Proving that sequentially compact spaces are compact.

Question

I remember seeing this proof somewhere (perhaps here, but I don't remember where) that goes something like this.

Suppose $X$ is sequentially compact, and by contradiction suppose $\{U_n\}$ is a countable open cover with no finite subcover. Then for any positive integer $n$, the set $\{U_i : i \le n\}$ is not an open cover, so there exists $x_n \notin \bigcup_{i \le n} U_i$. Hence, we obtain sequence, and by sequential compactness, there exists a subsequence $x_{n_j}$ that converges to $a \in X$. However, $ a \in U_k$ for some positive integer $k$ and by construction, $x_{n_j} \notin U_k$ if $n_j \ge k$. This is a contradiction.

Doesn't this only prove every countable open cover must have a finite subcover?

Answer 1

For metric spaces, here is a proof of the equivalence.

Let $ (X,d) $ be a metric space, and $ A \subseteq X $.

Thm: $ A $ is sequentially compact if and only if every open cover of $ A $ has a finite subcover.
Pf: $ \underline{\boldsymbol{\implies}} $ Say $ A $ is sequentially compact, and $ \{ U_i \}_{i\in I} $ is an open cover of $ A $.
Firstly there is a $ \delta > 0 $ such that any ball $ B(x, \delta) $ with $ x \in A $ is contained in some $ U_i $.

Suppose not. Then for every integer $ n > 0 $ there is an $ x_n \in A $ such that $ B(x_n , \frac{1}{n} ) $ isnt fully contained in any $ U_i $. Pass to a subsequence $ (x_j)_{j \in J} $, $ J \subseteq \mathbb{Z}_{>0} $ which converges to a point $ p \in A $ as $ j \in J, j \to \infty$.
Now $ p $ is contained in some $ U_k $, and there is an $ \epsilon > 0 $ such that $ B(p, \epsilon) \subseteq U_k $. Picking a $ j \in J $ such that $ \{ d(x_j, p) < \frac{\epsilon}{2} ; \frac{1}{j} < \frac{\epsilon}{2} \}$, we see $ B(x_j, \frac{1}{j}) \subseteq B(p, \epsilon) \subseteq U_k $, a contradiction.

There are finitely many open balls, with radius $ \delta $ and centres in $ A $, whose union contains $ A $.

[Corrected] Suppose not. Pick $ x_1 \in A $. As $ B(x_1, \delta) $ doesnt cover $ A $, pick $ x_2 \in A \setminus B(x_1, \delta) $. As $ B(x_1, \delta) \cup B(x_2, \delta) $ doesnt cover $ A $, let $ x_3 \in A \setminus (B(x_1, \delta) \cup B(x_2, \delta)) $, and so on. Pass to a subsequence $ (x_j)_{j \in J} $, $ J \subseteq \mathbb{Z}_{>0} $ convergent to a point $ p \in A $ as $ j \in J, j \to \infty $.
But ${ d(x _{j}, x _{j'}) \geq \delta }$ for all distinct ${ j, j' }$ in ${ J }.$ Especially ${ (x _j) _{j \in J} }$ isnt Cauchy, a contradiction.

So there are finitely many balls $ B(x_1, \delta), \ldots, B(x_n, \delta) $ with centres in $ A $, whose union contains $ A $. Also each of these balls is contained in some $ U_i $, giving a finite subcover.
$ \underline{\boldsymbol{\impliedby}}$ Say every open cover of $ A $ has a finite subcover, and let $ (x_n) $ be a sequence in $ A $.
Seq $ (x_n) $ has an accumulation point in $ A $.

Suppose not. So for every $ y \in A $ there is a $ \delta_y > 0 $ such that $ x_n \in B(y, \delta_y) $ for only finitely many $ n $. Now open cover $ \{ B(y, \delta_y) : y \in A \} $ of $ A $ has a finite subcover $ \{ B(y_1, \delta_{y_1}), \ldots, B(y_k, \delta_{y_k}) \} $. But there are only finitely many indices $ n $ such that $ x_n \in B(y_1, \delta_{y_1}) \cup \ldots \cup B(y_k, \delta_{y_k}) $, a contradiction.

Answer 2

Maybe you are unclear for the definitions of the compactness and countably compactness.

1. compactness = for any open cover, there is a finite open subcover which covers the whole space.
2. countably compactness = for any countable open cover, there is a finite open subcover which covers the whole space.

By your proof we only see that every sequentially compact space is countably compact, which you can see the Theorem 3.10.30 of the Engelking's book:) However not every sequentially compact space is compact.

---

For example, the topological space $\omega_1$ with the order topology is a sequentially compact space but not a compact space. In the first countable space (in fact, it's only need sequential space), sequentially compace = countably compact. As we know, the space is a first countable space and countably compact, therefore, it is a sequentially compact. But, it is not a compact space:)

---

By your method ( we use it much as a topologist ), the sequence your obtained is a closed discrete subspace. It hasn't a cluster point:) So it's a contradiction with sequential compactness.

Answer 3

In addition to the condition that metrizability ensures the equivalence between sequential compactness and compactness, there is another sufficient condition: the space being second countable and Hausdorff. It is important to note that metrizability is equivalent to the combination of paracompactness, the Hausdorff property, and local metrizability. In other words, if we eliminate local metrizability and strengthen paracompactness to second countability, the equivalence between sequential compactness and compactness is still preserved.

To prove, let a countable basis for the topology be denoted by $\{V_i\}_{i \in \mathbb{N}}$, and let $\{W_\alpha\}_{\alpha \in J}$ be an open cover of the space $X$. For each point $x \in X$, there exist corresponding elements $V_{i_x}$ from the basis and $W_{\alpha_x}$ from the open cover such that $x \in V_{i_x} \subseteq W_{\alpha_x}$. Clearly, the set $\{V_{i_x}\}_{x \in X}$ forms a countable open cover of $X$ that refines the original open cover.