Proof verification of compactness

Question

Let $K$ be the set $\{0\} \cup \{1/n : n \text{ is an element of the positive integers}\} $ Prove that $K$ is compact.

In my head, it seems that what they are asking in this question to prove is that every cover of the set $K$ can be broken down into finitely many pieces. For instance if I give the cover $(-2,2)$, that can be broken down to 1 piece(namely $(-2,2)$ or also broken down to $(-1, 1.5)$ getting rid of the excess I don't need. Is this a correct interpretation?

Let $C$ be any open cover the the set above. I prove that whatever cover you give me can be broken up into finitely many open sets whose union form an open cover:

Given that the set $C$ is a covering of the set $K$, then that means it contains all the elements of $K$. Even if you try to cover $K$ in an infinite way, say $(-\infty, \infty)$ then that set contains some neighborhood around zero for some $\epsilon$ that covers an infinite amount of points between 0 and that $\epsilon$. This will leave only a finite number of elements left, no matter what $\epsilon$ you pick.

Is my logic sound?

Answer 1

The main argument is there. The key is, as you say, there is (at least) one set in the cover that contains $0$ and since that set is open it will contain some neighboorhood of every point in it, thus in particular of $0$. So, that set contains all but finitely many elements of $K$. And so on. Just, you express it strangely.

Here is a start of an argument:

Let $\mathcal{O}$ be an open cover of $K$, that is $\mathcal{O}$ is a collection of open sets $C$ such that $$K \subset \bigcup_{C \in \mathcal{O}} C.$$

Since $0 \in K \subset \bigcup_{C \in \mathcal{O}} C$ there exists some $C_0 \in \mathcal{C}$ with $0 \in C_0$. Since $C_0$ is open, there exists some $\epsilon > 0$ such that $B_{\epsilon}(0) \subset C_0$.

I leave the rest to you.

Answer 2

It seems correct to me. Compactness of a set means that given $any$ open cover $\exists$ a finite subcover for that set, which is being guranteed because of $0$. Note that this question can be generalised in the following way : Given a sequence $(a_n)$ in a metric space, the set $\{a\}$ $\cup$ $\{a_n |n\in \mathbb{Z}\}$ is compact, where $(a_n)\rightarrow a$.