\cos(a)+\sin(b)\cos(b)}{\sin(a)\cos(a)-\sin(b)\cos(b)}$
QUESTION [3 upvotes]: I've been asked to verify the following identity but I don't know how to do it.
$$\frac{\tan(a+b)}{\tan(a-b)} = \frac{\sin(a)\cos(a)+\sin(b)\cos(b)}{\sin(a)\cos(a)-\sin(b)\cos(b)}$$
When I try I get
$$\frac{\tan(a+b)}{\tan(a-b)} = \frac{\dfrac{\sin(a+b)}{\cos(a+b)}}{\dfrac{\sin(a-b)}{\cos(a-b)}} = \frac{\dfrac{\sin(a)\cos(b)+\cos(a)\sin(b)}{\cos(a)\cos(b)-\sin(a)\sin(b)}}{\dfrac{\sin(a)\cos(b)-\cos(a)\sin(b)}{\cos(a)\cos(b)+\sin(a)\sin(b)}}$$  But I don't know really where to go from here.

REPLY [2 votes]: This can be somewhat lengthy and can take more time but give it a try :

In the  last step you know what you have to apply.
Hope it helps.


TITLE: Change of Variables and Splitting fields
QUESTION [0 upvotes]: A comment in Size of Galois Group based on roots and minimal polynomial says:
"By a change of variables a polynomial $f(X)$ in $\mathbb{Q}[X]$ can be made monic in $\mathbb{Z}[X]$ without changing its splitting field over $\mathbb{Q}$."
I guess the standard way to convert a polynomial $f(X)\in \mathbb{Q}[X]$ into a monic polynomial over $\mathbb{Z}$ is: First make it monic by multiplying it by a rational number and then replace it by $D^{ \, \text{deg}(f)} f(\frac{X}{D})$, where $D$ is a common denominator of the coefficients.
I don't see why the splitting field of the thus obtained polynomial is the same as that of the original polynomial? For example, the splitting of $5X^2 - 2$ is $E=\mathbb{Q}(\sqrt{\frac{2}{5}})$, the corresponding monic polynomial over $\mathbb{Z}$ is $X^2 - 10$, whose splitting field is $F=\mathbb{Q}(\sqrt{10})$. Are $E$ and $F$ isomorphic, and why so?

REPLY [1 votes]: To state everything clearly:
We have $E = \Bbb Q[X]/(5X^2 - 2)$ and $F = \Bbb Q[Y]/(Y^2 - 10)$.
We define a homomorphism $f$ from $\Bbb Q[X]$ to $F$ by sending $X$ to the class of $\frac 1 5 Y$.
It is clear that the image of $5X^2 - 2$ under $f$ is zero, i.e. the ideal $(5X^2 - 2)$ lies in the kernel of $f$.
Therefore $f$ induces a quotient homomorphism $\phi: E \rightarrow F$, sending the class of $X$ to the class of $\frac 1 5 Y$.
In the same way, we may define a homomorphism $\psi: F \rightarrow E$, sending the class of $Y$ to the class of $5X$.
We then verify that $\phi \circ \psi$ is the identity, and $\psi \circ \phi$ is the identity.
This shows that $E$ and $F$ are isomorphic to each other.


TITLE: evaluate the side of a triangle?
QUESTION [1 upvotes]: $ABC$ equilateral triangle
$M$ point inside the triangle , let $D, E, F$ be the vertical projection (orthogonal) of M on $AB, BC, AC$  respectively, where $MD=1,ME=2,MF=3$

Then evaluate the side of a triangle ?

I'm thinking calculated $MA,MB,MC$ then esay i use rotational $R(A,\frac{π}{3})$
 But how I find it ?
I'm try
Connect $D,E$ but $BED$ not Isosceles
So I have already to see your solution

REPLY [1 votes]: As shown above, join $M$ to each of the equilateral triangle vertices of $A$, $B$ and $C$. The line segments $MD$, $ME$ and $MF$ are the heights of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$.
Note the area of the surrounding equilateral triangle must equal the sum of the areas of these $3$ smaller triangles. Let the length of the equilateral triangle side be $s$. The height of $\triangle ABC$ would be $\frac{\sqrt{3}}{2}\left(s\right)$ (you can see this by using the side length ratios of $1:\sqrt{3}:2$ of a $30^{o}\text{ - }60^{o}\text{ - }90^{o}$ triangle). Using that the area of a triangle is $\frac{1}{2}\left(bh\right)$ formula, where $b$ is the base length and $h$ is the height, you then get
$$\begin{equation}\begin{aligned}
\frac{1}{2}\left(s\right)\left(\frac{\sqrt{3}}{2}\left(s\right)\right) & = \frac{1}{2}\left(s\right)\left(1 + 2 + 3\right) \\
\frac{\sqrt{3}}{2}s & = 6 \\
s & = 4\sqrt{3}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$


TITLE: Perfect square method
QUESTION [0 upvotes]: I didn't get this method.
\begin{equation*}
3a^2-9a+6=3(a^2-3a+2) \\
=3(a^2-1a*3+3^2-3^2+2)= \\
= 3[(a-3)^2-7],
\end{equation*}
then?

