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