I think that I understand why compactness implies limit point compactness:

Suppose $A \subseteq X$ has no limit points. Then $A^{\prime} \subseteq A$. Thus, $A$ is closed. Then for all $a \in A$, there exists some neighborhood $U_a$ of $A$ s.t. $U_a \cap A= \{a\}$. But then since $X$ is compact and $A \cup (X-A)=X$, it is an open covering. [$X-A$ has to be open, since $A$ is closed]. Then by compactness, there is a finite open covering and since $(X-A) \cap A = \emptyset$, It is clear that $A$ is finite. The result follows

However, the converse of the statement is supposedly incorrect. This definitely seems weird, particularly in $R$ (equipped with usual topology.)

Can somebody provide me with a counter-example?

  • $\begingroup$ You mean whose intersection with $A$ is $\{a\}$. For points not in $A$ we can even choose it to be empty. $\endgroup$ – Henno Brandsma Jan 27 '16 at 21:57
  • $\begingroup$ Look here: math.stackexchange.com/questions/512393/… $\endgroup$ – MotylaNogaTomkaMazura Jan 27 '16 at 22:01
  • $\begingroup$ "Say $Y=\{a,b\}$. If $S$ is a subset of $\Bbb Z_+\times Y$ and $(n,a)\in S$, then $(n,b)$ is a limit point of $S.$" I do not understand why $(n,b)$ is a limit point of $S$. Is it because $(Y,\tau)=\{\emptyset, Y\}$ so if $S$ is open about $(n,a)$, then it also contains $(n,b)$?. $\endgroup$ – Andres Mejia Jan 28 '16 at 0:00
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    $\begingroup$ @AndresMejia That's exactly the reason, yes. A basic subset of the product has $Y$ as its second factor, as that is the only open set that is non-empty. So if $(n,a) \in O$ or $(n,b) \in O$, then the open set $\{n\} \times Y$ must be a subset of $O$. So the other point is in $O$ as well. Every point has a "sticky neighbour". $\endgroup$ – Henno Brandsma Jan 28 '16 at 21:11

Your proof is not correct as it stands, as $U_a$ is not equal to $\{a\}$ but we know that $U_a \cap A \subseteq \{a\}$, in general.

We also don't really need the fact that $A$ is closed (though this is true). Pick for every $x \in X$ some open $U_x$ that contains $x$ and such that $U_x \cap A \subseteq \{x\}$. This is, by definition almost, an open cover of $X$. Compactness tells us that there is a finite subset $F$ of $X$ such that $X = \cup_{x \in F} U_x$. But then $A = X \cap A = A \cap (\cup_{x \in F} U_x) = \cup_{x \in F} (U_x \cap A) \subseteq F$, by how the $U_x$ were chosen, and this contradicts that $A$ is infinite.

A countably compact space that is not compact is the first uncountable ordinal, $\omega_1$, in the order topology. Or $\{0,1\}^\mathbb{R} \setminus \{\underline{0}\}$ and many more.

You are right that for metric spaces the reverse does hold, but this is atypical.

  • $\begingroup$ I don't have any working definition for "countably compact." $\endgroup$ – Andres Mejia Jan 28 '16 at 0:01
  • $\begingroup$ also, I have edited my proof accordingly. $\endgroup$ – Andres Mejia Jan 28 '16 at 0:02
  • $\begingroup$ A space is countably compact provided every countable open cover admits a finite subcover. This is strictly weaker than compactness, which requires that every open cover admits a finite subcover. $\endgroup$ – Austin Mohr Jan 28 '16 at 1:00
  • $\begingroup$ @AndresMejia limit point compact is equivalent to countable compact for $T_1$ spaces. $\endgroup$ – Henno Brandsma Jan 28 '16 at 5:11

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