# Proof of a compact set without using Heine-Borel

I'm struggling with a question about compact set, but I can't figure out it on my mind.

Is this set compact or not? Prove it without using Heine-Borel Theorem.

a)$$\mathbb{R}^n$$ such that $$B$$ doesn't contain at least one of its accumulation point.

I know that it's a closed set because a $$\mathbb{R}^n$$ subset is closed if and only if contains all accumulation point, then the question's subset is obviously open. But, without Heine-Borel I just can't think of an example that a open cover doesn't have a finite subcover to prove it. Could someone help me? Thanks!

Let $$x$$ be the missing accumulation point in $$B$$. Let $$D_{\tfrac{1}{n}}(x)$$ denote the open ball of radius $$\tfrac{1}{n}$$ around $$x$$, where $$n \in \mathbb{N}$$. Take the open cover $$\{{D_{\tfrac{1}{n}}(x)}\bigcap B\}_{n \in \mathbb{N}} \bigcup B \setminus \overline{D_{\frac{1}{2}}(x)}$$.
It's true in any metric space, without using Heine-Borel (which is specific to $$\mathbb{R}^n$$ in the Euclidean metric), that a compact set contains all its limit points. So a $$B$$ that does not satisfy it, cannot be compact.
Heine-Borel is the opposite: if $$B$$ is closed and (Euclidean)-bounded, $$B$$ is compact;it doesn't help you to show non-compactness of a set.
One other way to see this is that if $$p \notin B$$ is a limit point of $$B$$, the $$x \to \frac{1}{d(x,B)}$$ is a continuous well-defined function on $$B$$ that has no maximum. (Continuous functions on a compact set are bounded.)