I know that "compact intervals" are closed and bounded intervals and that "compact sets" are sets with every infinite sequence of points having an infinite subsequence that converges to some point of the set

But are Compact Intervals also Compact Sets?

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    $\begingroup$ Your definition of compact set is actually the definition of a sequentially compact set. It just happens that a metric space is sequentially compact if and only if it is compact, so in metric spaces the two concepts coincide. In general they do not: there are topological spaces that are compact but not sequentially compact, and topological spaces that are sequentially compact but not compact. $\endgroup$ Nov 26, 2016 at 19:39

2 Answers 2


Yes! A compact interval is just a connected compact set. In $\mathbb R$ the two characterizations for compactness are the same. This it the Heine Borel theorem, which states that every compact set is closed and bounded.

For clarity's sake there are three definitions of compactness that are worth mentioning here, and all are (nontrivially) equivalent for $\mathbb R$:

Sequential Compactness: A space $X$ is said to be sequentially compact if every infinite sequence $(a_n) \subseteq X$ has a convergent subsequence.

Compactness: A space $X$ is said to be compact if every open cover admits a finite subcover. In other words, if $X \subseteq \bigcup_{\alpha \in A} U_{\alpha}$ where each $U_{\alpha}$ is open, then there exists a finite collection from this family so that $$X \subseteq \bigcup_{i=1}^{N} U_{i}.$$

Closed and Bounded: this is standard.

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    $\begingroup$ +1. For the OP, it's worth pointing out that it is not the case that every compact subset of $\mathbb{R}$ is a union of finitely many compact intervals - e.g. consider the Cantor set. However, every compact subset of $\mathbb{R}$ is a union of possibly infinitely many compact intervals (maybe of the form $[a, a]$). This is a good exercise: show that if $x\in C$ is an element of a compact set, there is some compact interval $[a, b]$ with $x\in[a, b]\subseteq C$ such that no larger compact interval containing $x$ is a subset of $C$. $\endgroup$ Nov 26, 2016 at 20:17
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    $\begingroup$ Yes, this is true as well. I think I will edit my post to contain also the standard definition for compactness. $\endgroup$ Nov 26, 2016 at 20:20

You might also look at the theorem of Bernard Bolzano/Karl Weierstraß that every bounded sequence in $\Bbb R^n$ has a convergent sub-sequence, which is the same as to say that all bounded sets in $\Bbb R^n$ are (sequentially) pre-compact.


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