Abstract Nonsense

Thoughts about compactness (Pt. V compactness misc.)

In this post we will discuss some points that may be put into one of the other categories, and it is just happenstance that I “forgot” to put it there. Topics that are indirectly related to compactness will also be included here. I also feel as though I should mention that a lot of things aren’t going to be discussed. It just isn’t feasible. Ascoli’s theorem for example, local compactness (at least not yet…maybe when we deal with the one-point compactification).

We begin with a few definitions.

Perfect set: We call a subset $E$ of a topological space $X$ perfect if $D(E)=E$ where $D(E)$ is the derived set or the set of all limit points of $E$.

Condensation point: If $X$ is a topological space $E$ is a subset of $X$ then $p$ is a condensation point of $E$ if for every neighborhood of $p$ there are uncountably many points of $E$.

Theorem: Let $X$ be a second countable topological space and latex $E$ an uncountable subset of $X$. a) Then the set of all condensation points of $E$, denoted $\mathfrak{C}$, is perfect and b) all but countably many points of $E$ lie in $\mathfrak{C}$

Proof: a) First suppose that $c\in\mathfrak{C}$ but $c\notin D\left(\mathfrak{C}\right)$. Then, there exists some neighborhood latex $N$ of $c$ such that $N\cap D\left(\mathfrak{C}\right)=\{c\}$. In particular, for each $x\in N$ there exists some $N_x$ such that $N_x\cap E$ is countable. We therefore see that $\left\{N_x\right\}_{x\in N}$ is an open cover of $N$ and since $X$ is second countable we see that Lindelof’s theorem we are guaranteed a countable subcover $\Omega$. Thus, $\displaystyle\bigcup_{\omega\in \Omega}\omega\implies N\cap E=\bigcup_{\omega\in\Omega}\left\{\omega\cap E\right\}$. But, by assumption we have that $\omega\cap E$ is countable and thus $N\cap E$ is the countable union of countable sets thus countable. This contradicts that $c$ is a condensation point of $E$, and so $c\in D\left(\mathfrak{C}\right)\implies \mathfrak{C}\subseteq D\left(\mathfrak{C}\right)$

Conversely, let $c\in\mathcal{D}\left(\mathcal{C}\right)$ then then for every neighborhood $N$ of $c$ we have that there exists some  $k\in N\cap\mathcal{C}\ne\varnothing$ such that $k\ne c$. But, since $N$ is open there exists some neighborhood $M$ of $k$ such that $M\subseteq N$, but by assumption every neighborhood of $k$ contains uncountably many points of $E$ and thus so does $M$, but since $M\subseteq N$ it follows that $N$ contains uncountably many points of $E$. Since $N$ was arbitrary we see that $c\in\mathfrak{C}\implies D\left(\mathfrak{C}\right)\subseteq\mathfrak{C}$. The conclusion follows.

b) Let $\mathfrak{B}$ the guaranteed countable base of $X$. Define $\mathcal{B}=\left\{B\in\mathfrak{B}:B\cap E\text{ is countable}\right\}$ and let $\displaystyle \mathfrak{U}=\bigcup_{B\in\mathcal{B}}B$. We claim that $\displaystyle \mathfrak{C}=\mathfrak{U}'=\bigcap_{B\in\mathcal{B}}B'\Leftrightarrow \mathfrak{C}'=\mathfrak{U}$.

Let $x\in \mathfrak{C}'$ then for there exists some $N$ of $X$ such that $N\cap E$ is countable, but since $\mathfrak{B}$ is an open base for $X$ there exists some $B\in\mathfrak{B}$ such that $x\in B\subseteq N$ and clearly then since $B\cap E\subseteq N\cap E$ we see that $x\in\mathfrak{U}\implies \mathfrak{C}'\subseteq\mathfrak{U}$

Conversely, let $x\in \mathfrak{U}$ then there exists some $B\in \mathfrak{B}$ such that $B\in x$ and $B\cap E$ is countable. Thus, $B$ is a neighborhood of $x$ which has only countably many points of $E$ and so $x\notin\mathfrak{C}\implies x\in\mathfrak{C}'\implies \mathfrak{U}\subseteq\mathfrak{C}'$

It follows that $\mathfrak{C}'=\mathfrak{U}$. Thus, $\displaystyle \mathfrak{C}'\cap E=\bigcup_{U\in\mathfrak{U}}\left\{U\cap E\right\}$ and since the right hand side is the countable union of countable sets the conclusion follows. $\blacksquare$

From this we can prove a nice little theorem.

Theorem: Let $X$ be a separable metric space, then every closed subset $C$ of $X$ may be written as $P\cup N$ where $P$ is perfect and $N$ is countable.

Proof: Since $C$ is closed we have that $D(C)\subseteq C$ but clearly $P\subseteq D(C)$ and so $P\subseteq C$. Thus, we have that $C=P\cup \left(C-P\right)$. But from the above we see that $P$ is perfect and $C-P$ countable. The conclusion follows. $\blacksquare$.

Corollary: Every compact subspace of a separable space mEay be written as the union of a perfect set and a closed set.

A theorem which has a cute little solution is the following.

Theorem: Let $X$ be a compact metric space and $E,G$ closed subsets of $X$. Then $d(E,G)=d(e,g)$ for some $e\in E$ and $g\in G$.

Proof: Notice that $\displaystyle d(E,G)=\inf_{e\in E}\inf_{g\in G}d(e,g)=\inf_{(e,g)\in E\times G}d(e,g)$ and since $E,G$ are closed subsets of a compact space we know that they are compact. And, by Tychonoff’s theorem we know that $E\times G$ is compact and so $d:X\times X\mapsto\mathbb{R}$ given by $(x,y)\mapsto d(x,y)$  is trivially continuous and so $d:E\times G\mapsto \mathbb{R}$ is continuous and by previous comments assumes its minimum on $E\times G$. The conclusion follows. $\blacksquare$

We can weaken the prerequisites of the sets involved slightly when dealing with $\mathbb{R}^n$

Theorem: Let $E,G$ be closed subsets of $\mathbb{R}^n$ such that $E$ is compact. Then $d(E,G)=d(e,g)$ for some $e\in E$ and some $g\in G$.

Proof: The idea here is that we could revert back to the previous theorem if $G$ was also compact. But, since every closed and bounded subset of $\mathbb{R}^n$ is compact and since $G$ is already closed we must merely find someone to bound it. But, this is pretty obvious in the sense that since $E$ is bounded that a lot of $G$ contributes nothing to the distance between the two sets. More formally, let $\delta=d(E,G)$. Since $E$ is bounded we know that $E\subseteq\left\{\bold{x}:\|\bold{x}\|\leqslant M\right\}$ for some $M\in\mathbb{R}$. So, consider the set $A=\left\{\bold{x}:\|\bold{x}\|\leqslant M+\delta+1\right\}$. Clearly this is a closed and bounded subset of $\mathbb{R}^n$. We pause for a lemma:

Lemma: Let $X$ be a topological space, and $\left\{C_j\right\}_{j\in\mathcal{J}}$ a class of closed subspaces of $X$ such that at least one of them, call it $C_\ell$, is compact. Then, $\displaystyle\bigcap_{j\in\mathcal{J}}C_j$ is a compact subspace of $X$.

Proof: Clearly we have that $\displaystyle \bigcap_{j\in\mathcal{J}}C_j\subseteq C_\ell$ and since they are all closed we see they are a closed subspace of a compact space. Thus, compact. Since $\displaystyle \bigcap_{j\in\mathcal{J}}C_j$ is the same as a subspace of $C_\ell$ as it is as  subspace of $X$ the conclusion follows. $\blacksquare$.

With this in mind we see that $G\cap A=K$ is a compact subspace of $\mathbb{R}^n$. Now, the conclusion follows from the last theorem if we can prove that $d(E,K)=d(E,G)$ but this is a relatively simple task. $\blacksquare$

February 27, 2010

Thoughts about compactness (Pt. IV (cont.) compactness and metric spaces)

Before we get to the main part of the post we prove one nice little theorem:

Theorem: A metric space $X$ is compact if and only if it’s complete and totally bounded.

Proof:

$\implies$: This is relatively simple. We know that since every compact metric space is sequentially compact and every sequentially compact metric space is totally bounded we already have the second part. But, for didactic purposes we can prove the second part without the use of sequential compactness (in fact it’s easier).

Lemma: Every compact metric space is totally bounded.

