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

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.

• What do you mean "contained by some metric space $(X,d$)?" Commented Sep 30, 2019 at 16:52
• If we are talking about the usual metric space in $\mathbb{R}$ then you can cover $(a,b)$ with $\cup_{n=1}^\infty (a+\frac{1}{n},b)$. Is there a chance to find a finite subcover?
– Mark
Commented Sep 30, 2019 at 16:55
• In Rudin's book he denotes a metric space by (X, d), where d is the distance metric. So I was just saying we're talking about being inside a metric space. Sorry if that was confusing I'm still learning how to ask questions about Real Analysis. @Randall Commented Sep 30, 2019 at 16:59
• @solidstatejake The answer would depend on which metric space. We assume you mean the ordinary metric on $\mathbb{R}$, but the question could feasibly make sense in another universe (with a different answer). Commented Sep 30, 2019 at 17:00
• Okay, thank you Randall, I will further specify the "ordinary" metric, like the one in R, if that's what I mean, from now on. Commented Sep 30, 2019 at 17:03

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 $$(a,b)$$.

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?

$$HINT$$

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