REPLY [1 votes]: If you desire to factor it by way of completing the square then it is simpler to first multiply by 4, namely
$$\rm\ \ 4\:(a^2-3\: a+2)\ =\ (2\:a-3)^2 - 1\ =\ (2\:a-2)\:(2\:a-4)$$
Finally, divide both sides through by $\rm\ \: 4\ =\ 2\cdot 2$


TITLE: Faulty proof in the claim "If a polynomial is irreducible in $Z_p [x]$ then it is irreducible in $Z_{p^k} [x]$?
QUESTION [2 upvotes]: I was reading a paper regarding the divisibility of polynomials modulo a composite number. The paper can be found here: https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.821.1301&rep=rep1&type=pdf.
In Propositions 2.7/2.8 of the paper, the author gives a proof of the claim that "If a polynomial $f(x) \in Z_{p^k} [x]$is irreducible in $Z_p [x]$ then it is irreducible in $Z_{p^k} [x]$". However, I don't quite understand why the proof given is correct. In fact, I suspect that it is faulty but I feel like I might be missing something here.
It is proven in Proposition 2.7 that $f$ can be written as $f'+pg$, where $p$ does not divide $f'$. I believe that the proof for this proposition is valid. However, it is worthy to note that there is no guarantee here that the degree of $f'$ is greater than 0.
The proof of 2.8 reads essentially as follows. Suppose for sake of contradiction $f$ is reducible in $Z_{p^k} [x]$, then $f \equiv gh \mod p^k \Rightarrow f \equiv(g'+pg_{1})(h'+ph_2)\mod p \Rightarrow f \equiv g'h' \mod p $, at which point the proof is ended. But how do we know that $g', h'$ are not constants? I don't believe they necessarily have to be, in which case the proof breaks down.
Am I missing something here?

REPLY [4 votes]: That paper has some problems.  I wouldn't spend much time on it unless you have a specific reason you need to do so.
For instance, 2.7 as written is trivially true, but then the proof is invalid.
And as you've pointed out, 2.8 is false.  As a simple counterexample, note that
$(x^2 + x + 1)(2x+1)$
is reducible in $\mathbb{Z}_4[x]$ but its image is irreducible in $\mathbb{Z}_2[x]$.
I would say it's possible we're restricting our attention to just monic $f$, but comments near the beginning of chapter 2 make me think that's not even the case.


text:
\begin{document}

\title[Hyperspaces of continua with connected boundaries]{Hyperspaces of continua with connected boundaries in $\pi$-Euclidean Peano continua}

\author{Pawe{\l}  Krupski}

\email{pawel.krupski@pwr.edu.pl}
\address{Department of Computer Science, Faculty of Fundamental Problems of Technology, Wroc\l aw University of Science and Technology, Wybrze\.{z}e Wyspia\'nskiego 27, 50-370 Wroc\l aw, Poland.}
\thanks{The author was partially supported by the Faculty of Fundamental Problems of Technology, Wroc\l aw University of Science and Technology, grant 0401/0017/17}

\date{\today}
\subjclass[2010]{Primary 57N20; Secondary 54H05, 54F15}
\keywords{absorber, Hilbert cube, hyperspace, $\pi$-Euclidean space}

\begin{abstract} Let $X$ be a nondegenerate Peano unicoherent continuum.  The  family $CB(X)$ of proper subcontinua of $X$ with connected boundaries is a  $G_\delta$-subset of the hyperspace $C(X)$ of all subcontinua of $X$. If every nonempty open subset of  $X$ contains an open subset homeomorphic to  $\mathbb R^n$ (such space is called $\pi$-$n$-Euclidean) and $2\le n<\infty$,  then $C(X)\setminus CB(X)$ is  recognized as an $F_\sigma$-absorber in $C(X)$; if additionally,  no one-dimensional subset separates $X$, then the family of all members of $CB(X)$ which separate $X$ is a $D_2(F_\sigma)$-absorber in $C(X)$, where $D_2(F_\sigma)$ denotes the small Borel  class of  differences of two $\sigma$-compacta.
\end{abstract}

\maketitle

All continua in the paper are metric. For a continuum $X$,
we consider the hyperspaces
$$2^{X}=\{A\subset X: \text{$A$ is closed and nonempty}\}$$ and $$C(X)= \{A\in  2^X: \text{$A$ is connected}\}$$
 with the Hausdorff metric. Define
 $$CB(X)= \{A\in C(X)\setminus \{X\}: \text{$Bd(A)$ is connected}\},$$
 where $Bd(A)$ denotes the boundary of $A$ in $X$.

