Sequential characterization and removability for uniform continuity

There is a useful sequential characterization of continuity in metric spaces. Let {f\colon X\to Y} be a map between metric spaces. If for every convergent sequence {x_n\to p} in {X} we have {f(x_n)\to f(p)} in {Y}, then {f} is continuous. And the converse is true as well.

Uniformly continuous functions also have a useful property related to sequences: if {f\colon X\to Y} is uniformly continuous and {\{x_n\}} is a Cauchy sequence in {X}, then {\{f(x_n)\}} is a Cauchy sequence in {Y}. However, this property does not characterize uniform continuity. For example, if {X=Y=\mathbb R}, then Cauchy sequences are the same as convergent sequences, and therefore any continuous function preserves the Cauchy-ness of sequences—it does not have to be uniformly continuous.

Let us say that two sequences {\{x_n\}} and {\{x_n'\}} are equivalent if the distance from {x_n} to {x_n'} tends to zero. The sequential characterization of uniform continuity is: {f\colon X\to Y} is uniformly continuous if and only if for any two equivalent sequences {\{x_n\}} and {\{x_n'\}} in {X}, their images {\{f(x_n)\}} and {\{f(x_n')\}} are equivalent in {Y}. The proof of this claim is straightforward.

In the special case when {\{x_n'\}} is a constant sequence, the sequential characterization of uniform continuity reduces to the sequential characterization of continuity.

A typical example of the use of this characterization is the proof that a continuous function on a compact set is uniformly continuous: pick two equivalent sequences with non-equivalent images, pass to suitable subsequences, get a contradiction with continuity.

Here is a different example. To state it, introduce the notation {N_r(p) = \{x\colon d(x, p)<r\}}.

Removability Theorem. Let {f\colon X\to Y} be continuous. Suppose that there exists {p\in X} such that for every {r>0}, the restriction of {f} to {X\setminus N_r(p)} is uniformly continuous. Then {f} is uniformly continuous on {X}.

This is a removability result because from having a certain property on subsets of {X} we get it on all of {X}. To demonstrate its use, let {X=[0, \infty)} with the standard metric, {p=0}, and {f(x)=\sqrt{x}}. The uniform continuity of {f} on {X\setminus N_r(p)} follows immediately from the derivative {f'} being bounded on that set (so, {f} is Lipschitz continuous there). By the removability theorem, {f} is uniformly continuous on {X}.

Before proving the theorem, let us restate the sequential characterization in an equivalent form (up to passing to subsequences): {f\colon X\to Y} is uniformly continuous if and only if for any two equivalent sequences {\{x_n\}} and {\{x_n'\}} there exist equivalent subsequences {\{f(x_{n_k})\}} and {\{f(x_{n_k}')\}}, with the same choice of indices {n_k} in both.

Proof of the theorem. Suppose {\{x_n\}} and {\{x_n'\}} are equivalent sequences in {X}. If {x_n\to p}, then {x_n'\to p} as well, and the continuity of {f} at {p} implies that both {\{f(x_n)\}} and {\{f(x_n')\}} converge to {p}, hence are equivalent sequences. If {x_n\not\to p}, then by passing to a subsequence we can achieve {d(x_n, p)\ge r } for some constant {r>0}. By the triangle inequality, for sufficiently large {n} we have {d(x_n', p)\ge r/2}. Since {f} is uniformly continuous on {X\setminus N_{r/2}(p)}, it follows that {\{f(x_n)\}} and {\{f(x_n')\}} are equivalent.

Continuous:Lipschitz :: Open:?

A map is continuous if the preimage of every open set is open. If the topology is defined by a metric, we can reformulate this as: the inverse image of an open ball B_R(f(x)) contains an open ball B_r(x). Like this:

Continuous map

But bringing these radii R and r into the picture will not serve any purpose unless we use them to quantify continuity. For example, if we insist that r\ge cR for a fixed constant c>0, we arrive at the definition of a Lipschitz map.

But why do we look at the inverse image; what happens if we take the direct image instead? Then we get the definition of an open map: the image of every open set is open. Recasting this in metric terms: the image of an open ball B_R(x) contains an open ball B_r(f(x)). Like this:

Open map

If we quantify openness by requiring r\ge cR for a fixed c>0, we arrive at the definition of a co-Lipschitz map. [Caution: some people use “co-Lipschitz” to mean |f(a)-f(b)|\ge c|a-b|, which is a different condition. They coincide if f is bijective.]

I don’t know if openness without continuity is good for anything other than torturing students with exercises such as: “Construct an open discontinuous map from \mathbb R to \mathbb R.” We probably want both. At first one can hope that open continuous maps will have reasonable fibers f^{-1}(x): something (m-n)-dimensional when going from m dimensions to n, with m\ge n. The hope is futile: an open continuous map f\colon \mathbb R^2\to\mathbb R^2 can squeeze a line segment to a point (construction left as an exercise).

A map that is both Lipschitz and co-Lipschitz is called a Lipschitz quotient; this is a quantitative analog of “open continuous”. It turns out that for any Lipschitz quotient f\colon \mathbb R^2\to\mathbb R^2 the preimage of every point is a finite set. Moreover, f factors as f=g\circ h where g is a complex polynomial and h is a homeomorphism.

This is encouraging… but going even one dimension higher, it remains unknown whether a Lipschitz quotient f\colon \mathbb R^3\to\mathbb R^3 must have discrete fibers. For an overview of the subject, see Bill Johnson’s slides.