3  Compactness

It can be helpful to know the conditions that a maximum or minimum exists. For any \(f: [a, b]\to \mathbb R\) continuous, there is an \(x \in [a, b]\) with \(f(x) = \sup \{f(u): a \le u \le b\}\) (and similarly for \(\inf\)). This property that a continuous real-valued function is bounded and attains its supremum and infimum can be extended to compact topological spaces.

3.1 Compactness

Definition 3.1 A topological space \((K, \mathcal T)\) is called compact iff whenever \(\mathcal U\subset \mathcal T\) and \(K = \bigcup \mathcal U\), there is a finite \(\mathcal V \subset \mathcal U\) such that \(K = \bigcup \mathcal V\).

Definition 3.2 Let \(X\) be a set, \(A \subset X\). A collection of sets whose union includes \(A\) is called a cover(ing) of \(A\). If the collection consists of open sets, it is called an open cover.

The definition of compactness is basically “Every open cover has a finite subcover”.

Example 3.1  

  • The open intervals \((-n, n)\) form an open cover of \(\mathbb R\) without a finite subcover. \(\mathbb R\) is not compact.
  • The intervals \((1/(n+2), 1/n)\) for \(n = 1,2,...\) form an open cover of \((0,1)\) without a finite subcover. \([0,1]\) is not compact.

Definition 3.3 A subset \(K\) of a topological space \(X\) is called compact iff it is compact for its relative topology. That is, \(K\) is compact if for any \(\mathcal U \subset \mathcal T\) such that \(\mathcal K \subset \bigcup \mathcal U\), there is a finite \(\mathcal V \subset \mathcal U\) such that \(K \subset \bigcup V\).

Non-empty set \(A\) of real numbers has an upper bound \(b\), so that \(x \le b\) for all \(x \in A\), then \(A\) has a least upper bound, or \(\text{supremum }c := \sup A\). That is, \(c\) is an upper bound of \(A\) such that \(c \le b\) for any other upper bound \(b\) of \(A\). Likewise, a non-empty set \(D\) of real numbers with a lower bound has a greatest lower bound, or \(\text{infimum}\), \(\inf D\). If a set \(A\) is unbounded above, let \(\sup A := +\infty\). If \(A\) is unbounded below, let \(\inf A := -\infty\).

Theorem 3.1 Any closed interval \([a, b]\) with its usual (relative) topology is compact.

We prove for the interval \([a, b] = [0, 1]\).

Let \(\mathcal U\) be an open cover of \([0,1]\). Let \(H = \{x \in [0, 1]: [0, x] \subset \bigcup \mathcal V\}\) where \(\mathcal V\subset \mathcal U\) finite.

Then since \(0 \in V\) for some \(V \in \mathcal U\), \([0, h] \subset H\) for some \(h > 0\). If \(H \ne [0,1]\), let \(y := \inf([0,1] \setminus H)\). Take \(y \in V\) for some \(V \in \mathcal U\), so for some \(c > 0\), \([y - c, y] \subset V\) and \(y-c \in H\). Taking a finite open subcover of \([0, y-c]\) and adjoining \(V\) gives an open cover of \([0, y]\), so \(y\in H\). If \(y=1\) we are done; otherwise, for some \(b > 0\), \([y, y+b] \subset V\), so \([0, y+b] \subset H\), contradicting the choice of \(y\).

Theorem 3.2 If \((K, \mathcal T)\) is a compact topological space, \(F \subset K\) closed, then \(F\) is compact.

Let \(\mathcal U\) be an open cover of \(F\), where we may take \(\mathcal U \subset \mathcal T\). Then \(\mathcal U \cup \{K \setminus F\}\) is an open cover of \(K\), so has a finite subcover \(\mathcal V\). Then \(\mathcal V \cup \{K \setminus F\}\) a finite cover of \(F\), included in \(\mathcal U\).

Theorem 3.3 If \((K, \mathcal T)\) is compact and \(f\) is continuous from \(K\) onto another topological space \(L\), then \(L\) is compact.

Let \(\mathcal U\) be an open cover of \(L\). Then \(\{f^{-1}(U): U \in \mathcal U\}\) is an open cover of \(K\), with a finite subcover \(\{f^{-1}(U): U \in \mathcal V\}\) where \(\mathcal V\) is finite. Then \(\mathcal V\) is a finite subcover of \(L\).

Example 3.2 If \(f\) is a continuous real-valued function on a compact space \(K\), then \(f\) is bounded, since any compact set in \(\mathbb R\) is bounded, consider the open cover by intervals \((-n, n)\).