\section{Evaluation of the Borel complexity  of $CB(X)$}

 K. Kuratowski observed in~\cite[p. 156]{Ku} that, for any compact nondegenerate $X$, the function $\alpha:2^X\setminus \{X\} \to 2^X$,    $\alpha(A)=\overline{X\setminus A}$ is lower semicontinuous, hence of the first Borel class, while the function $Bd: 2^X\setminus \{X\} \to 2^X$, $A\mapsto Bd(A)=A\cap \overline{X\setminus A} $, is of the second  Borel class for any nondegenerate continuum $X$.
 It means that, for each nondegenerate continuum $X$, the preimage $\alpha^{-1}(D)$ of a closed subset $D\subset 2^X$ is $G_\delta$  and the preimage $Bd^{-1}(D)$ is $F_{\sigma\delta}$ in $2^X\setminus \{X\}$. It follows that the preimages are also $G_\delta$ and  $F_{\sigma\delta}$ subsets of $2^X$.   Since $CB(X)=C(X)\cap Bd^{-1}(C(X))$, we have the following proposition.

 \begin{proposition}\label{p1}
 If $X$ is a nondegenerate continuum, then $CB(X)$  is an $F_{\sigma\delta}$-subset of $2^X$.
 \end{proposition}

Actually,  in many cases the Borel class of $CB(X)$ can be reduced. For example, one can easily see that
\begin{enumerate}
\item
if $X$ is the circle $S^1$ then  $CB(X)\overset{top}=S^1$ is   compact;
\item
 $CB(I)$    is homeomorphic to the real line $\mathbb R$, so it is an absolute $G_\delta$-set;
 \item
 if $X$ is an indecomposable nondegenerate continuum, then the family $CB(X)$ is also $G_\delta$ since it is equal to $C(X)\setminus \{X\}$.
 \end{enumerate}

\begin{lemma}\label{l1}
If $X$ is a nondegenerate  unicoherent continuum then   $CB(X)=C(X)\cap \alpha^{-1}(C(X))$.
\end{lemma}

\begin{proof}
If $C\in CB(X)$,  then $Bd(C)=C\cap \overline{X\setminus C}$ is connected and since $X=C\cup \overline{X\setminus C}$ is connected, the set $\overline{X\setminus C}$ is also connected by a well known fact~\cite[Corollary 5, p. 133]{Ku1}. Thus $C\in C(X)\cap \alpha^{-1}(C(X))$.

The converse implication follows directly from the unicoherence of $X$.
\end{proof}

Since $\alpha$ is of the first Borel class, Lemma~\ref{l1} implies the following proposition.

\begin{proposition}\label{p2}
If $X$ is a nondegenerate  unicoherent continuum then   $CB(X)$ is a $G_\delta$-subset of $C(X)$.
\end{proposition}

\section{Subcontinua with connected boundaries in $\pi$-Euclidean unicoherent Peano continua}

Recall that a family $\mathcal U$ of nonempty open subsets of a space $Y$ is a $\pi$-base of $Y$ if each nonempty open subset  of $Y$ contains some $U\in\mathcal U$.

\begin{definition}
A space $Y$ is said to be   $\pi$-Euclidean if there exist $n\in\mathbb N$ and a $\pi$-base of $Y$ whose elements are homeomorphic to $\mathbb R^n$;  such $Y$ will be also called $\pi$-$n$-Euclidean.
\end{definition}

In this section we assume that $X$ is a nondegenerate  unicoherent Peano continuum without free arcs.
Then the hyperspace $C(X)$ is a Hilbert cube~\cite{CS}.

\

One can immediately notice  that every subcontinuum of $X$ is approximated  (in the Hausdorff metric) by arcs  and the arcs  are nowhere dense in $X$, so they belong to $CB(X)$. Thus, we have
\begin{proposition}\label{p3}
$CB(X)$ is a dense $G_\delta$-subset of $C(X)$.
\end{proposition}

Imposing yet an additional structure on $X$ we can fully characterize the family $CB(X)$ in the next theorem which is our main result of this section.

Notice that if $n\ge 2$ then a  $\pi$-$n$-Euclidean space contains no free arcs.

\begin{theorem}\label{t1}
If $X$ is $\pi$-$n$-Euclidean, $n\ge 2$, then there is a homeomorphism $h:I^\infty \to C(X)$ such that $h((0,1)^\infty)=  CB(X)$.
\end{theorem}

 In proving the theorem we will rather concentrate on the complement $C(X)\setminus CB(X)$ and show that it is an $F_{\sigma}$-absorber  in  $C(X)$. For a basic terminology and facts on such absorbers the reader is referred to~\cite{M} and~\cite{Ca}.   Recall here that, given a class $\mathcal M$   of spaces which is topological (i.e., if $M\in \mathcal M $ then each homeomorphic image of $M$ is in $\mathcal M$) and closed hereditary (i.e., each closed subset of $M\in \mathcal M$ is in $\mathcal M$), a subset $A$ of a Hilbert cube $\mathcal Q$ is called an   $\mathcal M$-\emph{absorber}  in $\mathcal Q$ if
 \begin{enumerate}
\item $A\in \mathcal Q$,
\item $A$ is strongly $\mathcal M$-universal, i.e.  for each
$M\subset I^\omega$ from the class $\mathcal M$ and
each  compact set $K\subset I^\omega$, any embedding $f:I^\omega \to\mathcal Q$ such that
$f(K)$ is a $Z$-set in $\mathcal Q$
can be approximated arbitrarily closely
 by an embedding $g:I^\omega\to \mathcal Q$ such that
   $g(I^\omega)$ is a $Z$-set in $\mathcal Q$,
$g|K=f|K$ and $g^{-1}(A)\setminus K=M\setminus K$.
\item $A$ is contained in a $\sigma Z$-set in $\mathcal Q$.
\end{enumerate}
If $A\subset\mathcal Q$ is an $\mathcal M$-absorber in  $\mathcal Q$ and $B\subset I^\infty$ is an $\mathcal M$-absorber in $I^\infty$, then there is a homeomorphism $h:I^\infty\to \mathcal Q$ such that $h(B)=A$~\cite[Theorem 5.5.2]{M}.

