In my earlier post on Tychonoff’s theorem, I talked about the original proof, based on the following characterization of compactness which is due to Kuratowski.

**Definition** Let \(S\) be a subset of a topological space \(X\). A point \(x\in X\) is a *point of perfection* of \(S\) if, for every neighborhood \(U\) of \(x\), the set \(U\cap S\) has the same cardinality as \(S\).

**Lemma** (Kuratowski) A topological space \(X\) is compact if and only if every infinite subset has a point of perfection.

Now I will give the proof of this characterization (again following Wright).

Suppose \(X\) is compact, and let \(S\) be an infinite subset, with cardinality \(\frak a\). If \(S\) has no point of perfection, then every point of \(X\) has a neighborhood that meets \(S\) in fewer than \(\frak a\) points. By compactness, \(X\) has a finite cover by such neighborhoods. But a finite union of sets of cardinality \(<\frak a\) must itself have cardinality \(<\frak a\), contradiction.

In the other direction it seems simplest to use the well-ordering principle. Suppose that \(X\) is *not* compact, and let \(\mathscr U\) be an open cover without finite subcover; we assume that \(\mathscr U\) has the minimum cardinality \(\frak a\) with respect to this condition. Let \(\Omega\) be the least ordinal having cardinality \(\frak a\). Then the sets of the cover \(\mathscr U\) can be enumerated as \(\{U_\alpha: \alpha < \Omega \} \).

Let \(B = \{\alpha<\Omega: U_{\alpha}\setminus \bigcup_{\beta<\alpha} U_\beta\neq\emptyset \} \). Since the \(U_{\alpha}\) for \(\alpha\in B\) cover \(X\), the set \(B\) has cardinality \(\frak a\). Let \(L\subseteq X\) be constructed by choosing one point \(x_\alpha \in U_{\alpha}\setminus \bigcup_{\beta<\alpha} U_\beta \) for each \(\alpha\in B\). Then \(L\) has cardinality \(\frak a\); but each point of \(X\) lies in a neighborhood, namely some \(U_\alpha\), that contains at most \( \text{card}(\alpha) < {\frak a} \) points of \(L\). Thus \(L\) has no point of perfection.