Spaces with the property: Uniformly continuous equals continuous

Solution 1:

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.

Solution 2:

Such metric spaces are called UC-spaces (Google Scholar, Google Books) or Atsuji spaces. (Google Scholar, Google Books).

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