Proof: Let $\varepsilon>0$ be arbitrary and consider $\left\{B_{\varepsilon}(x)\right\}_{x\in X}$. This is clearly an open cover for $X$ and by assumption there exists a finite subcover $\left\{B_{\varepsilon}(x_1),\cdots,B_{\varepsilon}(x_n)\right\}$. Clearly then $\left\{x_1,\cdots,x_n\right\}$ will serve as a perfectly fine $\varepsilon$-net. The conclusion follows. $\blacksquare$

So it remains to show that $X$ is complete. But, we know that given any $\{x_n\}_{n\in\mathbb{N}}$ in $X$ which is Cauchy that it has some convergent subsequence (since every compact metric space is sequentially compact). So, consider the following lemma

Lemma: Let $\{c_n\}_{n\in\mathbb{N}}$ be a Cauchy sequence in a metric space $\left(C,d\right)$ then $\{c_n\}_{n\in\mathbb{N}}$ is convergent if and only if it has a convergent subsequence.

Proof: Clearly if $\{c_n\}_{n\in\mathbb{N}}$ is convergent it will serve as it’s own convergent subsequence.

Conversely, let $\{c_{\varphi(n)}\}_{n\in\mathbb{N}}$ be the guaranteed convergent subsequence of $\{c_n\}_{n\in\mathbb{N}}$ with limit $c$.  Since $c_{\varphi(n)}\to c$ there exists some $N_1\in\mathbb{N}$ such that $N_1\leqslant n\implies d(c_{\varphi(n)},c)<\frac{\varepsilon}{2}$ and since $\{c_n\}$ is Cauchy it is relatively easy to see there exists some $N_2\in\mathbb{N}$ such that $N_2\leqslant n\implies (d(c_{\varphi(n)},c_n)<\frac{\varepsilon}{2}$. Taking $N=\max\{N_1,N_2\}$ clearly gives that $n\leqslant N\implies d(c,c_n)\leqslant d(c_n,c_{\varphi(n)})+d(c,c_{\varphi(n)})<\varepsilon$. The conclusion follows. $\blacksquare$

With this it’s clear that since any Cauchy sequence has a convergent subsequence (by sequential compactness) that every Cauchy sequence is convergent.

$\Leftarrow$: Conversely, by previous argument we must only show that $X$ is sequentially compact. But, under the assumption of completeness it suffices to show that every sequence has a Cauchy subsequence. So let $\{x_n\}$ be sequence in $X$ it is clear that we could relabel this sequence as $\{x_{11},x_{12},x_{13},\cdots\}$. Since $X$ is totally bounded, there exists a finite class of open balls with radius $\frac{1}{2}$ which cover $X$. From this we see that $\{x_{11},x_{12},\cdots\}$ has a subsequence $\{x_{21},x_{22},\cdots\}$ all of which lie in some open ball of radius one-half. Applying the total boundedness of $X$ shows there exists a subsequence $\{x_{31},x_{32},\cdots\}$ of $\{x_{21},x_{22},\cdots\}$ which lies entirely in an open ball of radius $\frac{1}{3}$. Continuing in this way we see that we may define $\Delta=\left\{x_{11},x_{22},x_{33},\cdots\right\}\sqsubseteq\{x_n\}$  which is Cauchy. The conclusion follows $\blacksquare$

Now, I’ve looked through a bunch of books and found what I believe to be the most useful/interesting results of compactness.

The following is actually a GREATLY generalized result that has a compactness related corollary.

Quasi-totally bounded: Call a metric space $X$ quasi-totally bounded if for every $\varepsilon>0$ there exists some countable set $M_{\varepsilon}$ such that $\left\{B_{\varepsilon}(m)\right\}_{m\in M_{\varepsilon}}$ is an open cover for $X$.

Theorem: Every quasi-totally bounded metric space $X$ is separable.

Proof: Let $E_n$ be the guaranteed countable $M_{\varepsilon}$ discusssed above with $\displaystyle \varepsilon=\frac{1}{n}$ and let $\displaystyle E=\bigcup_{n=1}^{\infty}E_n$. Clearly we have $E$ is countable since it is the countable union of countable sets. We must merely prove that $E$ is dense in $X$. So, let $x\in X$ be arbitrary and $\varepsilon>0$ be given. We must find some $e\in E$ such that $e\in B_{\varepsilon}(x)$. But, by the Archimedean principle we have that there exists some $n\in\mathbb{N}$ such that $\frac{1}{n}<\varepsilon$. So considering $E_n$ we see that there exists some $e\in E_n$ such that $x\in B_{\frac{1}{n}}(e)$ and so $\displaystyle d(x,e)<\frac{1}{n}<\varepsilon$ and so $m\in B_{\varepsilon}(x)$. The conclusion follows. $\blacksquare$

Corollary: Every compact metric space being totally bounded and thus quasi-totally bounded is separable.

Theorem: Let $X$ be a metric space and Let $E$ and $G$ be disjoint subsets such that $E$ is compact and $G$ closed, then $d(E,G)=\delta>0$.

Proof: Define $\rho:E\mapsto\mathbb{R}$ by $\displaystyle \rho(e)=\inf_{g\in G}d(e,g)$. We see that $\displaystyle \rho(e)-\rho(e')=\inf_{g\in G}d(e,g)-\inf_{g\in G}d(e',g)\leqslant\inf_{g\in G}\left\{d(e,g)-d(e',g)\right\}\leqslant d(e,e')$ and similarly $\rho(e')-\rho(e)\leqslant d(e,e')$ and thus $\left|\rho(e)-\rho(e')\right|\leqslant d(e,e')$ and thus $\rho$ is Lipschitz and thus continuous. Since $E$ is compact we know then that $\displaystyle \inf_{e\in E}\rho(e)=\inf_{e\in E}\inf_{g\in G}d(e,g)=\inf_{g\in G}d(e',g)$ for some $e'\in E$. But, since $G$ and disjoint we have that $\displaystyle \inf_{g\in G}d(e',g)=\delta>0$. The conclusion follows. $\blacksquare$

Corollary: Let $E,G$ be disjoint compact subsets of $X$ then $d(E,G)=\delta>0$.

Corollary: Let $E$ be compact and latex $\{x\}\in X-E$ then $d(E,x)=\delta>0$

We can now prove the following

Theorem: Let $K$ be compact metric space and let $f:K\mapsto K$ be continuous. Suppose that $f$ is expanding in the sense that $d\left(f(x),f(y)\right)\geqslant d(x,y),\text{ }\forall x,y\in K$. Then a) $f$ is injective and $f^{-1}:f(K)\mapsto K$ is continuous and b) $f(K)=K$

Proof:

a) Injectivity follows immediately since if there existed $x,y\in K$ such that $x\ne y$ but $f(x)=f(y)$ then $0=d\left(f(x),f(y)\right)\geqslant d(x,y)>0$.

Also, we see that $f^{-1}$ is automatically uniformly continuous since it is Lipschitz with Lipschitz constant one since given any $f(x),f(y)\in f(K)$ we have that $d\left(f^{-1}(f(x))-f^{-1}(f(y))\right)=d\left(x-y\right)\leqslant d\left (f(x)-f(y)\right)$

b) Suppose there exists some $k\in K$ such that $k\notin f(K)$. Since $K$ is compact and $f$ continuous we know that $f(K)$ is compact and thus by our previous theorem we know that $d(f(K),k)=\delta>0$.

So, consider the sequence $\left\{f^n(k)\right\}_{n\in\mathbb{N}}$ where $f^n(k)=\underbrace{f(f(\cdots(f(k))\cdots)}_{n\text{ times}}$. Clearly this is a sequence in $K$ and by $K$‘s compactness we know that there exists some subsequence $\left\{f^{n_m}\right\}_{m\in\mathbb{N}}$ which converges, and is thus Cauchy. Consequently, there exists some $m\in\mathbb{N}$ such that $d\left(f^{n_{m+1}}(k),f^{n_m}(k)\right)<\frac{\delta}{2}$. But

\displaystyle \begin{aligned} d\left(f^{n_{m+1}}(k),f^{n_m}(k)\right) &\geqslant d\left(f^{n_m}(k),f^{n_{m-1}}(k)\right) \\& \vdots \\& \geqslant d(f(k),k)\geqslant \delta \end{aligned}

which is clearly a contradiction. $\blacksquare$

Corollary: Every isometry $f:X\mapsto X$ where $X$ is compact is a homeomorphism.

The next theorem is surprisingly powerful.

Theorem: Let $X$ be a metric space and $\{x_n\}_{n\in\mathbb{N}}$ a convergent sequence in $X$ with $x_n\to x$. Then, $E=\left\{x\right\}\cup\left\{x_n:n\in\mathbb{N}\right\}$ is compact.

Proof: Let $\Omega$ be an open cover for $E$, then there exists some $\omega\in \Omega$ such that $x\in\omega$ and since $\omega$ is open there exists some $B_{\delta}(x)\subseteq\omega$. But, since $x_n\to x$ we have that all but finitely many points of $\left\{x_n:n \in\mathbb{N}\right\}$ are in $B_{\delta}(x)$ and consequently $\omega$. Thus, taking $\omega$ and an open set containing each of the finitely many points not in $\omega$ procures the necessary finite subcover of $\Omega$. The conclusion follow. $\blacksquare$

Corollary: Let $X$ and $E$ be as above, then $E$ is closed.

