Do compact sets always contain a convergent sequence?

Question

I was looking at a proof for the Heine-Boral theorem which goes:

1st. Suppose $ A \in \mathbb{R} $ is closed and bounded

therefore, any subsequence $ a_n{_k} \rightarrow L $ (bolzano weierstrass theorem)

For $ a_n \in A $ and we know A is closed, which implies $ L \in A $ which implies A is compact.

2nd. Suppose $ A \in \mathbb{R} $ is compact

Let $ a_n \in A$ be a sequence s.t. $ a_n \rightarrow L $ ***

we know $ \exists\ a_n{_k} \ of \ a_n : a_n{_k} \rightarrow M \in A $ (As A is compact) By uniqueness of limits this implies L = M and therefore $ L \in A$.

(Then we go on to prove compactness implies boundedness)

My issue is with the line ***, when we just assume that there is a sequence in this set that converges. I was wondering if there were any theorems or lemmas that say something along the lines of, we can always find a sequence in a compact set that converges? I am just worried because I know we cannot use the fact the set is closed/bounded yet because that is what we are trying to prove.

I am just very confused about this assumption, any help would be greatly appreciated!

Answer 1

To prove that a set $C$ is close you assume that a sequence $\{c_n\}$ in $C$ converges to a point $C$ and prove that $c \in C$. Of course there is always a constant seqeuence in any non-empty set that is convergent but that is irrelevant.

Answer 2

In fact, your doubt is a logic question. You are trying to prove an implication

$$P\implies Q.$$

An implication is false only when $P$ is true and $Q$ false. Consequence: is irrelevant if there is some sequence $a_n\in A$ s.t. $a_n\to L$. You only want that any hypothetical sequence verifies that $L\in A$.

In fact, when $P$ is false, $P\implies Q$ is true. In your example this happens when $A = \emptyset$.