# Abstract Nonsense

## Thoughts about compactness (Pt. II)

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, $\displaystyle \bigcap_{n=1}^{\infty}[a_n,b_n]\ne\varnothing$.

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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 $X$. $\blacksquare$