Proof of a compact set without using Heine-Borel

Question

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!

Answer 1

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 $\{{D_{\tfrac{1}{n}}(x)}\bigcap B\}_{n \in \mathbb{N}} \bigcup B \setminus \overline{D_{\frac{1}{2}}(x)}$.

Answer 2

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.)