# Abstract Nonsense

## Crushing one theorem at a time

Point of post: In this post we discuss the notion of compactness for topological spaces.

Anyone who has spent more than twenty minutes doing analysis has surely notice that all subsets of $\mathbb{R}$ are not created equally. For example, a lot more can be said about a function $f:[0,1]\mapsto\mathbb{R}$ than a function $g:(0,1)\mapsto\mathbb{R}$. We know for example that $f$ is bounded and uniformly continuous. In fact, we know that $f$ assumes a maximum value on $I$ in the sense that $\left\|f\right\|_{\infty}=f(c),$ for some $c\in [0,1]$. None of these things can be necessarily said about $g$. In fact, $g(x)=\frac{1}{x}$ has none of these properties.  It is not surprising that analysts wanted to find a quality that differentiates these “special” sets, and this quality is compactness.

Before we define compactness for subsets of $\mathbb{R}$ we first need some terminology.

Open cover- A class of sets $\Omega$ is said to be an open cover of a set $E\subseteq\mathbb{R}$ if each $\omega\in\Omega$ is open $\displaystyle E\subseteq\bigcup_{\omega\in\Omega}\omega$

Subcover- If $\Omega$ is an open cover of a set $E\subseteq \mathbb{R}$ then $\Sigma$ is called a subcover of $\Omega$ if $\Sigma\subseteq\Omega$ and $\Sigma$ is an open cover of $E$

With this in mind we say a set $E\subseteq \mathbb{R}$ is compact if every open cover of $E$ admits a finite subcover.

Example: The set $\displaystyle E=\{0\}\cup\left\{\frac{1}{n}\right\}_{n\in\mathbb{N}}$ is compact.

Proof: Let $\Omega$ be an open cover of $E$. Clearly by definition there exists some open set $O\in\Omega$ such that $0\in O$. But, by the formulation of open sets in $\mathbb{R}$ this means there exists some $\delta>0$ such that $(-\delta,\delta)\subseteq O$. The Archimedean principle then furnishes us with some $\mathfrak{n}\in\mathbb{N}$ such that $\displaystyle \frac{1}{\mathfrak{n}}<\delta$ which of course implies that $\displaystyle \mathfrak{n}\leqslant n\implies \frac{1}{n}\in(-\delta,\delta)\implies \frac{1}{n}\in O$. So taking some open set $E_k\in\Omega,\text{ }1\leqslant k <\mathfrak{n}$ such that $\displaystyle \frac{1}{k}\in E_k$ we see that $\Sigma=\left\{O,E_1,\cdots,E_{\mathfrak{n}-1}\right\}$ is a finite subcover of $\Omega$. Since $\Omega$ was arbitrary the conclusion follows. $\blacksquare$

Non-example: The set $(0,1)$ is not compact.

Proof: Let $\displaystyle \Omega=\left\{\left(0,1-\frac{1}{n}\right):\text{ }n\in\mathbb{N}-\{1\}\right\}$. This is an open cover for $(0,1)$ since given any $x\in(0,1)$ we have that, by the Archimedean principle, there exists some $\mathfrak{n}\in\mathbb{N}$ such that $\displaystyle \frac{1}{\mathfrak{n}}<1-x$ and so $\displaystyle x<1-\frac{1}{\mathfrak{n}}$ and so $\displaystyle x\in\left(0,1-\frac{1}{\mathfrak{n}}\right)$ and since $\displaystyle \left(0,1-\frac{1}{\mathfrak{n}}\right)\in\Omega$ it follows that $\displaystyle x\in\bigcup_{\omega\in\Omega}\omega$ and so $\displaystyle (0,1)\subseteq\bigcup_{\omega\in\Omega}\omega$ and so $\Omega$ is an open cover for for $(0,1)$. But given any finite subclass of $\Omega$ say $\displaystyle \left\{\left(0,1-\frac{1}{n_1}\right),\cdots,\left(0,1-\frac{1}{n_m}\right)\right\}$ we see that $\displaystyle 1-\frac{1}{n_{\text{max}}+1}\in(0,1)$ but $\displaystyle 1-\frac{1}{n_{\text{max}}+1}\notin\bigcup_{j=1}^{m}\left(0,1-\frac{1}{n_j}\right)$ where $\displaystyle n_{\text{max}}=\max\left\{n_1,\cdots,n_m\right\}$. And thus we have exhibited an open cover for $(0,1)$ which admits no finite subcover, thus by definition $(0,1)$ is not compact. $\blacksquare$

Of course, there was nothing in the above definition that limits us to considering $\mathbb{R}$. The exact same formulation works for any metric space.

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We now wish to  generalize this a tad further and talk about when a topological space $\left(X,\mathfrak{J}\right)$ is compact.

We call such a space $X$ compact if given any collection of open sets $\Omega$ such that $\displaystyle \bigcup_{\omega\in\Omega}\omega=X$ (an open cover)there exists a finite subclass $\Sigma$ such that $\displaystyle \bigcup_{\sigma\in\Sigma}\sigma=X$ (a finite subcover).

Some things are immediately obvious

Theorem: A topological space $\left(X,\mathfrak{J}\right)$ such that $\mathfrak{J}$ is finite is compact.

Proof: By necessity any open cover $\Omega$ must be finite (since $\Omega\subseteq\mathfrak{J}$ and thus serves as a finite subcover. $\blacksquare$

Corollary 1: Any finite space is compact.