Before we get to the next theorem we point something out that is true in any topological space.

Theorem: Let $X$ be a topological space and let $E$ and $F$ be a compact and a closed subspace of $X$ respectively. Then, $E\cap F$ is a compact subspace.

Proof: This follows since $E\cap F$ is closed subspace of $E$. $\blacksquare$

Theorem: Let $X$ and $Y$ be metric spaces and $f:X\mapsto Y$ be continuous. Also, let $\left\{K_n\right\}_{n\in\mathbb{N}}$ be a decreasing sequence of non-empty compact subsets of $X$. Then $\displaystyle f\left(\bigcap_{n\in\mathbb{N}}K_n\right)=\bigcap_{n\in\mathbb{N}}f(K_n)$.

Prove: It is well-known that $\displaystyle f\left(\bigcap_{n\in\mathbb{N}}K_n\right)\subseteq \bigcap_{n\in\mathbb{N}}f(K_n)$. So it remains to prove the reverse inclusion.

Firstly, it is easily proved that $\left\{f(K_n)\right\}_{n\in\mathbb{N}}$ is a decreasing sequence of compact sets with the F.I.P, and thus we know that $\displaystyle \bigcap_{n\in\mathbb{N}}f(K_n)\ne\varnothing$. Thus, let $\displaystyle k\in\bigcap_{n\in\mathbb{N}}f(K_n)$ since every metric space is Hausdorff (thus $T_1$) we know that $\{k\}$ is closed and since $f$ is continuous we know that $f^{-1}(\{k\})$ is closed as well. But, by the above theorem we see then that $f^{-1}(\{k\})\cap K_n$ is compact for every $n$. Also, $f^{-1}(\{k\})\cap K_{n+1}\subseteq f^{-1}(\{k\})\cap K_{n}$ and thus $\left\{f^{-1}(\{k\})\cap K_n\right\}_{n\in\mathbb{N}}$ is decreasing sequence of non-empty compact sets. Thus, we have that $\displaystyle \bigcap_{n\in\mathbb{N}}\left(f^{-1}(\{k\})\cap K_n\right)=f^{-1}(\{k\})\cap\bigcap_{n\in\mathbb{N}}K_n\ne\varnothing$ and thus $\displaystyle k\in f\left(\bigcap_{n\in\mathbb{N}}K_n\right)$. The conclusion follows. $\blacksquare$

Theorem: Let $X$ be a compact metric space and $E\subseteq X$. Prove that if for every $f:X\mapsto\mathbb{R}$ which is continuous that $f\mid E:E\mapsto\mathbb{R}$ attains a maximum then $E$ is compact.

Proof: Since $X$ is compact it suffices to show that $E$ is closed. If $E=X$ this is trivial, so assume otherwise and let $c\in X-E$. Define $\rho:X\mapsto\mathbb{R}$ by $\rho(x)=-d(x,c)$. It is trivial to prove that this is a continuous mapping and so $\rho\mid E$ attains a maximum on $E$, let’s say at $e$. Thus, $-d(x,e')\leqslant -d(x,e),\text{ }\forall e\in E$. But, $-d(x,e)<0$ since $d(x,e)>0$ and so $0 and the conclusion follows.$\blacksquare$

February 23, 2010

Thoughts about compactness (Pt. IV compactness in metric spaces)

Up until this point we’ve talked about compactness mainly in the context of general topological spaces, with only brief allusions to metric spaces. Now, while topologically spaces may be a more fruitful study than metric spaces it is admittedly easier to apply facts about metric spaces. That said, we devote a few sections specifically to the study of compactness in metric spaces. Since metric spaces are in general much nicer than general topological spaces it isn’t surprising that there are other, more applicable, forms of compactness. So, we need to start with some definitions:

Sequential Compactness: A metric space $\left(X,d\right)$ is said to be sequentially compact if every sequence has a convergent subsequence.

Bolzano-Weierstrass Property (B.W.P): A metric space $(X,d)$ is said to have the Bolzano-Weierstrass Property (B.W.P) if every infinite subset has a limit point.

We begin by showing that these are equivalent statements.

Theorem: A metric space $(X,d)$ is sequentially compact if and only if it has the B.W.P.

Proof:

$\implies$: Suppose that $X$ is sequentially and let $E$ is an infinite subset of $X$. It is clear that we can extract an infinite sequence $\{e_n\}_{n\in\mathbb{N}}$ of distinct points from $E$ and by assumption there exists some subsequence $\{e_k\}_{k\in\mathcal{K}}\sqsubseteq \{e_n\}_{n\in\mathbb{N}}$ which is convergent. The conclusion follows by considering the following lemma

Lemma: Let $x_n\to x$ where $\{x_n\}_{n\in\mathbb{N}}$ has an infinite number of points, then $x$ is a limit point of $\left\{x_n:n\in\mathbb{N}\right\}$.

Proof: Let $\varepsilon>0$ be arbitrary. Since $x_n\to x$ we know that $B_{\varepsilon}(x)$ contains all but finitely many points of $\{x_n\}_{n\in\mathbb{N}}$ and in particular it contains more than two distinct points of $\{x_n:n\in\mathbb{N}\}$. Since both of these can’t be equal to $x$ it follows that there exist some point of $\{x_n:n\in\mathbb{N}\}$ in $B_{\varepsilon}(x)$ distinct from $x$. The conclusion follows $\blacksquare$

$\Leftarrow$: Conversely, let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence in $X$. If $\{x_n\}_{n\in\mathbb{N}}$ has a value infinitely repeated then that constitutes a convergent subsequence. Otherwise, the set $\{x_n:n\in\mathbb{N}\}$ is infinite and by assumption has a limit point. It is easy to find a convergent subsequence from this. (consider taking $y_n$ to be any point in $B_{\frac{1}{n}}(x)$ where $x$ is the limit point afforded by the B.W.P).

The conclusion follows. $\blacksquare$

Our ultimate goal is to prove that a metric space being sequentially compact, having the B.W.P. , and being compact are all equivalent. We have laid the ground-work for this above.  The rest will actually uncover some equally fruitful concepts.

Theorem: A compact metric space $X$ has the B.W.P.

Proof: Suppose not. Then there exists some infinite subset $E$ which has no limit point in $X$. Thus, for every $x\in X$ there exists some $\varepsilon_x$ such that $B_{\varepsilon_x}(x)\cap E\subseteq \{x\}$. Clearly then $\left\{B_{\varepsilon_x}(x)\right\}_{x\in X}$ is an open cover for $X$ and by assumption there exists a finite subcover. But, since $E$ is contained in the set of centers of these open balls it follows that $E$ is finite. This is clearly a contradiction, and the conclusion follows. $\blacksquare$

To complete our goal we must now prove that sequential compactness implies compactness. This, fortunately, is not as simple as the previous theorem. It is fortunate because the sequence of theorems needed to ultimately prove this fact ends up being equally, if not more so, useful than the theorem itself. We begin by discussing the concept of a Lebesgue number.

Lebesgue number: Let $X$ be a metric space and let $\Omega$ be an open cover for $X$. Then, we call $\mathfrak{l}>0$ a Lebesgue number for $\Omega$ in $X$ if whenever $\text{diam }E\leqslant \mathfrak{l}$ there exists some $\omega\in\Omega$ such that $E\subseteq\omega$.

In other words, an open cover has a Lebesgue number if whenever the diameter of a set gets small enough it is contained in at least one of the sets in the open cover.

We now prove that every sequentially compact set has a Lebesgue number.

Theorem: Let $X$ be a sequentially compact metric space. Then $\Omega$ has a Lebesgue number for any open cover $\Omega$.

Proof: Call a subset of $X$ big if it is not contained in any $\omega\in \Omega$ and define $\mathfrak{l}=\inf\left\{\text{diam }B:B\text{ is a big set}\right\}$. If there are no big sets, then $\mathfrak{l}=\infty$ and thus any real number will serve as a Lebesgue number. So, assume that there are big sets. Clearly we must have that $0\leqslant\mathfrak{l}\leqslant\infty$. Once again, if $\mathfrak{l}=\infty$ any real number will do and if $\mathfrak{l}>0$ then $\displaystyle \frac{\mathfrak{l}}{2}$ will work. So, it is only problematic if $\mathfrak{l}=0$. So, for a contradiction assume that it does.

