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I am self-studying functional analysis and have just learned the definition of compact operators. However, it isn't clear to me why the name "compact operator" was chosen.

The operator itself could be considered a one-point subset of a topological vector space of operators, but the "compactness" doesn't seem to refer to any such topology (every operator would constitute a compact subset of such a space, I think).

The definition involves compactness, but only in an indirect way (bounded sets get mapped to relatively compact subsets). Thus it is unclear to me if there is a good reason for the term.

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    $\begingroup$ Well, the reason is precisely that bounded subsets map to relatively compact sets. There isn't any more reason to it. $\endgroup$
    – Wojowu
    Commented Feb 5, 2021 at 21:30
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    $\begingroup$ math.stackexchange.com/questions/3802255/… See the answer to this question. $\endgroup$
    – ferhenk
    Commented Feb 6, 2021 at 0:12
  • $\begingroup$ What's the motivation behind the term "bounded linear operator"? Bounded sets get mapped to bounded sets. I think "compact linear operator" just strengthens "bounded linear operator" by saying that bounded sets get mapped not only to merely bounded sets, but to (pre)compact sets. $\endgroup$
    – gtoques
    Commented Mar 4, 2021 at 22:51

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Well, as my professor says, compact operators are small and Fredholm operators are big since cokernel of a Fredholm operator should be finite dimensional while the image of a compact operator between Banach spaces can not contain an infinite dimensional closed subspace.

There is also a deeper connection between two of those types of operators which is a subject of Riesz--Schauder theory.

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  • $\begingroup$ Image of a compact operator can be infinite-dimensional. Take any integral operator with bounded kernel. $\endgroup$
    – Wojowu
    Commented Feb 5, 2021 at 21:31
  • $\begingroup$ @Wojowu To be more formal, the image of a compact operator between Banach spaces can not contain an infinite dimensional closed subspace. $\endgroup$ Commented Feb 5, 2021 at 21:34
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    $\begingroup$ Then you should edit the answer in order to clarify that. As written it is straight up wrong. $\endgroup$
    – Wojowu
    Commented Feb 5, 2021 at 21:37
  • $\begingroup$ @Wojowu Sure, no problem $\endgroup$ Commented Feb 5, 2021 at 21:41
  • $\begingroup$ I think now that this has been corrected it offers some useful intuition. The answer in the linked comment also has good intuition. Going to leave this question open for a bit to see if others have more input. $\endgroup$
    – WillG
    Commented Feb 6, 2021 at 17:06

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