Prove the set $K = \{0\} \cup \{\frac{1}{n} \in \mathbb{R} : n \in \mathbb{N} \}$ is compact without Heine-Borel

Question

Prove the set $K = \{0\} \cup \{\frac{1}{n} \in \mathbb{R} : n \in \mathbb{N} \}$ is compact without Heine-Borel

I have completed the question and used the same procedure that was done in this version: Prove the set $K = \{\frac{1}{n} \mid n\in \mathbb{N}\}\cup \{0\}$ is compact. .

My question though is why this proof is valid? If we are showing the set is compact it means that there must exist a finite sub-cover for all open covers. Is the all condition captured by generalizing $U$ to represent any open cover thus meaning all open covers?

Answer 1

Note the beginning of the proof:

Let $\mathcal{G} = \{G_\alpha \mid \alpha \in A\}$ be any open cover for $K$

(emphasis mine). Since $\mathcal{G}$ is a completely arbitrary open cover, anything we can prove about $\mathcal{G}$ must in fact be true of every open cover.

This is something we do all the time. E.g. to prove the infinitude of primes:

Let $n$ be any natural number. Let $\{p_1, p_2,..., p_k\}$ be the set of all prime numebrs $<n$. It's easy to check that $p_1\cdot p_2\cdot ...\cdot p_k+1$ is not divisible by any $p_i$ ($i<k$), and hence must be divisible by some prime $>n$. This means for every natural number, there is a larger prime number.

I suspect in the current situation it seems more complicated since the subject matter is more abstract, but it's exactly the same underlying logic.