Corollary 2: Any indiscrete space is compact.

Theorem: A discrete space $X$ is compact if and only if it is finite.

Proof:

$\implies$: Let $\displaystyle \Omega=\left\{\{x\}:x\in X\right\}$. This is clearly an open cover for $X$ with no proper subcover (i.e. the removal of any element results in a non-cover). It follows from $X$‘s compactness that $\Omega$ itself is a finite subcover, which obviously implies that $X$ is compact. $\blacksquare$

$\Leftarrow$: Follows from an earlier problem.

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We now give some alternative, and very useful, conditions that are necessary and sufficient for a topological space $X$ is compact.

Theorem: A topological space $X$ is compact if and only if every closed class of closed subsets of $X$ $\displaystyle \left\{C_j\right\}_{j\in\mathcal{J}}$ with empty intersection has a finite subclass with finite intersection.

Proof:

$\implies$: We see that $\displaystyle \bigcap_{j\in\mathcal{J}}C_j=\varnothing\implies \bigcup_{j\in\mathcal{J}}C'_j=X$ and so $\displaystyle \left\{C'_j\right\}_{j\in\mathcal{J}}$ is an open cover for $X$ so so by assumption there exists some $\displaystyle \left\{C'_1,\cdots,C'_n\right\}$ which is a finite subcover. Thus, $\displaystyle \bigcup_{j=1}^{n}C'_j=X\implies \bigcap_{j=1}^{n}C_j=\varnothing$ and the conclusion follows.

$\Leftarrow$: Let $\displaystyle \left\{O_j\right\}_{j\in\mathcal{J}}$ be an open cover of $X$, it then follows (using the same logic as the previous problem) that $\displaystyle \bigcap_{j\in\mathcal{J}}O'_j=\varnothing$ and so there exists some $\displaystyle \left\{O'_1,\cdots,O'_n\right\}$ such that $\displaystyle \bigcap_{j=1}^{n}O'_j=\varnothing\implies\bigcup_{j=1}^{n}O_j=X$ and the conclusion follows. $\blacksquare$

We now introduce the term finite intersection property (F.I.P) to mean a class of sets such that every finite intersection is non-empty.

Theorem: A topological space $X$ is compact if and only if every class of closed subsets with the finite intersection property has non-empty interseciton

Proof:

$\implies$: Let $\displaystyle \left\{C_j\right\}_{j\in\mathcal{J}}$ be class of closed subsets of $X$ with the F.I.P. and suppose that $\displaystyle \bigcap_{j\in\mathcal{J}}C_j=\varnothing$ then $\displaystyle \bigcup_{j\in\mathcal{J}}C'_j=X$ and so $\displaystyle \left\{C'_j\right\}_{j\in\mathcal{J}}$ is an open cover for but given any finite subcover $\displaystyle \left\{C'_{j_1},\cdots,C'_{j_m}\right\}$ we have that $\displaystyle \bigcup_{k=1}^{m}C'_{j_k}=\left(\bigcap_{k=1}^{m}C_{j_k}\right)'$ and since $\displaystyle \varnothing\subset\bigcap_{k=1}^{m}C_{j_k}$ we see that $\displaystyle\left(\bigcap_{k=1}^{m}C_{j_k}\right)'\subset X$ and so $\displaystyle \left\{C'_j\right\}_{j\in\mathcal{J}}$ is an open cover of $X$ with no finite subcover, which of course contradicts $X$‘s compactness.

$\Leftarrow$: Suppose that $\displaystyle \left\{O_j\right\}_{j\in\mathcal{J}}$ was an open cover for $X$ which admitted no finite subcover. Then clearly, following the logic of the last problem (DeMorgan’s law), that $\displaystyle \bigcup_{j\in\mathcal{j}}O_j=X\implies \bigcap_{j\in\mathcal{J}}O'_j=\varnothing$ but $\displaystyle \bigcup_{k=1}^{m}O_{j_k}\ne X\implies \bigcap_{k=1}^{m}O'_{j_k}\ne \varnothing$ and so $\displaystyle \left\{O'_j\right\}_{j\in\mathcal{J}}$ is a class of closed subsets of $X$ with the F.I.P but that has empty intersection. This is a contradiction. $\blacksquare$

From this we can derive kind of a nice little theorem:

Theorem: Call a sequence of sets $\left\{F_n\right\}_{n\in\mathbb{N}}$ decreasing if $F_{n+1}\subseteq F_n$. Then, if $\{F_n\}_{n\in\mathbb{N}}$ is a decreasing sequence of non-empty closed sets in a compact topological space $X$, then $\displaystyle \bigcap_{n=1}^{\infty}F_n\ne\varnothing$.

Proof: Clearly given any finite subclass $\left\{F_{n_1},\cdots,F_{n_m}\right\}$ of $\left\{F_n\right\}_{n\in\mathbb{N}}$ we have (since the subclass forms a finite chain) that $F_{n_1}\cap\cdots\cap F_{n_m}=F_N$ where $N=\max\{n_1,\cdots,n_m\}$ and since $F_N\in\left\{F_{n}\right\}_{n\in\mathbb{N}}$ it follows that $F_N\ne \varnothing$. Thus, $\left\{F_n\right\}_{n\in\mathbb{N}}$ is a class of closed sets with the F.I.P. and by the assumption that $X$ is compact we see that $\displaystyle\bigcap_{n=1}^{\infty}F_n\ne\varnothing$. $\blacksquare$