Here is Theorem 28.1 in the book Topology by James R. Munkres, 2nd edition:

Compactness implies limit point compactness, but not conversely.

And, here is Munkres' proof:

Let $X$ be a compact space. Given a subset $A$ of $X$, we wish to prove that if $A$ is infinite, then $A$ has a limit point. We prove the contrapositive---if $A$ has no limit point, then $A$ must be finite.

So suppose $A$ has no limit point. Then $A$ contains all its limit points, so that $A$ is closed. Furthermore, for each $a \in A$ we can choose a neighborhood $U_a$ of $a$ such that $U_a$ intersects $A$ in the point $a$ alone. The space $X$ is covered by the open set $X-A$ and the open sets $U_a$; being compact, it can be covered by finitely many of these sets. Since $X-A$ does not intersect $A$, and each set $U_a$ contains only one point of $A$, the set $A$ must be finite.

This proof is OK, but I thought I would like to give a slightly different proof, which runs as follows:

Suppose that $A$ has no limit points in $X$. Then for each point $x \in X$, there is an open set $U_x$ containing $x$ such that $U_x$ does not intersect $A$ in any point other than $x$ itself, that is, either $U_x \cap A = \emptyset$ or $U_x \cap A = \{x\}$.

Now the collection $$ \left\{ \, U_x \, \colon \, x \in X \, \right\} \tag{1} $$ is an open covering of the compact topological space $X$, and so there exists a finite subcollection of the collection (1) that also covers $X$; let this finite subcollection be $$ \left\{ \, U_{x_1}, \ldots, U_{x_n} \, \right\}, $$ for some points $x_1, \ldots, x_n \in X$.

Now as $$ A \subset X = \bigcup_{i=1}^n U_{x_i}, $$ and as either $U_{x_i} \cap A$ is either empty or a singleton set, for each $i = 1, \ldots, n$, so we can conclude that the set $A$ must be finite.

Is my proof correct? If so, is this proof simpler (or clearer) than the one in Munkres? If not, then where are problems in my reasoning?

  • 1
    $\begingroup$ Your argument is fine. Though the presentation is different, I’d argue that it’s essentially the same as the one in Munkres — his $X\setminus A$ is $\bigcup\{U_x:U_x\cap A=\varnothing\}$ in your notation. I think that it’s probably a matter of taste which one is clearer. $\endgroup$ – Brian M. Scott Mar 29 '20 at 18:01
  • $\begingroup$ @BrianM.Scott how are you? Hope you've managed to successfully evade COVID-19. How're things as regards COVID-19 around where you are? Can you please share with me the materials (i.e. lecture notes, assignments and exams with solutions) from your general topology classes? Or, do you have any webpage(s)? $\endgroup$ – Saaqib Mahmood Mar 29 '20 at 18:15
  • $\begingroup$ I’m doing okay; there are cases of COVID-19 in my area, but I’ve avoided it so far and have no need to leave my house. I hope that you’re also doing okay. I wish that I could help with general topology material, but I’ve been retired since 2011 and hadn’t had a chance to teach topology in many years before that, so I’m afraid that I don’t have any material in electronic form: what little I still have is all handwritten or, in a few cases, typed hard copy. $\endgroup$ – Brian M. Scott Mar 29 '20 at 18:37

The argument is finte, though the final part could be clearer and I'd present it more directly as

$$A = A \cap X =A \cap (\bigcup_i^n U_{x_i}) = \bigcup_i^n (A \cap U_{x_i})\tag{1}$$

and then $A$ would be written as a finite union fo finite sets, implying $A$ is finite.

This also makes the generalisation obvious: if $A$ is an infinite subset of a space $X$, $x \in X$ is called a point of total condensation of $A$ iff for every open neighbourhood $O$ of $x$ we have $|O \cap A|=|A|$. Even for the countable case this is stronger than a "mere" limit point (Counterexamples calls it an $\omega$-limit point in that case) and in fact

$X$ is compact iff every infinite subset $A$ of $X$ has a point of total accumulation.

This is one of the oldest equivalent formulations of compactness, in fact. And for the countable case (i.e. we only ask for every countably infinite set to have an $\omega$-limit point) we get an exact equivalence with countable compactness (in the open covering sense). But the forward direction in the general case is a straightforward modification of your proof idea, using my $(1)$, and using a cover of open $U_x$ with $|U_x \cap A| < |A|$, of course.


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