We will be interested in two classes $\mathcal M$: the Borel class of $F_{\sigma}$-subsets of the Hilbert cube and the small Borel class $D_2(F_\sigma)$ of all subsets of the Hilbert cube that are differences of two $F_\sigma$-sets.
In the first case, property (3) above is redundant in presence of (1) and (2) (see~\cite[Theorem 5.3]{BGM}).
The pseudo-boundary $\partial(I^\infty)= \{(t_i)\in I^\infty: \text{$t_i\in\{0,1\}$ for some $i$}\}$ is a standard $F_{\sigma}$-absorber in $I^\infty$, while  $\partial(I^\infty)\times (0,1)^\infty$ is a $D_2(F_\sigma)$-absorber in  $I^\infty\times I^\infty$.

 \begin{proposition}\label{p4}
If $X$ contains an open subset homeomorphic to $\mathbb R^n$, $2\le n<\infty$, then $C(X)\setminus CB(X)$ is $F_{\sigma}$-universal, i.e., for each $F_{\sigma}$-subset $M$ of the Hilbert cube $I^\infty$,  there is an embedding $\varphi:I^\infty\to C(X)$ such that
\begin{equation}\label{varphi}
\varphi^{-1}\bigl(C(X)\setminus CB(X)\bigr)=M.
\end{equation}
\end{proposition}

\begin{proof}
For convenience,  assume that $X$ contains $(-3,3)^n$ as an open subset.
Let us construct an embedding
$\Theta:I^\infty\to C(X)$ such that
\begin{equation}
 \Theta\bigl((t_k)\bigr)\in C(X)\setminus CB(X)\quad\text{if and only if}\quad (t_k)\in \partial(I^\infty).
\end{equation}

Denote $D:=I^n\setminus (\frac13,\frac23)^n$ and, for $t\in [0,1]$,  $$D(t):=D\setminus \bigl(\frac13,\frac13(1+t)\bigr)\times \bigl[0,\frac13\bigr]\times \bigl(\frac13,\frac23\bigr)^{n-2}$$
(let us agree that $D(t)=  D\setminus \left(\frac13,\frac13(1+t)\right)\times \left[0,\frac13\right]$ if $n=2$ and $D(0)=D$),
$$D_k(t):= \left(\frac1{2^k}, 0,\dots, 0\right) + \frac1{2^k}D\bigl(t(1-t)\bigr),\quad k\in\mathbb N,$$ and consider segments

 $L_k(t)$ in  $(-3,3)^n$ from $$\left(\frac1{2^k}, \frac{-t}{2^k}, 0,\dots, 0\right) \quad\text{to}\quad \left(\frac1{2^k}, \frac{1+t}{2^k},0,\dots,0\right).$$

So, $D$ is the cube $I^n$ with the smaller open cube $D'=(\frac13,\frac23)^n$ subtracted; its boundary in  $(-3,3)^n$ has two components.   $D(t)$ is obtained from $D$ by removing a cube $D'(t)=(\frac13,\frac13(1+t))\times [0,\frac13]\times (\frac13,\frac23)^{n-2}$ adjacent to the both components; the size of  $D'(t)$ continuously depends on parameter $t$ so that the boundary of $D(t)$ is connected if and only if $t>0$.
The sets $D_k(t_k)$ are copies of $D\bigl(t_k(1-t_k)\bigr)$ scaled by
 factors $\frac1{2^k}$ and shifted by vectors $\left(\frac1{2^k}, 0,\dots, 0\right)$, correspondingly.

 The union of all $D_k(t_k)$'s compactified by the point $(0,\dots, 0)$  is a continuum whose  boundary is  connected if and only if each parameter $t_k$ is strictly between 0 and 1;  moreover, the continuum continuously depends on sequence $(t_k)$ but not in a one-to-one fashion. In order to get the one-to-one correspondence, we attach  segments $L_k(t_k)$ (Figure~\ref{fig}). Formally,
we define an embedding
\begin{equation}\label{e:theta}
\Theta\bigl((t_k)\bigr)=\{(0,\dots, 0)\} \cup \bigcup_{k=1}^\infty \bigl(L_k(t_k) \cup D_k(t_k)\bigr)\subset  (-3,3)^n.
\end{equation}

\begin{figure}[ht]
\setlength{\unitlength}{1mm}
\begin{picture}(135,40)
\thicklines

\put(80,0){\line(1,0){9}}
\put(95,0){\line(1,0){12}}
\put(80,27){\line(1,0){27}}
\put(107,0){\line(0,1){27}}
\put(80,-8){\line(0,1){43}}