Clearly we must have a sequence $\left\{B_n\right\}_{n\in\mathbb{N}}$ such that $\text{diam }B_n<\frac{1}{n}$. So let $\{b_n\}_{n\in\mathbb{N}}$ be an arbitrary but fixed sequence such that $b_n\in B_n$. By assumption we have that $\{b_n\}_{n\in\mathbb{N}}$ has a convergent subsequence which converges to a point $b\in X$. Since $\Omega$ covers $X$ there exists some $\omega\in\Omega$ such that $b\in\omega$ and since $\omega$ is open we have that there exists some $\delta>0$ such that $B_{\delta}(x)\subseteq\omega$. Consider $B_{\frac{\delta}{2}}(b)$, since $b_n\to b$ we have that $B_{\frac{\delta}{2}}(b)$ contains all but finitely many points of $\left\{b_n\right\}_{n\in\mathbb{N}}$, in particular there exists some $N\in\mathbb{N}$ such that $\displaystyle \frac{1}{N}<\frac{\delta}{2}$. Since $\displaystyle \text{diam }B_N<\frac{\delta}{2}$ and $B_N\cap B_{\frac{\delta}{2}}(b)\ne\varnothing$ it can be easily seen that $B_N\subseteq B_{\delta}(b)$ which contradicts that $B_N$ is big. The conclusion follows. $\blacksquare$

To discuss the next topic we need to define an $\varepsilon$-net

$\varepsilon$-net: Let $X$ be a metric space. We see that a finite point set $E=\{e_1,\cdots,e_n\}$ is an $\varepsilon$-net for $X$ if $\left\{B_{\varepsilon}(e_k)\right\}_{k=1}^n$ is an open cover for $X$.

Totally bounded: A metric space $X$ is totally bounded if it has an $\varepsilon$-net for ever $\varepsilon>0$

Theorem: Every sequentially compact metric space $X$ is totally bounded.

Proof: Let $\varepsilon>0$ and $x_1\in X$ be arbitrary. If $B_{\varepsilon}(x)$ covers $X$ then $\{x_1\}$ is an $\varepsilon$-net and we’re done, so assume not. Let $x_2\in X-B_{\varepsilon}(x_1)$. If $\left\{B_\varepsilon(x_1),B_{\varepsilon}(x_2)\right\}$ cover $X$ we are done, so assume not. Continuing on this way we get a sequence $\{x_n\}$. If $\{x_n\}$ is infinite then it clearly has no convergent subsequence which contradicts the sequentially compactness of $X$. It follows that $\{x_n\}$ is finite and thus an $\varepsilon$-net for $X$. The conclusion follows. $\blacksquare$

Using this we can prove what we’re really after.

Theorem: If $X$ is sequentially compact then it is compact.

Proof: Let $\Omega$ be an arbitrary open cover of $X$. Since $X$ is sequentially compact it has a Lebesgue number $\mathfrak{l}$. Let $\varepsilon=\frac{\mathfrak{l}}{3}$. Once again, by virtue of $X$‘s sequential compactness, we may find an $\varepsilon$-net $\{e_1,\cdots,e_n\}$ of $X$. Since $\text{diam }B_{\varepsilon}(e_k)=\frac{2\mathfrak{l}}{3}<\mathfrak{l}$ we know that $B_{\varepsilon}(e_k)\subseteq \omega_k$ for some $\omega\in \Omega$. Doing this for $1\leqslant k\leqslant n$ produces a finite subclass of $\Omega$ such that $\omega_1\cup\cdots\cup\omega_n\supseteq B_\varepsilon(e_1)\cup\cdots\cup B_\varepsilon(e_n)=X$. The conclusion follows. $\blacksquare$

We can summarize this by saying that $X\text{ is compact }\Leftrightarrow\text{ }X\text{ is sequentially compact }\Leftrightarrow \text{ }X\text{ has the B.W.P}$

From this we can prove one of the most useful theorems in real analysis.

Theorem: Let $\left(X,d\right),\left(Y,d'\right)$ be metric spaces and $f:X\to Y$ be continuous, then $f$ is uniformly continuous.

Proof: Let $\varepsilon>0$ be arbitrary and consider $\left\{B_{\frac{\varepsilon}{2}}(f(x))\right\}_{x\in X}$, clearly then $\left\{f^{-1}\left(B_{\frac{\varepsilon}{2}}(f(x))\right)\right\}_{x\in X}$ is an open cover for $X$. And since $X$ is compact we know that this open cover has a Lebesgue number $\delta$.  In particular if $d(x,y)<\delta\implies \text{diam }\{x,y\}<\delta$ and thus $\{x,y\}\subseteq f^{-1}\left(B_{\frac{\varepsilon}{2}}(f(x))\right)\implies f(\{x,y\})=\{f(x),f(y)\}\subseteq B_{\frac{\varepsilon}{2}}(f(x))$ and thus $|f(x)-f(y)|<\text{diam }B_{\frac{\varepsilon}{2}}(f(x))=\varepsilon$. The conclusion follows $\blacksquare$

February 22, 2010

The dreaded product topology (Pt. III Tychonoff’s theorem and odds and ends)

In this post we will discuss one of the most important in point-set topology, Tychonoff’s theorem, and prove up some miscellaneous facts that are either interesting or useful.

Theorem (Tychonoff): Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty class of non-empty compact topological spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is compact under the product topology.

Proof: It follows from an earlier theorem that we must merely show that every class of closed sets in the defining closed subbase with the F.I.P have non-empty intersection. So, let $\mathcal{S}=\left\{S_k\right\}_{k\in\mathcal{K}}$ be such a class. Consider some arbitrary $j\in\mathcal{j}$, since $S_k$ is subbasic closed for ever $S_k\in\mathcal{S}$ we know that $\pi_j(S_k)$ is closed, and so $\left\{\pi_j(S_k)\right\}_{j\in\mathcal{J}}$ is a class of closed subsets of $X_j$, and noticing that $\pi_{j}(S_{k_1})\cap\cdots\cap\pi_j(S_{k_n})\supseteq\pi_j\left(S_{k_1}\cap\cdots\cap S_{k_n}\right)\supseteq\varnothing$ and so by assumption we have that $\displaystyle \bigcap_{k\in\mathcal{K}}\pi_j(S_k)\ne \varnothing$. In other words, we have there exists some $\displaystyle x_j\in\bigcap_{k\in\mathcal{K}}\pi_j(S_k)$ and so every element of $\mathcal{S}$ contains an element which has $x$ as it’s $j$th coordinate. Doing this for each $j\in\mathcal{J}$ produces a set $\left\{x_j\right\}_{j\in\mathcal{J}}$ such that $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\bigcap_{k\in\mathcal{K}}S_k$. In light of our initial comments the conclusion follows. $\blacksquare$

From this we can prove a nice little result that is used incessantly in multivariate analysis. But, before we continue we need a little terminology

n-cell: An $n$-cell is a multi-dimensional generalization of   a closed interval in $\mathbb{R}^n$ . In other words, we call $E$ an $n$-cell if $\pi_k(E)=[a,b]$ for every $1\leqslant k\leqslant n$. In other words, $E$ is an $n$-cell if there exists some $[a,b]\subseteq\mathbb{R}$ such that $E=\underbrace{[a,b]\times\cdots\times[a,b]}_{n\text{ times}}=[a,b]^n$.

Theorem (Generalized Heine-Borel): Every closed and bounded subspace of $\mathbb{R}^n$ is compact.

Proof: Let $E$ be such a set. Since $E$ is bounded we know that $E$ is contained in some $n$-cell $[a,b]^n$. So, if we prove that every $n$-cell is compact we are done (since a closed subspace of a compact space is compact).

But, we know that $[a,b]^n=[a,b]\times\cdots\times[a,b]$ and since $[a,b]\subseteq\mathbb{R}$ is compact we have by Tychonoff’s theorem that the product of compact spaces is compact. The conclusion follows. $\blacksquare$

