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Let $(X, \mathcal{T})$ be a space that is both first countable and countably compact (every countable open cover has a finite cover). Show that $X$ is sequentially compact (every sequence has a convergent subsequence.

Related theorems I can use:

(1) X countably compct => X limit point compact

(2) In a first countable space is $x \in cl A \iff \exists (a_n)_n \in A^\mathbb{N}: a_n \to x$

I was able to prove the result if the space is $T_1$, but I don't think we need this hypothesis.

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    $\begingroup$ The point is that in a first countable space, for every accumulation point of a sequence there is a subsequence converging to it. $\endgroup$ – Daniel Fischer Jun 8 '18 at 13:48
  • $\begingroup$ Where do you use $T_1$ in your proof? And your question is: how to get rid of it? $\endgroup$ – Henno Brandsma Jun 8 '18 at 15:59
  • $\begingroup$ Yes, that's my question. I don't have time to write down the proof though. $\endgroup$ – user370967 Jun 8 '18 at 16:04
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Theorem: $X$ is countably compact, then $X$ is strongly limit compact: every countably infinite subset $A$ has an $\omega$-limit point, i.e. a point $x$ such that for every neighbourhood $U$ of $x$ we have $U \cap A$ is infinite.

Proof: suppose not, then every $x \in X$ has a neighbourhood $O_x$ such that $O_x \cap A$ is finite. Now a nice trick: for each finite subset $F$ of $A$ (and there are countably many finite subsets of $A$) define

$$O(F) = \bigcup\{O_x: O_x \cap A = F\}$$

As every $O_x$ is a subset of one of the $O(F)$ (namely that with $F = O_x \cap A$), and the $O_x$ cover $X$, the $O(F)$ form a countable cover for $X$. Hence there is a finite subcover $O(F_1), \ldots, O(F_N)$ but then there is some $a_0 \in A \setminus \bigcup_{i=1}^N F_i$ (as the $F_i$ are finite subsets of $A$) and this $a_0$ is not covered by any of the $O(F_i)$ for $i \le N$, and this is a contradiction. So $A$ does have an $\omega$-limit point.

BTW the reverse also holds, as you can see here, from which I also borrowed the above argument.

But now the sequential compactness can be proved: let $(x_n)$ be a sequence in $X$. If $A = \{x_n : n \in \mathbb{N} \}$ is finite, there is a constant (hence convergent) subsequence. So we can assume $A$ is infinite. So $A$ has an $\omega$-limit point $p \in X$, as we saw above. Let $U_n, n \in \mathbb{N}$ be a countable local base at $p$. Then pick $n_1$ with $x_{n_1} \in U_1 \cap A$. Then having chosen all $x_{n_k} \in (U_1 \cap \ldots U_k) \cap A$, for some $k \ge 1$, where we also have $n_1 < n_2 <\ldots < n_k$, we note that $\cap_{i=1}^{k+1} U_i$ is an open neighbourhood of $p$, so contains infinitely many points of $A$, so in particular we can pick $n_{k+1} > n_k$ such that $x_{n_{k+1}} \in \cap_{i=1}^{k+1} U_i$. Continue this recursion.

It's now standard to check that $x_{n_k} \to p$ is a convergent subsequence of $(x_n)$.

Because I used $\omega$-limit points, there was no need for $T_1$-ness (which you can have with just limit points).

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Let $(x_n)_{n\in\mathbb N}$ be a sequence of elements of $X$. There are two possibilities:

  1. There is some $x\in X$ such that $x_n=x$ for infinitely many $n$'s. Then it is obvious that $x$ is a limit of a subsequence of the sequence $(x_n)_{n\in\mathbb N}$.
  2. For every $x\in X$, the equality $x_n=x$ holds for finitely many $n$'s only. In that case, the set $\{x_n\,|\,n\in\mathbb N\}$ is infinite. Since $X$ is limit point compact, there is some $x\in X$ which is a limit of a sequence of elements of $\{x_n\,|\,n\in\mathbb N\}$. Therefore, $x$ is a limit of some subsequence of the sequence $\{x_n\,|\,n\in\mathbb N\}$.
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  • $\begingroup$ I don't see how x is a limit of some subsequence of the sequence $(x_n)$. Why is it true that if you make a sequence $(y_n)$ converging to $x$ with elements of $\{x_n \mid n\}$, then there is a subsequence of $(x_n)$ that converges to $x$? For example, what if $(y_n)$ contains finitely many elements? $\endgroup$ – user370967 Jun 8 '18 at 14:26
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    $\begingroup$ Let $(U_n)_{n\in\mathbb N}$ be a countable set of neigbourhoods of $x$ such that every neigbourhood of $x$ contains one of them. Let $n_1\in\mathbb N$ be such that $x_{n_1}\in U_{n_1}$, let $n_2\in\mathbb N$ be such that $x_{n_2}\in U_{n_1}\cap U_{n_2}$, let $n_3\in\mathbb N$ be such that $x_{n_3}\in U_{n_1}\cap U_{n_2}\cap U_{n_3}$, and so on. Them $x$ is limit of the sequence $(x_{n_k})_{k\in\mathbb N}$. $\endgroup$ – José Carlos Santos Jun 8 '18 at 14:35
  • $\begingroup$ What do you mean with "one of them"? $\endgroup$ – user370967 Jun 8 '18 at 14:45
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    $\begingroup$ Every neigbourhood of $x$ contains some $U_n$. $\endgroup$ – José Carlos Santos Jun 8 '18 at 14:58
  • $\begingroup$ Thanks. That makes a lot of sense. Here again you used that the space is first countable right? $\endgroup$ – user370967 Jun 8 '18 at 15:04

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