Show that for a T1 space X, if X is limit point compact then X is countably compact.

general-topology

I was hoping someone could review my proof for correctness. Thanks in advance!

Let $X$ be a $T_1$ space.
Let $X$ be limit point compact.
Suppose $X$ is not countably compact.

Then take the countable open covering without a finite subcover : $\mathcal{U} = \{ U_n \mid n \in Z_+ \}$

Then since no finite subcollection of $\mathcal{U}$ can cover $X$ we can define a sequence $(x_n)$, that obeys:

$$\forall n \in Z_+: x_{n+1} \notin U_1, U_2, \ldots U_n$$

Again, each $x_n$ exists since if we were unable to choose such an $x_n$ then we have reached a finite subcover $U_1, \ldots , U_n$. And by hypothesis, no finite subcover can be found for our open covering $\mathcal{U}$

Now $B = \{x_n: n \in Z_+\}$ is an infinite set. Therefore $B$ has a limit point, call it $z$.

Since $\mathcal{U}$ is an open covering, some $U_n$ must contain $z$.

(1) Since $X$ is a $T_1$ space and $z$ is a limit point of $B$, $U_n$ must intersect $B$ in an infinite number of points.

But for any $m > n$ we have $x_m$ isn't a member of $U_n$ by construction of the sequence $(x_n)$. So then we are left with $U_n$ only able to intersect B in a finite number of points — namely at most n points. A contradiction since $X$ is a $T_1$ space and $U_n$ must intersect $B$ in infinitely many points.

Therefore, $X$ must be countably compact.

Best Answer

This proof is fine (after the formatting I did); it's a bit easier to assume that $U_n \subseteq U_{n+1}$ for all $n$, which can easily be achieved WLOG, using the cover $V_n = \bigcup_{i=1}^n U_i$ instead: if the $U_n$ have no finite subcover, the same holds for the increasing open cover $V_n$; this also guarantees that $x_n \neq x_m$ for $n \neq m$, making $B$ infinite trivially.

Note that this proof shows that in any space $X$ (not necessarily $T_1$)

$X$ is countably compact iff every (countably) infinite subset $A$ of $X$ has an $\omega$-limit point.

where $x$ is a strong limit point of $A$ when every open neighbourhood of $x$ contains infinitely many (hence the $\omega$) points of $A$. The proof of the forward direction is not too hard, but nice, see the proof I gave here, e.g.

The latter characterisation via limit points can also be done for "plain" compactness:

$X$ is compact iff every infinite subset $A$ of $X$ has a point $p$ such that for every neighbourhood $O$ of $p$: $|O \cap A| = |A|$ (where the bars denote cardinality); such a $p$ is called a point of total condensation for $A$, it's a special kind of limit point.

(it's a somewhat old-fashioned characterisation that Tychonoff used to show his theorem on the product of compact spaces being compact).

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