Intuition for why $(a, b)$ is not compact, but $[a, b]$ is compact, where $[a, b]$ is contained by some metric space $(X, d)$.

Question

I understand why the closed interval $[a, b]$ is compact, but am having a hard time understanding why $(a, b)$ is not compact. Any help would be appreciated.

Answer 1

The set $(a,b)$ is not compact because the set$$\left\{\left(a+\frac1n,b-\frac1n\right)\,\middle|\,n\in\mathbb N\right\}$$is an open cover of $(a,b)$ (since $(a,b)=\bigcup_{n\in\mathbb N}\left(a+\frac1n,b-\frac1n\right)$) with no finite subcover.

Or you can say that it is not compact because the sequence $\left(a+\frac1n\right)_{n\in\mathbb N}$ has no subsequence which converges to an element of $(a,b)$.

Answer 2

The open interval is not compact. Consider the open cover $\left(a+\frac{1}{n},b-\frac{1}{n}\right)$ for $n$ large enough. How to find the finite subcover?

Answer 3

$HINT$

Take the covering $\{(a+\frac{1}{n},b-\frac{1}{n}):n \in \Bbb{N}\}$ of $(a,b)$ and prove that there is not a finite subcovering.

Or use sequential compactness.

Take the sequence $a_n=a+\frac{b-a}{2n} \in (a,b)$