Be careful about taking this and attempting to try to make a generalization of the extreme value theorem, namely “$If $f:X\mapsto\mathbb{R}^n$ is continuous and $X$ compact then $\sup\text{ }f\left(X\right),\inf\text{ }f\left(X\right)\in f\left(X\right)$“. It fails for an easily over-looked and decidedly non-topological property of $\mathbb{R}^n,text{ }n\geqslant 2$. Particularly, we have that $\mathbb{R}$ has a canonical ordering (the one we know and love), but what is the ordering on $\mathbb{R}^n$? No ordering no concept of infimums and supremums. We now take a look at some of the odds and ends that were previously mentioned. It should probably be mentioned for those that are reading this and already know topology that, for now, we omit mention of the majority (save Hausdorfness AKA $T_2$) of the separability axioms and things like local compacntess, connectedness, paracompactness, and path-connectedness. We shall mention them in due time, from where we will discuss which are invariant under product. Theorem: Let $X_1,\cdots,X_n$ be a finite number of discrete spaces, then the product set $X=X_1\times\cdots\times X_n$ is the same topological space under both the product and discrete topology. Proof: Let $\mathcal{J}$ be the product topology on $X$ and, of course, let $\wp\left(X\right)$ be the discrete topology (I use the $\wp$ for power set). It is clear that $\mathcal{J}\subseteq\wp\left(X\right)$ and so we must merely show the reverse inclusion to finish. So, let $E\in\wp\left(X\right)$ then $E=U_1\times\cdots\times U_n$ for some subsets $U_k\subseteq X_k$, but since $X_k$ is discrete we know that each $U_k$ is open. Thus, $E$ is the finite product of open sets and thus open in the product (box) topology. The conclusion follows. $\blacksquare$ The fact that I noted that product topology and box topology were the same here was meaningful. We know that the two coincide for the product of finitely many spaces, but the next remark shows (once again) how they differ for infinite. Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of discrete spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ under the box topology coincides with $X$ under the discrete topology. The same cannot be said about the product topology. Proof: Follows the exact same reasoning as in the last theorem. To see why the infinite product of discrete spaces need not be discrete under the product topology, one need only consider $\displaystyle X=\prod_{j=1}^{\infty}X_j$ where $X_j=[0,1]$ under the discrete topology. Clearly then $(0,1)\times\cdots\times(0,1)\times\cdots$ is open in $X$ under the discrete topology but NOT the product topology. $\blacksquare$ The last thing we will talk about is the product of maps. Product of maps: Let $f:X\mapsto Y$ and $f':X'\mapsto Y'$ be two maps, then define the product of the maps to be $f\times f':X\times X'\mapsto Y\times Y'$ by $(x,x')\mapsto (f(x),f'(x'))$. We extend this definition to an arbitrary number of topological spaces as follows: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ and $\left\{Y_j\right\}_{j\in\mathcal{J}}$ be two sets of topological spaces and let $\left\{f_j\right\}_{j\in\mathcal{J}}$ be a class of mappings such that $f_j:X_j\mapsto Y_j$ (this can obviously be formulated in the guise of ordered triples as well), we then define $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{j}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ by $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\mapsto\prod_{j\in\mathcal{j}}\{f_j(x_j)\}$ The reader may recall that this concept was used to prove that if $f:X\mapsto Y$ is continuous and $Y$ Hausdorff then $\Gamma_f$ is a closed subset of $X\times Y$. We begin by proving the product of continuous maps is continuous. This is the first of a sequence of three proofs which will lead to a very satisfying corollary. Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ and $\left\{Y_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of topological spaces and let $\left\{f_j\right\}_{j\in\mathcal{J}}$ be a corresponding set of continuous mappings such that $f_j:X_j\mapsto Y_j$. Then, $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is continuous, where $\displaystyle \prod_{j\in\mathcal{J}}X_j,\prod_{j\in\mathcal{J}}Y_j$ are under the product topology. Proof: Lemma: $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j^{-1}\left(O_j\right)$ Proof: Let $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\left( \prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)$ then, $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}\in\prod_{j\in\mathcal{J}}O_j$ which clearly means that $f_j(x_j)\in O_j,\text{ }\forall j\in\mathcal{J}$ and so $x_j \in f^{-1}(O_j)$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$ Conversely, let $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$. Then, $x_j\in f^{-1}\left(O_j\right)\implies f(x_j)\in O_j,\text{ }\forall j\in\mathcal{J}$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)\in \prod_{j\in\mathcal{J}}O_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)$. The conclusion follows. $\blacksquare$ Now, as was previously proven we need only prove that the inverse image of each open basic set is open. So, let $B\in\mathfrak{B}$ (the defining open base for $\displaystyle \prod_{j\in\mathcal{J}}Y_j$). Then, $\displaystyle B=\prod_{j\in\mathcal{J}}O_j$ where $O_j=Y_j$for all but finitely many $j$, and thus $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(B\right)=\left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$. But, we see that since $O_j$ is an open set which equals $Y_j$ for all but finitely many $j$‘s that $f^{-1}(O_j)$ is going to be an open set (since each $f_j:X_j\mapsto Y_j$ is continuous) which equals $X_j$ for all but finitely many $j$‘s. The conclusion follows. $\blacksquare$ Theorem: Let everything be as in the previous theorem, except instead of each $f_j:X_j\mapsto Y_j$ being continuous, let it be bijective. Then $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is bijective. Proof: Injectivity: Suppose that $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{y_j\}\right)$. Then, by definition $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{f_j(y_j)\}$ and since two functions (remember we think of arbitrary tuples as functions) are equal if and only if they take the same value at every point we see that $f_j(x_j)=f_j(y_j),\text{ }\forall j\in\mathcal{J}$ and since each $f_j$ is bijective (and thus injective) it follows that $x_j=y_j,\text{ }\forall j\in\mathcal{J}$, from where injectivity follows. Surjectivity: Let $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}Y_j$ be arbitrary. For each $j\in\mathcal{J}$ we have that there exists some $x_j\in X_j$ such that $f_j(x_j)=y_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}$ from where surjectivity immediately follows. $\blacksquare$ Theorem: Let everything be as before, except now instead of continuous or bijective assume that each $f_j:X_j\mapsto Y_j$ is surjective and open. Then, $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is open. Proof: Lemma: $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j(O_j)$. Proof: Let $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)$ then $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}x_j\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}$ for some $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}(O_J)$. Clearly then we have that $x_j\in O_j$ and so $f_j(x_j)\in O_j$ and so $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j(O_j)$. Conversely, let $\displaystyle\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}\{f_j(O_j)\}$ then $y_j=f_j(x_j),\text{ }\forall j\in\mathcal{J}$ for some $x_j\in O_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)$ The conclusion follows. $\blacksquare$. Now, let $E$ be open in $\displaystyle \prod_{j\in\mathcal{J}}X_j$ then $\displaystyle E=\prod_{j\in\mathcal{J}}O_j$ where $O_j=X_j$ for all but finitely many $j$ it follows then that $f_j(O_j)$ is an open set which equals $Y_j$ for all but finitely many $j$ (since each $f_j:X_j\mapsto Y_j$ is surjective and open). Thus $\displaystyle \prod_{j\in\mathcal{J}}f_j(E)=\prod_{j\in\mathcal{J}}f_J\left(\prod_{j\in\mathcal{J}}\right)=\prod_{j\in\mathcal{J}}f_j(O_j)$ which by prior comment is the product of open sets in $\displaystyle \prod_{j\in\mathcal{J}}Y_j$ which equals the full space for all but finitely many $j$. It follows that $f(E)$ is open in $\displaystyle \prod_{j\in\mathcal{J}}Y_j$ under the product topology. $\blacksquare$ Theorem: Combining these three we see that if $X_j\approx Y_j$ for all $j\in\mathcal{J}$, then there exists some homeomorphism (a bijective continuous open map) $f_j:X_j\mapsto Y_j$ and so $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is a bijective continuous open map and thus a homeomorphism. It follows that $\displaystyle \prod_{j\in\mathcal{J}}X_j\approx\prod_{j\in\mathcal{J}}Y_j$ where each are under the product topology. February 21, 2010 Thoughts about compactness (Pt. III, compactness and mappings) Up until now we’ve talked quite extensively about necessary and sufficient conditions for a topological space to be compact, but we’ve neglected to speak about how compactness interacts with mappings of topological spaces into another