Compact sets in Banach spaces

In a Euclidean space, a set is compact if and only if it is closed and bounded. This fails in all infinite-dimensional Banach spaces (and in particular in Hilbert spaces) where the closed unit ball is not compact. However, one still has a simple description of compact sets:

A subset of a Banach space is compact if and only if it is closed, bounded, and flat.

By definition, a set is flat if for every positive number r it is contained in the r-neighborhood of some finite-dimensional linear subspace.


  • The r-neighborhood of a set consists of all points whose distance to the set is less than r.
  • In a finite-dimensional subspace every subset is vacuously flat.

Necessity: Suppose K is a compact set. Every compact set is closed and bounded, this is true in all metric spaces. Given a positive number r, let F be a finite set such that K is contained in the r-neighborhood of F; the existence of such F follows by covering K with r-neighborhoods of points and choosing a finite subcover. Then the linear subspace spanned by F is finite-dimensional and demonstrates that K is flat.

Sufficiency: to prove K is compact, we must show it’s complete and totally bounded. Completeness follows from being a closed subset of a complete space, so the issue is total boundedness. Given r > 0, let M be a finite-dimensional subspace such that K is contained in the (r/2)-neighborhood of M. For each point of K, pick a point of M at distance less than r/2 from it. Let E be the set of all such points in M. Since K is bounded, so is E. Being a bounded subset of a finite-dimensional linear space, E is totally bounded. Thus, there exists a finite set F such that E is contained in the (r/2)-neighborhood of F. Consequently, K is contained in the r-neighborhood of F, which shows its total boundedness.

It’s worth noting that the equivalence of compactness with “flatness” (existence of finite-dimensional approximations) breaks down for linear operators in Banach spaces. While in a Hilbert space an operator is compact if and only if it is the norm-limit of finite-rank operators, some Banach spaces admit compact operators without a finite-rank approximation; that is, they lack the Approximation Property.



2 thoughts on “Compact sets in Banach spaces”

  1. Nice observation – I was thinking to myself “is there a way to characterize compactness in Hilbert spaces using something about finite dimensional subspaces?” And I found your post.

    Is “flat” the standard language here? I feel a little weird about it, because the whole space is flat (geometrically, i.e. there’s no curvature). I’d prefer something like “almost finite dimensional.”

    1. “Flat” is just the word I happened to use in this post; I do not know if there is a standard one. “Almost finite dimensional’ sounds good.

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