A characterization of continuity

Yesterday, Randall Holmes mentioned to me the following nice characterization of continuity for functions between Euclidean spaces.

Theorem. A function f:{mathbb R}^nto{mathbb R}^m is continuous iff it preserves path-connectedness and compactness.

This is an easy exercise, but I don’t remember having seen the characterization before, so I figured I could as well write down the argument I found. It is clear that the result holds for a much wider class of spaces than the {mathbb R}^n, but to keep this post simple, I’ll just leave it this way.

Proof. For Euclidean spaces, continuity and sequential continuity coincide. Towards a contradiction, assume x_i converges to x but f(x_i) does not converge to f(x).

  • Case 1. The range of {f(x_i): ige 0} is infinite.

This quickly leads to a contradiction since {f(x_i):ige 0}cup{f(x)} is a compact set: We may as well assume that the map imapsto f(x_i) is injective, and since f(x_i) does not converge to f(x) we may in fact assume that all the f(x_i) stay away from f(x). The set {f(x_i):ige 0} has an accumulation point, which cannot be f(x) so it must be f(x_m) for some m. But then the set {f(x_i):i>m}cup{f(x)} is both compact and lacks one of its accumulation points, contradiction.

  • Case 2. The range is finite.

We may as well assume all x_i have the same image. Fix paths A_i=[x_i,x_{i+1}] in {mathbb R}^n that we can combine to get a path bigcup_i A_icup{x} (for example, A_i could simply be a segment). By preservation of path connectedness, any A_i with range of size at least 2 in fact has range of size continuum. If infinitely many of the A_i have infinite range, one easily reduces to Case 1. So we may assume all the A_i have constant range, but then bigcup_i A_icup{x} has a disconnected image. {sf QED}

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