## 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).

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:

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:

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.