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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.

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  • $\begingroup$ What do you mean "contained by some metric space $(X,d$)?" $\endgroup$
    – Randall
    Commented Sep 30, 2019 at 16:52
  • $\begingroup$ 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? $\endgroup$
    – Mark
    Commented Sep 30, 2019 at 16:55
  • $\begingroup$ 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 $\endgroup$ Commented Sep 30, 2019 at 16:59
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    $\begingroup$ @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). $\endgroup$
    – Randall
    Commented Sep 30, 2019 at 17:00
  • $\begingroup$ 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. $\endgroup$ Commented Sep 30, 2019 at 17:03

3 Answers 3

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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 $(a,b)$.

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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?

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$HINT$

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

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