\put(82,32){\makebox(0,0){\scriptsize$\}t_1$}}
\put(92,-3){\makebox(0,0){\scriptsize$\underbrace{}$}}
\put(92,-5){\makebox(0,0){\scriptsize$t_1(1-t_1)$}}

\put(89,0){\line(0,1){18}}
\put(95,0){\line(0,1){9}}
\put(89,18){\line(1,0){9}}
\put(95,9){\line(1,0){3}}
\put(98,9){\line(0,1){9}}

\put(62,18){\line(1,0){18}}
\put(62,12){\line(0,1){13}}
\put(62,0){\line(1,0){6}}
\put(71,0){\line(1,0){9}}
\put(68,0){\line(0,1){12}}
\put(68,12){\line(1,0){6}}
\put(74,6){\line(0,1){6}}
\put(71,0){\line(0,1){6}}
\put(71,6){\line(1,0){3}}
\put(62,-7){\line(0,1){7}}

\put(64,22){\makebox(0,0){\scriptsize$\}t_2$}}
\put(70,-3){\makebox(0,0){\scriptsize$\underbrace{}$}}
\put(70,-5){\makebox(0,0){\scriptsize$t_2(1-t_2)$}}

\put(50,0){\framebox(12,12)}
\put(54,4){\framebox(4,4)}

\put(42,-2){\line(0,1){12}}
\put(42,0){\line(1,0){3}}
\put(46,0){\line(1,0){4}}
\put(46,0){\line(0,1){3}}
\put(46,3){\line(1,0){1}}
\put(47,3){\line(0,1){2}}
\put(42,8){\line(1,0){8}}
\put(45,0){\line(0,1){5}}
\put(46,0){\line(1,0){4}}
\put(45,5){\line(1,0){2}}

\put(54,14){\makebox(0,0){\tiny$t_3=0$}}

\put(45,9){\makebox(0,0){\tiny$\}t_4$}}

\put(32,0){\makebox(0,0){$\cdots$}}

\put(20,0){\makebox(0,0){\textbf{$\cdot$}}}

\put(19,-2){\makebox(0,0){\scriptsize$(0,0)$}}

\end{picture}
\vspace{.5cm}
\caption{ $\Theta\bigl((t_1,t_2,0, t_4 \dots)\bigr)$ for $n=2$.}\label{fig}
\end{figure}

Since  the pseudo-boundary
 $\partial(I^\infty)$ is strongly $F_\sigma$-universal,  it is $F_\sigma$-universal, in particular. So,   there exists, for each $ F_\sigma$-set $M\subset I^\infty$,  an embedding $\chi: I^\infty \to I^\infty$ such that
$\chi^{-1}(\partial(I^\infty))=M$. Hence, the composition $\varphi=\Theta\chi:I^\infty \to C(X)$  satisfies~\eqref{varphi}.

\end{proof}

\newpage

 \begin{lemma}\label{l5}
If $X$ is $\pi$-$n$-Euclidean, $2\le n<\infty$, then $C(X)\setminus CB(X)$ is strongly $F_{\sigma}$-universal.
\end{lemma}

\begin{proof}
The proof is based on a  technique developed in~\cite{BC,Ca}. For our purpose, we  closely follow its rough description given in~\cite[Section 3.2]{KS}.
Without loss of generality, we can assume that
  $\varphi$ from Proposition~\ref{p4}  satisfies
  \begin{equation}\label{e:varphi}
\varphi(I^\infty)\subset C\bigl((0,1)^n\bigr)\subset C(I^n)\subset C(X).
\end{equation}

Given an $F_{\sigma}$-subset  $M$ of $I^\omega$,
  for each open non-empty subset $U$ of $X$ and an open copy of $(0,1)^n$ in $U$, let $\varphi_U:I^\omega \to C(U)$ be a composition of $\varphi$ with an embedding of $C(I^n)$ into the hyperspace of that copy. Then
  $$\varphi_U^{-1}\bigl(C(X)\setminus CB(X)\bigr)=M$$
   which means that

  \emph{$\varphi_U(q)\in C(X)\setminus CB(X)$ if and only if  $\varphi(q)\in C(X)\setminus CB(X)$ if and only if $q\in M$.}

   Notice that the equivalences remain valid if, given $q\notin K$, we add to  the one-dimensional part $A(q)$,  appearing in the construction of embedding $g(q)$, finitely many pairwise disjoint sets $\varphi_{U_i}(q)$, $i< m$ for some $m\in\mathbb N$,  such that  $A(q)\cap \varphi_{U_i}(q)$ is a singleton for each $i$.  It follows that $g^{-1}\bigl(C(X)\setminus CB(X)\bigr)\setminus K=M\setminus K$.
  Now, the construction of embedding $g$, as presented in~\cite[Section 3.2]{KS}, satisfies all the required conditions.

