Abstract Nonsense

Crushing one theorem at a time

Thoughts about compactness.

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.


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.


\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.


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.


\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


\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


February 18, 2010 - Posted by | General Topology, Topology | , ,

1 Comment »

  1. […] this normal topology one can prove that has the Heine-Borel Property that while in general compact subspaces of metric spaces are closed and bounded the converse is true in (this can be proven […]

    Pingback by Structure of Euclidean Space « Abstract Nonsense | May 22, 2011 | Reply

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