topological spaces. Probably the most widely used theorem is the following: Theorem 1: The continuous image of a compact topological space is compact, or phrased differently if $f:X\mapsto Y$ be continuous and $X$ compact, then $f(X)$ is a compact subspace of $Y$. Proof: Let $\left\{O_j\right\}_{j\in\mathcal{J}}$ be an open cover of $f(X)$ then $O_j=E_j\cap f(X)$ for some open set $E_j$ in $Y$. Clearly we have that $\left\{E_j\right\}$ covers $f(X)$ and so $\displaystyle X=f^{-1}\left(\bigcup_{j\in\mathcal{J}}E_j\right)=\bigcup_{j\in\mathcal{J}}f^{-1}\left(E_j\right)$ and since $f$ is continuous and $E_j$ open in $Y$ we see that $\left\{f^{-1}\left(E_j\right)\right\}_{j\in\mathcal{J}}$ is an open cover of $X$ and thus by assumption we have that there exists some $\left\{f^{-1}\left(E_{j_1}\right),\cdots,f^{-1}\left(E_{j_m}\right)\right\}\subseteq\left\{f^{-1}\left(E_j\right)\right\}_{j\in\mathcal{J}}$ which is also an open cover. Therefore,$\displaystyle X=f^{-1}\left(E_{j_1}\right)\cup\cdots f^{-1}\left(E_{j_m}\right)$ $\displaystyle =f^{-1}\left(\bigcup_{k=1}^{m}E_{j_k}\right)\implies f(X)=ff^{-1}\left(\bigcup_{k=1}^{m}E_{j_k}\right)\subseteq\bigcup_{k=1}^{m}E_{j_k}$. Therefore, $\left\{E_{j_1}\cap f(X),\cdots,E_{j_m}\cap f(X)\right\}=\left\{O_{j_1},\cdots,O_{j_m}\right\}\subseteq\left\{O_{j}\right\}_{j\in\mathcal{J}}$ is an open cover for $X$. The conclusion follows. $\blacksquare$ Corollary : If $f:X\mapsto Y$ is a surjective continuous mapping and $X$ is compact, then $Y$ is compact. Corollary : If $X\approx Y$ (that’s what I use for homeomorphic) then $X$ is compact if and only if $Y$ is compact. With this theorem we can prove one of the most highly used, and least thought about, theorems in elementary calculus. Theorem 2: Let $f:X\mapsto\mathbb{R}$ be continuous, where $X$ is a compact topological space. Then $\sup\left\{f(x):x\in X\right\},\inf\left\{f(x):x\in X\right\}\in f(X)$. Proof: Lemma: A compact subspace of a metric space $X$ is bounded. Proof: Let $E\subseteq X$ be compact. Consider the class $\Omega=\left\{B_{1}(e)\right\}_{e\in E}$. Clearly $\Omega$ is an open cover for $X$ and so there exists some finite subcover $\left\{B_{1}(e_1),\cdots,B_{1}(e_n)\right\}$. Clearly then, $\displaystyle \text{diam }E\leqslant\text{diam }\bigcup_{j=1}^{n}B_1(e_j)\leqslant \sum_{j=1}^{n}2=2n$. The conclusion follows. $\blacksquare$. Now, we know that $f(X)$ is compact, and combining the above with the earlier proof that ever compact subspace of a Hausdorff space is closed we see that $f(X)$ is closed and bounded. Thus, we have that $\sup\left\{f(x):x\in X\right\},\inf\left\{f(x):x\in X\right\}$ exist and it’s a relatively easy exercise in the basic topology of the real line that if $C\subseteq\mathbb{R}$ is closed then $\inf\text{ }C,\sup\text{ }C\in C$. It follows that $\sup\left\{f(x):x\in X\right\},\inf\left\{f(x):x\in X\right\}\in f(X)$. $\blacksquare$ Corollary 1: A continuous function $f:[a,b]\mapsto\mathbb{R}$ assumes a maximum and minimum value on $[a,b]$ Corollary 2: If $f:X\mapsto\mathbb{R}^+$ is continuous and $X$ compact, then $\inf\text{ }f(X)\geqslant 0$ Also, using corollary two you can prove the Heine-Borel theorem in more elementary, albeit more complex, terms. Theorem (Heine-Borel): If $E\subseteq\mathbb{R}$ is closed and bounded, then $E$ is compact. Proof: Just as last time we note that since $E$ is bounded that it is contained in some interval $[a,b],\text{ } a. And so, if we can prove that $[a,b]$ is compact then $E$‘s compactness follows since it is a closed subspace of a compact space. So now we need to prove that $[a,b]$ is compact, but we even have to prove something simpler thanks to the above theorem and the following lemma. Lemma: Let $[a,b],[c,d]\subseteq\mathbb{R}$ and $a then $[a,b]\approx[c,d]$ Proof: If we can find a linear function $\ell:[a,b]\mapsto[c,d]$ we are done. So assume that one exists and is of the form $\ell(x)=mx+n$. Then we need $\ell(a)=ma+n=c,\ell(b)=mb+n=d$. Solving this gives $\displaystyle \ell(x)=\frac{d-c}{b-a}x+\frac{bc-ad}{b-a}$ which clearly satisfies the conditions of a homeomorphism. $\blacksquare$ With this lemma in mind it remains to prove that any closed interval is compact. We choose, of course, for convenience $I=[0,1]$. So let $\Omega$ be an open cover of $[0,1]$, and assume that it admits no finite subcover. Clearly then either no finite subclass of $\Omega$ can cover either $\displaystyle \left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right]$. Let $[a_1,b_1]$ be one of these subintervals. Clearly $[a_1,b_1]$ has length $\displaystyle \frac{1}{2}$. Now, using the same logic we see that no finite subclass of $\Omega$ can cover either $\displaystyle \left[a_1,\frac{a_1+b_1}{2}\right],\left[\frac{a_1+b_1}{2},b_1\right]$. Continuing in this manner we have a sequence of closed intervals $\left\{[a_n,b_n]\right\}_{n\in\mathbb{N}}$. It is clear that $a_n\leqslant a_{n+1}\leqslant b_n\leqslant b_{b+1}$. Clearly then each $b_n$ is an upper bound for the set $\{a_n\}_{n\in\mathbb{N}}$ and so let $\displaystyle \alpha=\sup_{n\in\mathbb{N}}a_n$ and $\displaystyle \beta=\sup_{n\in\mathbb{N}}b_n$. Clearly by prior comment $\alpha\leqslant\beta$ but by definition $a_n\leqslant \alpha\leqslant\beta\leqslant b_n \text{ }(1)$ and by the contrivance of our sequence $(1)$ implies that $\beta-\alpha\leqslant\frac{1}{2^n}$ for all $n$ and so $\alpha=\beta$. Since $\Omega$ covers $I$ and $\alpha=\beta\in I$ where exists some $\omega\in \Omega$ such that $a\in\omega$. Now, since $\omega$ is open there exists some $\varepsilon>0$ such that $B_{\varepsilon}(a)\subseteq \omega$. The Archimedean principle now furnishes us with some $N\in\mathbb{N}$ such that $\displaystyle \frac{1}{2^N}<\varepsilon$. Thus, $b_N-a_N<\varepsilon$. Now, $\alpha=\beta\in\left[a_N,b_N\right]$. Therefore $\alpha-a_N<\frac{1}{2^N}<\varepsilon$ and $b_N-\beta<\frac{1}{2^N}$. Consequently, $[a_N,b_N]\subseteq B_{\varepsilon}(q)\subseteq\omega$. Therefore, we have that $[a_N,b_N]$ may be covered by a finite subclass of $\Omega$. Boom. $\blacksquare$ Another theorem which proves useful has to do with mappings from compact spaces into Hausdorff spaces, in particular: Theorem: Let $X$ be compact and $Y$ Hausdorff, then if $f:X\mapsto Y$ is continuous and bijective, then it is a homeomorphism. Proof: We must merely show that the mapping is open to conclude that it’s a homeomorphism, but since $f$ is bijective we know that given any $E\subseteq X$ that $f\left(E'\right)=\left[f\left(E\right)\right]'$ and so $f$ is open if and only if it’s closed. Thus, we must merely show that $f$ is closed. To do this we merely note that since $X$ is compact, that any closed $E\subseteq X$ must necessarily be compact. And by our first theorem then $f(E)$ is a compact subspace of $Y$. But, by previous problem compact subspaces of Hausdorff spaces are closed and so $f(E)$ is closed. The conclusion follows. $\blacksquare$ This theorem has some awfully amazing applications when trying to prove that two spaces are homeomorphic. It saves a lot of hassle. Our next theorem has to do with the graph of a continuous mapping from a compact space into a Hausdorff space, and although they’re are easier methods to prove this, I believe that a little elbow grease is worth the reward. That said, we need to prove some theorems not related to compactness. Theorem: Let $f:X\mapsto Y$ be continuous and define $f\times\iota_Y=X\times Y\mapsto Y\times Y$ by $(x,y)\mapsto (f(x),y)$. Then, this function is continuous. Proof: Lemma: $\left(f\times\iota_Y\right)^{-1}\left(U\times V\right)=f^{-1}(U)\times V$ Proof: Let $x\in \left(f\times \iota_Y\right)^{-1}(U\times V)$, then $\left(f\times \iota_Y\right)(x,y)=(f(x),y)\in U\times V$ and so $f(x)\in U$ and $y\in V$ and so $x\in f^{-1}(U)$ and $y\in V$ and so $(x,y)\in f^{-1}(U)\times V$. Conversely, let $(x,y)\in f^{-1}(U)\times V$, then $f(x)\in U$ and $y\in V$. Therefore, $(f(x),y)=\left(f\times \iota_Y\right)(x,y)\in U\times V$ and so $(x,y)\in\left(f\times\iota_Y\right)^{-1}\left(U\times V\right)$. The conclusion follows.$\blacksquare$ Now, since we must only prove that the inverse image of any basic open set in $Y$ is open in $X\times Y$, but this follows immediately since any basic open set is of the form $U\times V$ where both $U$ and $V$ are open in $Y$. And since $f:X\mapsto Y$ is continuous we see that $f^{-1}(U)$ is open and so $\left(f\times \iota_Y\right)^{-1}\left(U\times V\right)=f^{-1}(U)\times$ is the product of two open sets in $X$ and $Y$ and thus basic open in $X\times Y$ and so clearly open. The conclusion follows. $\blacksquare$ Next, we prove a fact that is used incessantly as a counter example when dealing with Hausdorff spaces. Theorem: Let $\Delta=\left\{(y,y):y\in Y\right\}$ (this is called the diagonal in $Y\times Y$). Then $\Delta$ is closed in $Y\times Y$. Proof: We prove that $\Delta$ is closed by proving that $\Delta'$ is open. So, let $(x,y)\in\Delta'$, then $x\ne y$ and since both $x$ and $y$ are in the Hausdorff space $Y$ there exists open sets $O_x$ and $O_y$ such that $x\in O_x$ and $y\in O_y$ and $O_x\cap O_y=\varnothing$. Clearly then$latext (x,y)\in O_x\times O_y\$ which is open in $Y\times Y$ and since $O_x\cap O_y=\varnothing$ we see that $O_x\times O_y\cap\Delta=\varnothing$. The conclusion follows. $\blacksquare$

