I was reading the book Elements of the theory of functions and functional analysis (Volume 1) by Kolmogorov. More precisely, the part of compact sets in metric spaces:

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The underlined text is so confusig for me. First, we know that in $\mathbb{R}$ the compact sets are the closed and bounded sets (Heine-Borel), then, can be a set that is only bounded compact? I don't think so. I know that the definition that Kolomorov use is, really, the definition of sequentially compact, and I know that the compactness and the sequentially compactness are equivalent in metric spaces. Then, am I losing something of context in the book?

Moreover, the book says that an arbitrary subset of a compact set is compact, but, is it true? I know that $[0,1]\subseteq\mathbb{R}$ is compact but $\left\{\displaystyle\frac{1}{n}:n\in\mathbb{N} \right\}\subseteq[0,1]$ is not compact. Again, Kolmogorov use some definition that I don't have? Am I misunderstanding? I will be really grateful if someone can explain me this. I really appreciate any help you can provide me.

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    $\begingroup$ It looks to me that what Kolmogorov is calling "compact" is in fact what is now more commonly called "precompact" or "relatively compact", meaning that the closure is compact. $\endgroup$ Dec 29, 2017 at 0:30

1 Answer 1


No, it is not wrong. It is just outdated in terms of terminology used.

The definition given is for what one would nowadays probably call relatively sequentially compact. A set is relatively (sequentially) compact if its closure is (sequentially) compact. The way Kolmogorov defines compact sets, all these examples are correct. Note that under the definition given, the limit of the subsequence need not lie in the set itself. In the next section, Kolmogorov defines a compactum as what we nowadays would call a (sequentially) compact set.


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