Given a topological space $X$ and a subset of it $S$, $S$ is compact iff for every open cover of $S$, there is a finite subcover of $S$.

Just curiosity: I've done some search in Internet why compact sets are called compact, but it doesn't contain any good result. For someone with no knowledge of the topology, Facing compactness creates the mentality that a compact set is a compressed set!

Does anyone know or have any information on the question?

  • $\begingroup$ Maybe you can try looking up a dictionary - the definition of compact. I personally like Google's definition: closely and neatly packed together. $\endgroup$ Jan 4, 2014 at 16:21
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    $\begingroup$ Actually, I think "compact" is one of the best names for a concept in mathematics. The word really captures the intuitive feel of this fairly abstract definition quite well. (Really, the only other word I can think of that does so much conceptual legwork is "smooth".) $\endgroup$
    – Jim Belk
    Jan 4, 2014 at 18:53

3 Answers 3


Frechet oiginally coined the term in 1904 to refer to a space where every sequence had a limit point. Thus, the space was 'compact' because there was no room for the sequence to escape (my interpretation).


I like to think of compact spaces as those which admit a "compact" description. So then the usual characterization of compactness can be interpreted along the lines of: whenever you have an infinite amount of information to describe a space, you really only need some finite reduction of that information.


If you are curious about the history of compact sets (the definition of which dates back to Fréchet as mentioned in another response) and you can read French, then I suggest checking out the following historical article:

Pier, J. P. (1980). Historique de la notion de compacité. Historia mathematica, 7(4), 425-443. Retrieved from http://www.sciencedirect.com/science/article/pii/0315086080900063.

I came across this article when posting a response on MO here.


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