We prove our second to last theorem before we get to the relevant part.

Theorem: Let $f\times\iota_Y$, $\Delta$,  $Y$, and $X$ be as above, then $\left(f\times\iota_Y\right)^{-1}\left(\Delta\right)=\Gamma_f$ where ($\Gamma_f=\left\{(x,y)\in X\times Y:y=f(x)\right\}$).

Proof: Let $(x,y)\in \left(f\times \iota_Y\right)^{-1}\left(\Delta\right)$, then $\left(f\times \iota_Y\right)(x,y)=(f(x),y)\in \Delta$ but this clearly implies that $f(x)=y$. Therefore, $(x,y)\in\Gamma_f$

Conversely, let $(x,f(x))\in\Gamma_f$, then $\left(f\times\iota_Y\right)(x,f(x))=(f(x),f(x))\in\Delta$. The conclusion follows. $\blacksquare$

From this we quickly derive the following fact:

Theorem: Let $f:X\mapsto Y$ be continuous and $Y$ Hausdorff, then $\Gamma_f$ is a closed in $X\times Y$.

Proof: Since $f:X\mapsto Y$ is continuous we know that $f\times\iota_Y:X\times Y\mapsto Y\times Y$ is continuous, and since $\Delta$ is closed in $Y\times Y$ it follows that $\left(f\times\iota_Y\right)^{-}\left(\Delta\right)=\Gamma_f$ is closed. The conclusion  follows. $\blacksquare$

Theorem: If $f:X\mapsto f(X)$ is continuous and $X$ is compact, then $\Gamma_f$ is a compact suspace of $X\times f(X)$.

Proof: It follows from out last theorem that $\Gamma_f$ is a closed subspace of $X\times f(X)$, but since $X$ is compact we know that $f(X)$ is compact and since (see next post) the product of compact spaces is compact, we know that $X\times f(X)$ is compact. Thus, $\Gamma_f$ is a closed subspace of a compact space, and thus compact. The conclusion follows. $\blacksquare$

February 20, 2010

In the previous post we looked at the real number motivation for the concept of compactness and generalized it to topological spaces. We also gave some useful alternative formulations of compactness. We continue briefly here:

Basic open cover: If $X$ is a topological space and $\left\{B_{k}\right\}_{k\in\mathcal{K}}$ an open base for $X$ then $\Omega$ is a basic open cover for $X$ if it is an open cover for $X$ and composed entirely of basic sets.

Theorem: A set is compact if and only if every basic open cover has a finite subcover.

Proof:

$\implies$: Let $\left\{O_j\right\}_{j\in\mathcal{J}}$ be an open cover for $X$, clearly we have that each $\displaystyle O_j=\bigcup_{k\in\mathcal{K}}B_k$ where $\left\{B_k\right\}_{k\in\mathcal{K_j}}$ is some class of open basic sets. Clearly then $M=\displaystyle \bigcup_{j\in\mathcal{J}}\left\{B_k\right\}_{k\in\mathcal{K}_j}$ is a basic open cover for $X$ and by assumption there exists some set $\left\{B_1,\cdots,B_m\right\}\subseteq M$ which covers $X$ clearly taking one of the $O_k$‘s that contain $B_k$ for $1\leqslant k\leqslant m$ finishes the argument.

$\Leftarrow$: This is obvious since a basic open cover is an open cover.

$\blacksquare$

Subbasic open cover: Similarly to before, a subbasic open cover is an open cover consisting entirely of sets from a given open subbase.

Theorem: A topological space $X$ is compact if and only if every subbasic open cover has a finite subcover.

Proof: It is clear from the last post that it is sufficient to show that $X$ is compact if and only if every class of closed subbasic sets with the F.I.P. have non-empty intersection.

So, let $\mathcal{S}$ be the subbase in question. By the previous theorem it suffices to show that every set of closed basic sets with the F.I.P has non-empty intersection.

So, let $\left\{B_i\right\}_{i\in\mathcal{I}}$ be the open base generated by $\mathcal{S}$ and let $\left\{B_k\right\}_{k\in\mathcal{K}}\subseteq\left\{B_i\right\}_{i\in\mathcal{I}}$ have the F.I.P. We first show that $\left\{B_k\right\}_{k\in\mathcal{K}}$ is contained in some maximal collection of open basic sets in the sense that if $\left\{B_k\right\}_{k\in\mathcal{K}}\subset\mathcal{B}$ then $\mathcal{B}$ does not have the F.I.P.

To see this we use Zorn’s lemma. Let $\Lambda$ be the collection of all superset of $\left\{B_k\right\}_{k\in\mathcal{K}}$ which have the F.I.P. It is clear that this class along with set inclusion induced a partially ordered set. Now, it is realtively easy to show that if $C\subseteq\Lambda$ is a chain, then $\displaystyle \bigcup_{c\in C}c$ is a class with the finite intersection property, and thus an upper bound. It follows by Zorn’s lemma that $\Lambda$ has a maximal element, call it $\left\{B_j\right\}_{j\in\mathcal{J}}$ and since $\displaystyle \bigcap_{j\in\mathcal{J}}B_j\subseteq\bigcap_{k\in\mathcal{K}}B_k$ it only remains to show that $\displaystyle \bigcap_{j\in\mathcal{J}}B_j\ne \varnothing$

Since each $B_j\in\left\{B_j\right\}_{j\in\mathcal{J}}$ is a basic open set we know that $B_j=S_1^j\cup\cdots\cup S_{m_j}^j$ where each $S_{m_k}^j$ is a subbasic closed set. And so if we can show that $\left\{S_{m_k}^j\right\}_{j\in\mathcal{J}}\subseteq\left\{B_j\right\}_{j\in\mathcal{J}}$ we are done because $\displaystyle \bigcap_{j\in\mathcal{J}}S_{m_k}^j\subseteq\bigcap_{j\in\mathcal{J}}B_j$.

It merely remains to show that at least one of $\left\{S_1^j,\cdots,S_{m_j}^j\right\}$ is in $\left\{B_j\right\}_{j\in\mathcal{J}}$ for each $j\in\mathcal{J}$. So now assume that none of the aforementioned sets are in $\left\{B_j\right\}_{j\in\mathcal{J}}$. Since $S_1$ is a subbasic open set is is necessarily a basic closed set and so to assume that $S_1\notin\left\{B_j\right\}_{j\in\mathcal{J}}$ would mean that $\left\{B_j\right\}_{j\in\mathcal{J}}\subset\left\{B_j\right\}_{j\in\mathcal{J}}\cup\left\{S_1\right\}$ and by the maximality of $\left\{B_j\right\}_{j\in\mathcal{J}}$ we see that $\left\{B_j\right\}_{j\in\mathcal{J}}\cup\left\{S_1\right\}$ does not have the finite intersection property. This then implies that $S_1\cap B_{j_1}\cap\cdots\cap B_{j_n}$ for some $\left\{B_{j_1},\cdots,B_{j_n}\right\}\subseteq\left\{B_j\right\}_{j\in\mathcal{J}}$. Doing this for each of $\left\{S_1,\cdots,S_{m_j}^j\right\}$ gives us a class whose union is disjoint from $S_1\cup\cdots\cup S_{m_j}^j=B_j$ which contradicts $\left\{B_j\right\}_{j\in\mathcal{J}}$ having the F.I.P. Doing this for each $B_j\in\left\{B_j\right\}_{j\in\mathcal{J}}$ finishes the proof. $\blacksquare$

These will be extremely useful later in the proof of some very powerful theorems. Here is one of them.

Theorem (Heine-Borel): A subset $E\subseteq\mathbb{R}$ is compact if$E$ is closed and bounded.

Proof: Since $E$ is bounded we know that $\in [a,b]$ for some finite closed interval $[a,b]$ and if we can prove that $[a,b]$ is compact it our theorem will follow since $E$ will be a closed subspace of a compact space.