\end{proof}

\begin{proof}[Proof of Theorem~\ref{t1}]
By Proposition~\ref{p2} and Lemma~\ref{l1}, $C(X)\setminus CB(X)$ is an $F_\sigma$-absorber in $\mathcal Q=C(X)$. Hence, there exists a homeomorphism $h:I^\infty \to C(X)$ such that $h\bigl(\partial(I^\infty)\bigr) = C(X)\setminus CB(X)$.
\end{proof}

\subsection*{Separators with connected boundaries}

\

Now, let
$$S(X):=\{C\in 2^X: \text{$C$ separates $X$}\},\quad N(X):= \{C\in 2^X: \inte C=\emptyset\}.$$

By~\cite[Proposition 5.1]{KS} and Proposition~\ref{p2}, we get

\begin{proposition}\label{p5}
 $S(X)\cap CB(X)\in D_2(F_\sigma)$.
 \end{proposition}

 It is proved in~\cite[Theorem 5.8]{KS} that  $S(X)\cap N(X)\cap C(X)$ is a $D_2(F_\sigma)$-absorber in $C(X)$ if $X$ satisfies hypotheses of Theorem~\ref{t1}, $n\ge3$ and no subset of dimension $\le 1$ separates $X$.
In a similar way, we will show that $S(X)\cap CB(X)$ is also a $D_2(F_\sigma)$-absorber in $C(X)$ for such $X$.

\begin{proposition}\label{p6}
If $X$ contains an open subset homeomorphic to  $\mathbb R^n$, $2\le n<\infty$, then $S(X)\cap CB(X)$ is $D_2(F_\sigma)$-universal in $C(X)$, i.e., for each $D_2(F_\sigma)$-subset $M$ of  $I^\infty$,  there is an embedding $f:I^\infty\to C(X)$ such that
\begin{equation}\label{f}
f^{-1}\bigl(S(X)\cap CB(X)\bigr)=M.
\end{equation}
\end{proposition}

\begin{proof}
Without loss of generality, we can again  assume  that  $(-3,3)^n$ is an open subset of  $X$.
In~\cite[Proposition 5.2]{KS}, an embedding $\Psi:I^\infty\to C(X)$ is constructed such that
\begin{equation}\label{e:psi}
\Psi\bigl((q_k)\bigr)\in  S(X)\quad\text{if and only if}\quad (q_k)\in \partial(I^\infty).
\end{equation}

For each $(q_k)$, continuum $\Psi\bigl((q_k)\bigr)$ can be located in $[-1,0]\times [-1,1]^{n-1}$  as a subset consisting of the segment $L'=[-1,0]\times \{0,\dots,0\}$ and of a subset of the union of  cubes $[\frac{-1}{2k},\frac{-1}{2k+1}]\times [\frac{-1}{2k},\frac1{2k}]^{n-1}$, $k\in \mathbb N$.

We can now combine both embeddings $\Theta$~\eqref{e:theta} and $\Psi$~\eqref{e:psi} to define an embedding $\Phi: I^\omega\times I^\omega \to C(X)$ satisfying
\begin{multline}\label{Phireduction}
\Phi\bigl((q_k), (t_k)\bigr) \in   S(X)\cap CB(X) \quad\text{if and only if}\\
 \bigl((q_k), (t_k)\bigr) \in \partial(I^\infty)\times (0,1)^\infty,
  \end{multline}
by putting
\begin{equation}\label{Phi}
\Phi\bigl((q_k), (t_k)\bigr)=\Psi\bigl((q_k)\bigr)\cup \Theta\bigl((t_k)\bigr).
\end{equation}
Since $\partial(I^\infty)\times (0,1)^\infty$ is strongly $D_2(F_\sigma)$-universal in $I^\infty\times I^\infty$,  there exists, for each $D_2(F_\sigma)$-set $M\subset I^\infty$,  an embedding $\tau: I^\infty \to I^\infty\times I^\infty$ such that
$\tau^{-1}\bigl(\partial(I^\infty)\times (0,1)^\infty\bigr)=M$. Hence, the composition $f=\Phi\tau:I^\infty \to C(X)$  satisfies~\eqref{f}.
 Similarly as in~\eqref{e:varphi}, we can additionally assume that $(0,1)^n$ is an open subset of $X$ and
 \begin{equation}\label{eq:f}
 f(I^\infty)\subset C(I^n)\subset C(X).
 \end{equation}
\end{proof}

 \begin{theorem}\label{t2}
If  $X$ is $\pi$-$n$-Euclidean and no subset of dimension $\le 1$ separates $X$, then  there is a homeomorphism $h:I^\infty\times I^\infty \to C(X)$ such that $h\bigl(\partial I^\infty\times (0,1)^\infty\bigr) = S(X)\cap CB(X)$.
\end{theorem}

\begin{proof}
Since  subsets of dimension $\le 1$ do not separate $X$, $n=\dim X\ge 3$.

(1) By Proposition~\ref{p5}, family $S(X)\cap CB(X)$ is $D_2(F_\sigma)$.

