# Metric space on which every function is uniformly continuous

“Characterise the metric spaces in which every function is uniformly continuous”

The first thing I observed is that every subset of such metric spaces (let us call them X for the sake of semplicity) must be open for otherwise there would be some metric spaces Y and some function $$f:X \rightarrow Y$$ which may not be continuous (hence, also not uniformly continuous). Then, I conjectured that X must be compact, in fact, by the Heine-Cantor theorem this is a sufficient condition; however I can not manage to find a counter-example for when X is not compact in order to conclude that it is also necessary.

Any help or hint is much appreciated!

• Can you prove your first observation? If yes it means any such space is discrete. Feb 25, 2022 at 15:48
• Nice answer here: math.stackexchange.com/questions/1292465/… Feb 25, 2022 at 16:34

As you rightly point out, since every function from $$X$$ to any other metric space $$Y$$ is continuous, every subset of $$X$$ is an open set.

(By the way, are you sure that the question doesn't mean "every function from $$X$$ to $$X$$ is continuous"?)

In the study of general topology, when every subset of $$X$$ is an open set, we say that $$X$$ has the discrete topology.

Let's now apply the condition about uniform continuity. My first instinct is that the infimum of $$d(x,y)$$ for $$x\neq y$$ has a role to play. And indeed it does. Let's call the infimum $$r$$.

If $$r>0$$, then every function is of course uniformly continuous. The image of each $$s$$-ball is a single point, for $$s.

What if $$r=0$$?

Then, with the help of the axiom of choice, choose sequences $$x_i,x'_i$$ such that $$d(x_i,x'_i)$$ decreases strictly to $$0$$, and such that all these points are distinct, i.e. no point is used more than once here.

Now, you can construct a function from $$X$$ to $$X$$, which sends $$x_i$$ to $$x_1$$, and $$x'_i$$ to $$x'_1$$. It wouldn't matter where the function sends any other point. The function is not uniformly continuous.

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In conclusion, for metric spaces $$X$$,

"every function on $$X$$ is uniformly continuous" is equivalent to

"there is some $$c>0$$ such that all pairs of points have distance at least $$c$$".

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EDIT: I forgot to explain a subtle technical point.

We know that a subset containing only one point $$p$$ is open. This means that for each $$p$$, there is some radius $$s_p$$ such that the open ball $$B(p,s_p)$$ contains only $$p$$.

This point is relevant to the following question. When we were dealing with the case $$r=0$$, how can we be sure that we can choose $$x_i,x'_i$$, so that we do not use any point more than once? (We want to ensure this, because we want all the $$x_i$$ to be mapped to $$x_1$$, and all the $$x'_i$$ to be mapped to a point $$x'_1$$ which is different from $$x_1$$. If $$x'_6=x_4$$, for example, then $$x'_6$$ will map to both $$x'_1$$ and $$x_1$$, which we don't want.)

We can do this, as shown by induction. Suppose you have already chosen $$x_1,\ldots,x_n$$ and $$x'_1,\ldots,x'_n$$. We want to choose $$x_{n+1}$$ and $$x'_{n+1}$$ so that neither equals any of those. Just draw an open ball around each of those such that each open ball contains only that one point.

Take the minimum $$m$$ of those radii. And declare that $$x_{n+1}$$ and $$x'_{n+1}$$ will have distance less than $$m$$. Such a choice is always possible, because the infimum of $$d(x,y)$$ for $$x\neq y$$ is assumed to be $$0$$.

It follows that $$x_{n+1}$$ and $$x'_{n+1}$$ do not repeat the previous points, because the open ball of radius $$m$$ around $$x_{n+1}$$ contains $$x'_{n+1}$$, and vice versa. Yet an open ball of radius $$m$$ around each of the previous points contains only one point.

• Some call this condition for a metric space "uniformly discrete." Feb 25, 2022 at 16:53
• Man, thank you very much! It is one of those answers that really gives you insight! Feb 25, 2022 at 17:20
• No problem. I've included an addendum, a technical part of the proof which I didn't explain, if you're interested Feb 25, 2022 at 17:49