# Extracting a finite subcover from a countable subcover for a compact set

Everything is in the context of metric spaces.

My definition of compactness is: every sequence has a convergent subsequence.

Let $$B_1, B_2 ...$$ be a countable open cover of a compact set $$A$$. Suppose there is no finite subcover. Then, $$\exists$$ $$a_1 \in A \setminus B_1$$,

$$\exists$$ $$a_2 \in A \setminus (B_1 \cup B_2)$$,

$$\exists$$ $$a_3 \in A \setminus (B_1 \cup B_2 \cup B_3)$$,

$$\exists$$ $$a_n \in A \setminus (B_1 \cup \cdots \cup B_n)$$,

But $$A$$ is compact, so $$\exists$$ a subsequence $$a_{n_k}$$ such that $$\lim \limits_{k \to \infty} a_{n_k} = a \in A$$.

there exists $$n_0$$ such that $$a \in B_{n_0}$$ so all but finitely many elements in the subsequence $$a_{n_k}$$ are inside $$B_{n_0}$$.

I know there are infinitely many $$a_{n_k}$$'s inside $$B_{n_0}$$ and only finitely many $$a_{n_k}$$'s outside $$B_{n_0}$$.

• What's your definition of compactness? It's usually defined to mean that every open cover has a finite subcover. Then the contradiction is already in your second sentence. – joriki Feb 6 at 1:16
• What you write as your definition of compactness is the definition of sequential compactness. The two are not equivalent. They're equivalent for metric spaces. Are you only considering metric spaces? I think your question needs a bit more context. – joriki Feb 6 at 1:20
• done............. – ironX Feb 6 at 1:21

Note that $$a_i \not \in B_j$$ if $$i > j$$. Since $$(a_{n_k})_k$$ is a subsequence, there exists $$k_0$$ such that $$n_{k_0} > n_0$$ and in particular, we have that $$n_l > n_0$$ for all $$l > k_0$$. This is absurd since it shows that infinitely many terms of $$(a_{n_k})_k$$ lie outside of $$B_{n_0}$$.
• Agreed, but I'm not seeing the mistake. Maybe we're using that $A$ is metric in an intermediate step? – qualcuno Feb 6 at 1:33