(2) To prove the strong $D_2(F_\sigma)$-universality, we proceed similarly as in the proof of   Lemma~\ref{l5}. We just replace $\varphi$ with embedding $f$ from Proposition~\ref{p6} satisfying~\eqref{eq:f} and  family $C(X)\setminus CB(X)$ with $S(X)\cap CB(X)$.  Then, for an arbitrary fixed $D_2(F_\sigma)$-set $M\subset I^\infty$,  we have

 \emph{$\varphi_U(q)\in S(X)\cap CB(X)$ if and only if  $f(q)\in S(X)\cap CB(X)$ if and only if $q\in M$.}

For $q\notin K$, attaching finitely many pairwise disjoint sets $\varphi_U(q)$,  to the one-dimensional part $A(q)$ at single points does not change the above equivalences  (no one-dimensional set separates $X$) which means that the required embedding $g$ satisfies
$$g^{-1}\bigl(C(X)\setminus CB(X)\bigr)\setminus K=M\setminus K.$$

(3) $S(X)\cap CB(X)$ is contained in the $F_\sigma$-absorber $S(X)\cap C(X)$  in $C(X)$ (see~\cite[Theorem 5.3]{KS}), so in  a  $\sigma Z$-set in $C(X)$.

Thus we conclude that  $S(X)\cap CB(X)$ is a $D_2(F_\sigma)$-absorber in $C(X)$ and the required  homeomorphism $h$ exists.
\end{proof}

\section*{Acknowledgments}
The author is indebted to the organizers of the 11th Workshop on Continuum Theory and Hyperspaces at UNAM, Cuernavaca, M\'{e}xico, 2017, for invitation and its participants for inspiring discussions on the  class $CB(X)$.
 Rusell-Aaron Qui\~{n}ones-Estrella  initiated the study of continua with connected boundaries (at the 10th Workshop in 2016) and Norberto  Ordo\~{n}ez asked about the Borel complexity and characterization of  $CB(X)$; they are also coauthors of  recent paper~\cite{EQOV} devoted to the continua.

\bibliographystyle{amsplain}

config: arxiv


TITLE: Evaluate integral for $\int \sin^2 (x+\frac{\pi}{6}) dx$
QUESTION [1 upvotes]: Can someone walk me through how to evaluate the integral $$\int \sin^2 (x+\frac{\pi}{6}) dx?$$
I get as far as
$$\int \frac {1 - \cos(x + \frac{\pi}{6})}2dx,$$
but I am not sure how to proceed.

REPLY [0 votes]: $$\int sin^2(x+\frac{\pi}{6})dx$$
Apply u-substitution: $u=x+\frac{\pi}{6}$
$$=\int sin^2(u)du$$
Use the following identity: $sin^2(x)=\frac{1-cos(2x)}{2}$
$$=\int \frac{1-cos(2u)}{2}du$$
$$=\frac12 \int 1-cos(2u)du$$
$$=\frac12(\int 1du-\int cos(2u)du)$$
$$=\frac12(u-\frac12)sin(2u)$$
Substitute back $u=x+\frac{\pi}{6}$
$$\int sin^2(x+\frac{\pi}{6})dx=\frac12(x+\frac{\pi}{6}-\frac12sin(2(x+\frac{\pi}{6})))+C$$


TITLE: Use linear algebra to find an $n$th-degree polynomial with a zero at $x = \sqrt{3} + \sqrt{5} + \sqrt{3 + 4\sqrt{5}}$.
QUESTION [4 upvotes]: It was easy enough (though tedious) to remove the radicals in the given zero algebraically. I got (and double-checked with software): $$x^8 - 44x^6 - 160x^5 + 254x^4 + 1600x^3 + 4740x^2 + 8800x - 5975$$But, I'm wondering how I might solve it with matrix methods. I can see a system of equations. Letting $a=\sqrt{3}, b=\sqrt{5}$, and $c =\sqrt{3 + 4\sqrt{5}}$, we get $$x-a-b-c = 0$$ $$ a^2 - 3 = 0$$ $$b^2 -5 = 0$$ $$c^2-4b-3 = 0$$But I'm stuck here. How can I set this up (and solve it) as a linear algebra problem?

REPLY [2 votes]: Here's a way to solve it using matrix methods, but it's very tedious.
The field $K$ generated over the rationals by $\sqrt3$, $\sqrt5$, and $\sqrt{3+4\sqrt5}$ is an 8-dimensional vector space over the rationals, with basis
$$
\{1,\sqrt3,\sqrt5,\sqrt{15},\sqrt{3+4\sqrt5},\sqrt{9+12\sqrt5},\sqrt{15+20\sqrt5},\sqrt{45+60\sqrt5}\}
$$
Express each of the nine numbers $1,x,x^2,\dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+\cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,\dots,a_7$.


