Spaces with the property: Uniformly continuous equals continuous

I found a nice book about functional analysis with a nice theorem in it: Continuity at 0 is equal to Lipschitz continuous for linear maps in normed spaces.

This fact inspires me to ask: Are there spaces with the property: uniformly continuous is the same as continuous (everywhere)?

My problem is to find (metric) spaces with this property, so that I'm able to find uniform structure on topological spaces.

Is there a name for this theory?

• Compact (at least Hausdorff, not sure about quasicompact non-Hausdorff spaces) spaces have this property. – Daniel Fischer Aug 7 '13 at 8:45
• @DanielFischer It's difficult to define uniform continuity without some notion of distance (so that "$\varepsilon$ away from $x$" makes sense for non-fixed $x$), which leads me to say you'd need a metric space. Those are always Hausdorff. But yeah, if you have a metric space where continuous $\Leftrightarrow$ uniformly continuous for all functions, then I believe it's compact. – Arthur Aug 7 '13 at 8:49
• @Arthur Uniform continuity can be defined for uniform spaces. Every compact (Hausdorff) spaces carries precisely one uniform structure that induces the topology. I don't know off-hand whether a non-Hausdorff quasicompact space is always uniformizable, and if it is, whether it also always has the property that all comtinuous maps are uniformly continuous. – Daniel Fischer Aug 7 '13 at 8:54

For a metric space $X$, the following are equivalent:

1. Every continuous map from $X$ to another metric space is uniformly continuous.
2. $X$ is complete and almost totally bounded.

Definition. $X$ is almost totally bounded if for every $r>0$ there is a finite set $\{x_1,\dots,x_n\}\in X$ and a number $\delta>0$ such that for any distinct points $a,b\notin \bigcup_{i=1}^n B(x_i,r)$ we have $d(a,b)\ge \delta$. (In other words, $X\setminus \bigcup_{i=1}^n B(x_i,r)$ is uniformly separated.) Here $B(x,r)=\{y:d(x,y)\le r\}$.

For example, the union of $\mathbb Z$ and of any compact subset of $\mathbb R$ satisfies 2. Another example is the set $\{0\}\cup \{k^{-1}e_n:k,n=1,2,\dots\}\in \ell^2$ where $\{e_n\}$ is an orthonormal basis of $\ell^2$. This example is interesting because the set cannot be written as a union of a compact and uniformly separated sets.

Proof. Suppose 2 holds but 1 fails. Let $f:X\to Y$ be a continuous map that is not uniformly continuous. Then there is $\epsilon>0$ and two sequences $p_n,q_n$ in $X$ such that $d_X(p_n,q_n)\to 0$ but $d_Y(p_n,q_n)\ge \epsilon$ for all $n$. Let $B(x_i,r)$ be as in the definition of almost totally bounded. Since $d_X(p_n,q_n)\to 0$, for all sufficiently large $n$ we have either $p_n$ or $q_n$ contained in $\bigcup_{i=1}^n B(x_i,r)$. Therefore, there is a ball $B(x_j,r)$ that contains, say, $p_n$ for infinitely many values of $n$. Consequently, $B(x_j,2r)$ contains both $p_n$ and $q_n$ for infinitely many values of $n$.

Since $B(x_j,2r)$ is also complete and almost totally bounded, we can repeat the above with a smaller value of $r$, and with $X$ replaced by $B(x_j,2r)$. The result is a nested sequence of shrinking closed balls $B_k$, each of which contains $p_n$ and $q_n$ for infinitely many values of $n$. By completeness, $\bigcap B_k$ contains a point $x$. The continuity of $f$ at $x$ contradicts the assumption that $d_Y(p_n,q_n)\ge \epsilon$ for all $n$.

Now suppose 2 fails. This can happen in two ways. (a) $X$ is not complete. Consider it as a subset of its completion $\overline{X}$. Pick $a\in \overline{X}\setminus X$. The function $f(x)=1/d_{\overline X}(x,a)$ is continuous on $X$, but not uniformly continuous. Indeed, there is a Cauchy sequence $x_n$ in $X$ such that $f(x_n)\to\infty$.

(b) $X$ is not almost totally bounded. Then there exists $r>0$ for which $X$ contains an infinite sequence of points $p_n$ such that $d(p_n,p_m)>r$ whenever $n\ne m$, and additionally $\operatorname{dist}(p_n,X\setminus \{p_n\})\to 0$. The latter condition allows us to choose $q_n\ne p_n$ so that $d(p_n,q_n)\to 0$. Define $$f_n(x) = \max\left(0, \;1- \frac{d(x,p_n)}{d(p_n,q_n)} \right),\quad n=1,2,\dots$$
and observe that $f_n(p_n)=1$, $f_n(q_n)=0$, and for large enough $n$ the supports of $f_n$ are disjoint. Thus, the function $\sum_n f_n$ is continuous on $X$, but by its construction it is not uniformly continuous.

Quote from G. Beer: Topologies on Closed and Closed Convex Sets, p.54:

2.3.1 Theorem. Let $\langle X,d \rangle$ be a metric space. The following are equivalent:
(1) Each continuous function on $\langle X,d\rangle$ with values in an arbitrary metric space $\langle Y.d'\rangle$ is uniformly continuous;
(2) Each real-valued continuous function on $\langle X,d\rangle$ is uniformly continuous;
(3) Whenever $A$ and $B$ are disjoint elements of $\operatorname{CL}(X)$, then there exits $\varepsilon>0$ such that $S_\varepsilon[A]\cap S_\varepsilon[B]=\emptyset$;
(4) The set $X'$ of accumulation points of $X$ is compact, and $\forall \varepsilon>0$ such that $S_\varepsilon[X']^c$ is uniformly discrete: $\exists\delta>0$ such that if $x\ne w$ and $\{x,w\}\subset S_\varepsilon[X']^c$ then $\delta>0$;
(5) Whenever $\langle x_n\rangle$ is a sequence in $X$ with $\lim_{n\to\infty} d(x_n,\{x_n\}^c)=0$, then $\langle x_n\rangle$ has a cluster point:
(6) Each open cover of X has a Lebesgue number: there exist a number $\lambda > 0$ such that each subset of $X$ of diameter at most $\lambda$ lies entirely in one member of the cover.

Here $\operatorname{CL}(X)$ denotes the set of all non-empty closed subsets of $X$ and $S_\varepsilon[A]=\{x\in X \colon d(x,A)<\varepsilon\}$.

Some authors also use name Lebesgue spaces, see the comments here.

Another related question is: Inverse of Heine–Cantor theorem