So I was working on a homework problem that boiled down to proving that a continuous map between compact spaces is uniformly continuous. Now I have only dealt with uniform continuity in metric spaces and proved the problem using the assumption that my spaces were metric spaces. I began to wonder if this is a safe assumption to make.

Continuity and compactness are defined on non-metrizable spaces, but is uniform continuity strictly in the realm of metric spaces? Is there some notion of 'uniform continuity' in non-metrizable spaces?

  • $\begingroup$ Uniform continuity can be defined for maps between uniform spaces. Every compact (quasicompact and Hausdorff) space carries a unique uniform structure inducing the topology, and every continuous map from a compact space to a uniform space is uniformly continuous. $\endgroup$ – Daniel Fischer Oct 13 '15 at 22:46
  • $\begingroup$ @DanielFischer Why don't you make your comment an answer, with a link to the wikipedia entry for uniform spces? $\endgroup$ – J.-E. Pin Oct 13 '15 at 23:03
  • $\begingroup$ @J.-E.Pin Since nobody posted an answer while I was sleeping, I now expanded my comment into an answer. $\endgroup$ – Daniel Fischer Oct 14 '15 at 10:59

For a map $f\colon X\to Y$ where $X,Y$ are topological spaces, we can - informally - characterise continuity at a point $x_0 \in X$ by requiring that "for any prescribed degree $d_Y$ of closeness to $f(x_0)$, we can find a degree $d_X$ of closeness to $x_0$ such that any point $d_X$-close to $x_0$ is mapped to a point $d_Y$-close to $f(x_0)$".

To speak of uniform continuity, we need a means to state these closeness degrees independently of the points $x_0$ and $f(x_0)$. Roughly, we need a way to say that two neighbourhoods of different points have "the same size". This idea is captured by the notion of uniform spaces.

Thus we have the concept of uniformly continuous maps between uniform spaces. Using the entourage definition, a map $f\colon X \to Y$ between uniform spaces is uniformly continuous if for every entourage $U$ in the uniform structure on $Y$, the set

$$(f\times f)^{-1}(U) = \{ (x_1,x_2) \in X\times X : (f(x_1),f(x_2)) \in U\}$$

is an entourage in the uniform structure on $X$.

If $X$ is a compact - quasicompact and Hausdorff - space, then there is a unique uniform structure on $X$ that induces the topology of $X$. The entourages of this uniform structure are exactly the neighbourhoods of the diagonal $\Delta_X$ in $X\times X$. This shows that any continuous map from a compact space to a uniform space is uniformly continuous. Since every entourage $U$ on $Y$ is a neighbourhood of the diagonal $\Delta_Y$ and $(f\times f) \colon X\times X \to Y\times Y$ is continuous, $(f\times f)^{-1}(U)$ is a neighbourhood of $\Delta_X$, and hence an entourage.

Note that Hausdorffness is important. Not every quasicompact space is uniformisable. An infinite space with the cofinite topology is quasicompact, but is not uniformisable. The cofinite topology is a $T_1$ topology, thus a fortiori $T_0$, but not Hausdorff if the space is infinite. A uniformisable topology is $T_{3\frac{1}{2}}$ (or completely regular, whichever is the weaker condition in the used nomenclature), and a topology that is $T_{3\frac{1}{2}}$ and $T_0$ is Hausdorff.


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