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I have already read A first countable, countably compact space is sequentially compact and wish not use limit point compactness in my proof. Please do not reference.

Let $X$ be first countable and suppose also that $X$ is countably compact. We need to show that every sequence in $X$ has a convergent subsequence. We know

1.) Since $X$ is first countable, every point has a countable local base

2.) Since $X$ is countably compact, every countable cover has a finite subcover

Suppose $X$ is first countable and countably compact. Let $\mathscr{U} = \{ U_i: i =1, 2,\dots, \}$ be a countable open cover of $X$. By hypothesis, there is a finite subcover $\mathscr{U}_0 = \{U_i: i =1, 2, \dots, k\}$. Let $\langle x_i\rangle_{i\in \mathbb{N}}$ be a sequence in $X$ satisfying $x_i \in U_i$ for every $i$.

I have this intuition that once we hit $x_k \in U_k$ we get a sort of boundedness for our sequence since the space is finally covered. But how do I utliize first countability here? What's this proof missing?

Grigor Hakobyan
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1 Answers1

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Fix a sequence $\left< x_n \right>_{n \in \mathbb{N}}$ in $X$ and for $n \in \mathbb{N}$ let

$$F_n = \operatorname{\text{cl}} \{ x_k : k > n \}.$$

We will show that the set $F := \bigcap_{n=0}^{\infty} F_n$ is non-empty. If it were empty, the family $\{ X \setminus F_n : n \in \mathbb{N} \}$ would be a countable open cover of $X$. By assumption we could find a finite subcover $\{ X \setminus F_{n_1}, \ldots, X \setminus F_{n_k} \}$. But since $X \setminus F_i \subseteq X \setminus F_j$ when $i \leqslant j$, one of the sets in the finite subcover contains all the others (namely: the one with the biggest index). It follows that $X \setminus F_n = X$ for some $n \in \mathbb{N}$, which means that $F_n = \varnothing$ which clearly is a contradiction.

Take any $a \in F$. We will show that $\left< x_n \right>_{n \in \mathbb{N}}$ has a subsequence convergent to $a$. Take a countable local base $\{ U_n : n \in \mathbb{N} \}$ at $a$. By a standard intersection argument we can assume that $U_0 = X$ and $U_i \supseteq U_j$ when $i \leqslant j$. We inductively define an increasing sequence $n_0, n_1, n_2, \ldots$ of natural numbers such that $x_{n_j} \in U_j$ as follows. Let $n_0 = 0$. Now assume we have already defined $n_0, n_1, \ldots, n_{j-1}$. Since $a \in F_{n_{j-1}}$ and $a \in U_j$, we can find $n > n_{j-1}$ with $x_n \in U_j$. Letting $n_j := n$ ends the inductive construction, which clearly has the declared properties.

Now it is routine to check that the subsequence $\left< x_{n_j} \right>_{j \in \mathbb{N}}$ converges to $a$.

Adayah
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