TITLE: Holomorphic vector fields on $\mathbb{P}^n$ that extend to the blow up
QUESTION [4 upvotes]: Hi, these are three questions regarding extendability of holomorphic vector fields on complex projective space to its blow up along a subvariety.
Let $\mathbb{P}^n$ be the complex projective space, with homogeneous coordinates of a point $p\in\mathbb{P}^n$ $$p=[z_0:\ldots:z_n]$$
1) suppose we blow up the point $p_0=[1:0:\ldots:0]$, which are the global holomorphic vector fields of $\mathbb{P}^n$ that extend to the blow up?
I made some computations in the affine chart $\lbrace z_0\neq0 \rbrace$ and modulo (fatal) errors i obtain that my vector fields are of the form $$l_0\frac{\partial}{\partial z_0}+\sum_{i=1}^n c_iz_i\frac{\partial}{\partial z_i}$$
with $l_0\in\mathbb{C}[z_0,\ldots,z_n]$ homogeneous of degree 1 and $c_i\in\mathbb{C}$.
So vector fields that extend to the blow up span a vector subspace of dimension $2n$ of $H^0(T\mathbb{P}^n)$. Am i right?
2)In a similar fashion i find that if i blow up a k-codimensional variety of type $$\lbrace z_k=\ldots=z_n=0 \rbrace$$ with $k>1$ my vector fields span a vector subspace of dimension
 $n(k+2)$ of $H^0(T\mathbb{P}^n)$. Again, am i right?
3) If i blow up a smooth curve of sufficiently high degree/genus do i still have holomorphic vector fields that extend to the blow up?
Thank you in advance.

REPLY [3 votes]: 1) No. There are many more vector fields. The vector fields you are looking for are precisely those which vanish at $p_0$. Since $h^0( \mathbb P^n, T \mathbb P^n) = (n+1)^2 -1$ and you are imposing $n$ linearly independent conditions you should get $n^2 + n - 1$ vector fields. In homegeneous coordinates they can be written as
$$
l_0 \frac{\partial}{\partial z_0} + \sum_{i=1}^n \sum_{j=1}^n c_{ij} z_i \frac{\partial}{\partial z_j} .
$$
Notice that the homogeneous vector field $\sum_{i=0}^n z_i \frac{\partial}{\partial z_i}$ does not contribute to the counting since it corresponds to the zero vector field on $\mathbb P^n$.
2) You should get $k^2 + (n+1-k) n -1$ vector fields. As pointed out in the comments the
result remains unchanged if we swap $k$ and $n-k+1$. The point is that $PGL(n+1)$ acts on $\mathbb P^n$  as well as on the dual projective space $\check{\mathbb P}^n$. If we look at the subgroup preserving a  linear subspace of codimension $k$ on $\mathbb P^n$, then the natural action on $\check{\mathbb P}^n$ will preserve its dual: a linear subspace of codimension $n-k+1$.
3) You will get a holomorphic vector field on the blow-up if and only if the curve is left invariant by the original vector field.
If your vector field vanishes on a linear subspace of dimension at least two then every curve contained in it will be invariant by the vector field. Thus on the blow-up along these curves you will still get holomorphic vector fields.
If instead you  restrict your attention to vector fields with zero set of dimension at most one then while you can find  curves of arbitrarily high degree invariant by this class of vector fields  (think on orbits of $\mathbb C^*$-actions on $\mathbb P^n$), they all have (geometric) genus zero.


TITLE: Electric Field from charged sphere within another charged sphere does not reinforce?
QUESTION [0 upvotes]: Lets say I have a positively charged conducting sphere of Q. The electric field outside the sphere is
$$E = \frac Q{4\epsilon\pi r^2}.$$
Now suppose this sphere is enclosed inside another hollow conducting sphere of -Q charge. By Gauss law, the electric field is still the same?
Won't the negatively charged outer sphere reinforce the electric field, causing it it be larger?
In another example,
The field from a infinite sheet of charge is
$$ E = \frac \sigma{2\epsilon}. $$
But the field between 2 oppositely charged infinite sheets is
$$ E = \frac \sigma{\epsilon}. $$
Why is this not the case above?

REPLY [3 votes]: It sounds like the math is not the problem. It is visualizing the geometry.
A point charge fills all space with an electric field. If you have multiple charges, they all create electric fields. To find the electric field at a point, add them up.

Do this with a sphere. If you are outside as shown above, all the E field contributions are more or less to the right. It works out just the same as a point charge E field. If you are inside, contributions from the left and right cancel perfectly. The total is 0. This property of the inverse square law has always struck me as one of the coolest things in physics.

With two spheres, the inner point has 0 field because contributions from the red sphere  cancel and so do contributions from the blue sphere.
For the middle point, the red sphere adds up to the same as a point charge, but the blue sphere contributions add to 0.
For the outer point, each sphere adds up to the equivalent of a point charge. In this case two opposite point charges at the same point would cancel.

For a plane, by symmetry the contributions add up to an E field perpendicular to the plane. The plane is infinite, and so the charge is infinite. But some is infinitely far away. The E field works out to be finite and uniform.
For two oppositely charged planes you get 0 on the outside and reinforcement in between. This is indeed similar to the two spheres, even though there is no charge off to the left.

Imagine the two spheres getting larger and larger, so the geometry looks more and more like two planes. The E field should approach the same solution as the two planes.
The left hand point is always inside two spheres. The right hand point is outside of two large spheres that always cancel. So the field is always 0 for them.
For the middle point, the red sphere matters. For the spheres to become like the planes, we must keep the charge density constant as they grow. This means the total charge grows larger. The total red E field is equivalent to a larger, but more distant point charge.
If the radius is r, the total charge, Q, is proportional to $r^2$, and the field is proportional to $Q/r^2$. So the field doesn't change. The E field becomes more uniform, and more like the plane.