Let $(X,\mathscr T)$ be a topological space. Given $A\subseteq X$, we say that $x$ is a complete limit point of $A$ if for every nbhd $N$ of $x$, $|N\cap A|=|A|$. I want to prove

Suppose $(X,\mathscr T)$ is compact. Then every infinite subset $A$ of $X$ has a complete limit point in $X$.

PROOF Assume the contrary. Then there exists an infinite subset $A$ of $X$ such that for any $x\in X$, there is a nbhd $N_x$ such that $|N_x\cap A|<|A|$. The collection $\{N_x:x\in X\}$ covers $X$. By compactness, there exists a finite subcover $\{N_{x_1},\ldots, N_{x_n}\}$. Then $A=\bigcup_{i=1}^n A\cap N_{x_i}$. But this is impossible, since $A$ is infinite, and $\bigcup_{i=1}^n A\cap N_{x_i}$ is a finite union of sets strictly smaller than $A$.

Is this correct? The problem is in "Introductory Real Analysis" by Kolmogorov.

Also, is a proof of the converse too complicated? Kolmogorov gives a reference as "P.S. Alexandroff, Einführung in die Mengenlehre und die Theorie der Reellen Funktionen (1956), pp. 250-251; J.L. Kelley, General Topology (1955), pp. 163-164" but I have no idea what this means. I am honestly clueless on how to approach it. Could you provide hints? I wouldn't mind a complete answer, if you think it is requires something too advanced.

  • 1
    $\begingroup$ Kelly offers it as an exercise, with an extended hint: "If $X$ is not compact choose an open cover $A$ with no finite subcover such that the cardinal number $c$ of $A$ is as small as possible. Let $C$ be a well-ordered set of cardinal $c$ such that the set of predecessors of each member has a cardinal less than $c$ (It is shown in the appendix that $c$ is such a set.) Let $f$ be a one-to-one map of $C$ onto $A$. Then for each member $b$ of $C$ the union $\cup\{ f(a):a<b\}$ does not cover $X$ and, in fact, the complement of this union must have cardinal number at least as great as $c$..." $\endgroup$ Jul 4, 2013 at 1:45
  • $\begingroup$ "...It is therefore possible to choose $x_b$ from the complement such that $x_a\ne x_b$ for $a<b$. Consider the set of all $x_b$." $\endgroup$ Jul 4, 2013 at 1:46
  • $\begingroup$ @DavidMitra Thank you. Why not put it as an answer? Now someone posted the same! =) $\endgroup$
    – Pedro
    Jul 4, 2013 at 3:48
  • $\begingroup$ I'm confused as to why $|N_{x_i}| < |A|$. Can you explain this? $\endgroup$
    – Ink
    Jul 4, 2013 at 5:25
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    $\begingroup$ @Ink Yes, it seems the last line is wrong. However, it is easy to repair. Note that the union of the sets $N_{x_1}\cap A,\dots,N_{x_n}\cap A$ is $A$; thus, by the pidgeonhole principle, $|N_{x_i}\cap A|=|A|$ for some $i$. $\endgroup$ Jul 4, 2013 at 13:51

1 Answer 1


Your proof is correct. For the converse, suppose $X$ is not compact. Let $\mathcal{O}$ be an open cover of $X$ of minimal cardinality [such that no finite subcover exists], and let $\{O_\alpha\}_{\alpha < |\mathcal{O}|}$ be an enumeration of $\mathcal{O}$. Construct a set $A$ as follows:

  • let $x_\alpha \in X$ be any element not covered by any of the $O_\beta$ for $\beta < \alpha$. Such $x_\alpha$ exists or else $\{O_\beta\}_{\beta < \alpha}$ would be an open cover, contradicting the minimality of $|\mathcal{O}|$.
  • let $A = \{x_\alpha : \alpha < |\mathcal{O}|\}$

We'll show that $A$ has no complete limit point. Indeed, for any $x \in X$ there is some $\alpha$ such that $O_\alpha$ is a neighbourhood of $x$, but $O_\alpha$ avoids all the $x_\beta$ for $\beta > \alpha$, and so:

$$\left |A \cap O_\alpha\right | \leq \left|\{x_\gamma : \gamma \leq \alpha\}\right| = |\alpha| < |\mathcal{O}| = |A|$$

  • $\begingroup$ $\{X\}$ is an open cover of minimal cardinality... Maybe you want to require that it doesn't have a finite subcover? $\endgroup$
    – Asaf Karagila
    Jul 4, 2013 at 19:15
  • $\begingroup$ @AsafKaragila Yes, he means the same as David. =) $\endgroup$
    – Pedro
    Jul 4, 2013 at 19:23
  • $\begingroup$ How do you know, that $A$ is an infinite set? $\endgroup$
    – el_tenedor
    Jan 4, 2016 at 16:34

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