# Tychonoff’s theorem II

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.