So, now if $a=b$ this is obvious, so assume $a. It is relatively easy to prove that the set $\left\{[a,d):a is an open subbase for $[a,b]$ so the set $\mathcal{S}=\left\{[a,c]:a is a closed subbase, we must merely show that any subclass of $\mathcal{S}$ with the F.I.P. has non-empty intersection. So let $S=\left\{[a,c_j]\right\}_{j\in\mathcal{J}}\cup\left\{[d_k,b]\right\}_{k\in\mathcal{K}}\subseteq\mathcal{S}$ be non-empty and posses the F.I.P. If $S$ contains only intervals of one of the two types then their intersection clearly contiains $a$ or $b$, so assume that $S$ contains intervals of both types. Let $d=\sup_{k\in\mathcal{K}}d_k$. We finish the argument by showin that $d\leqslant c_j$ for ever $j\in\mathcal{J}$. Suppose not, and there existed some $c_{j_1}. By definition then there exists some $c and so $[a,c_{j_1}]\cap[d_{k_1},b]=\varnothing$ contradicting $S$‘s F.I.P. The conclusion follows. $\blacksquare$

One immediately corollary of this is the following:

Theorem: Let $\left\{[a_n,b_n]\right\}$ be sequence of decreasing closed intervals. Then, $\displaystyle \bigcap_{n=1}^{\infty}[a_n,b_n]\ne\varnothing$.

——————————————————————————————————

To continue from here we need to define the following:

Compact subspace: If $\left(X,\mathfrak{J}\right)$ and $Y\subseteq X$ is compact with the relative topology we say that $Y$ is a compact subspace of $X$.

It is clear that not only is this definition necessary for many reasons one would be we might want to consider whether $X,Y$ is compact implies that $X\cup Y$, but as one can see we run into the problem of under what topology? As we will see later, in a generalization, if $X,Y$ are compact subspaces of some ambient space $Z$ then $X\cup Y$ is a compact subspace. But for now:

Theorem: If $X$ is a compact topological space and $E$ is a closed subset of $X$ then $E$ is a compact subspace.

Proof: Let $\left\{O_j\right\}_{j\in\mathcal{J}}$ be an open cover for $E$, by definition we have that $O_j=K_j\cap E$ for some open $K_j$ in $X$. Clearly $\left\{K_j\right\}_{j\in\mathcal{J}}$ covers $E$ and thus $\left\{K_j\right\}_{j\in\mathcal{J}}\cup\{Y'\}$ is an open cover for $X$. By assumption then there exists some finite subcover $\Sigma$ of $\{K_j\}_{j\in\mathcal{J}}\cup\{Y'\}$, if $Y'\in \Sigma$ we may discard it and be left with some finite set $\left\{K_{j_1},\cdots,K_{j_m}\right\}$ which covers $Y$. Clearly then taking the corresponding $\left\{K_{j_1},\cdots,K_{j_m}\right\}$ in $\left\{O_j\right\}_{j\in\mathcal{J}}$ produces the necessary finite subcover. $\blacksquare$

Of course any student worth their weight in sawdust is asking “Is the converse necessarily true? Do compact subspaces have to be closed?”. The answer is, no. Consider for example the topological space $\left(X,\mathfrak{J}\right)$ with $X=\{a,b,c\}$ and $\mathfrak{J}=\left\{\varnothing,\{a\},X\right\}$. Clearly $\{a\}$ is a compact subspace since it’s finite but it is clearly not closed since $X-\{a\}=\{b,c\}\notin\mathfrak{J}$. Fortunately, that’s not all she wrote. There is a partial converse:

Theorem: Let $X$ be a Hausdorff topological space and $E$ a compact subspace, then $E$ is closed in $X$.

Proof: We prove that $E$ is closed by proving that $X-E$ is open. Let $x\in X-E$. For each $e\in X$ let $O_e, U_e$ be the open sets containing $e,x$ respectively such that $O_e\cap U_e=\varnothing$. Clearly, if we let $K_e=O_e\cap E$ then $\left\{K_e\right\}_{e\in E}$ is clearly an open cover for $E$ and so there exists some $\left\{K_{e_1},\cdots,K_{e_m}\right\}\subseteq\left\{K_e\right\}_{e\in E}$ such that $\displaystyle E=\bigcup_{\ell=1}^{m}K_{e_\ell}$. Clearly then we have that $\displaystyle E\subseteq\bigcup_{\ell=1}^{m}O_{e_\ell}$ for the corresponding set $\displaystyle \left\{O_{e_1},\cdots,O_{e_m}\right\}$. So let $V=\bigcap_{\ell=1}^{m}V_{e_\ell}$ which is clearly an open set disjoint from $\displaystyle \bigcup_{\ell=1}^{m}O_{e_\ell}\supseteq E$. The conclusion follows. $\blacksquare$

We now proved what was promised earlier.

Theorem: Let $X$ be a topological space and let $X_1,\cdots,X_n$ be a finite class of compact subspaces of $X$. Then $X_1\cup\cdots\cup X_n$ is a compact subspace of $X$.

Proof: Let $\left\{O_j\right\}_{j\in\mathcal{J}}$ be an open cover for $X_1\cup\cdots\cup X_n$, then $O_j=E_j\cap\left(X_1\cup\cdots X_n\right)$ for some open set $E_j$ in $X$. Clearly then $\left\{E_j\right\}_{j\in\mathcal{J}}$ covers $X_1\cup\cdots\cup X_n$ and thus each $X_1,\cdots,X_n$. Clearly then $\left\{E_j\cap X_1\right\}_{j\in\mathcal{J}},\cdots,\left\{E_j\cap X_n\right\}_{j\in\mathcal{J}}$ are open covers for $X_1,\cdots,X_n$ respectively which by assumption means that for each $X_k,\text{ }1\leqslant k\leqslant n$ there exists some $E^k_1\cap X_k,\cdots,E_{m_k}^k\cap X_k$ which is a finite subcover of $X_k$. Let $\Omega=\left\{E_1^1,\cdots,E_{m_1}^1\right\}\cup\cdots\cup \left\{E_1^n,\cdots,E_{m_n}^n\right\}$. Then $\Omega$ is a class of open sets in $X$ that covers $X_1\cup\cdots\cup X_n$. Clearly then intersecting each element of $\Omega$ will result in the desired subclass of $\left\{O_j\right\}_{j\in\mathcal{J}}$. The conclusion follows. $\blacksquare$

Using this we can prove a nice little theorem. Suppose that you were a naive analyst and you encounter the following dilemma.  You have a function $f:E\mapsto\mathbb{R}$ where $E\subseteq \mathbb{R}$ but $E$ is not compact. You wonder, could I possibly make $E$ compact by removing a few points? The answer is no.

Theorem: Let $X$ be a topological space and let $E$ be a non-compact subspace of $X$, then $E-\{e_1,\cdots,e_n\}$

is non-compact.

Proof: Suppose that $E-\{e_1,\cdots,e_n\}$ was compact, then since each $\{e_k\},\text{ }1\leqslant k\leqslant n$ is compact (since they’re finite) it follows that $\displaystyle E-\{e_1,\cdots,e_n\}\cup\bigcup_{k=1}^{n}\{e_k\}=E$ is compact. Boom. $\blacksquare$

Also, one might ask “what about the infinite case?” The answer is…well no. It should be obvious intuitively that since the union of an infinite number of closed sets need not be closed and any compact subspace of a Hausdorff space is closed that this probably isn’t true. Just to sate your doubtlessly eager minds though, consider $\mathbb{N}$ with the discrete topology and consider that $\left\{\{n\}:n\in\mathbb{N}\right\}$ is an infinite class of compact spaces who’s union is the full space $\mathbb{N}$ and by previous discussion, this is not compact.

Now, what about the intersection of compact subspaces?

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty class of compact closed subspaces of a topological space $X$ then if $\displaystyle \bigcap_{j\in\mathcal{J}}X_j\ne \varnothing$ (we don’t ever want to talk about empty topological spaces) then $\displaystyle \bigcap_{j\in\mathcal{J}}X_j$ is compact.

Proof: Clearly we have that given any $X_\ell\in\left\{X_j\right\}_{j\in\mathcal{J}}$ that since $\displaystyle \bigcap_{j\in\mathcal{J},j\ne\ell}X_j$ is a closed subset of $X$ that $\displaystyle X_\ell\cap\bigcap_{j\in\mathcal{J},j\ne\ell}X_j=\bigcap_{j\in\mathcal{J}}X_j$ is a closed subset of $X_\ell$. And since $X_\ell$ is compact it follows that $\displaystyle\bigcap_{j\in\mathcal{J}}X_j$ is a compact subspace of $X_\ell$ and thus a compact subspace of $X$. $\blacksquare$

February 19, 2010