# Why are compact Hausdorff spaces divisible - and other uniformity related questions

My questions concern uniform spaces in specific cases involving compact spaces.

The first one is about how a compact Hausdorff space is a uniform space. I'm aware of indirect proofs using pseudometrics, but I'd like to see a direct proof that the set of neighbourhoods of the diagonal $\Delta_K$ of a compact Hausdorff space $K$ forms a basis for a uniformity on $K$ that induces its topology. How can I find such a proof ? (the only hard part being that for all such $U$ there is such a $V$ with $V\circ V\subset U$, i.e. $K$ is divisible)

Once this is solved, I'm wondering about the following question: let $X$ be a uniform space, $Z$ its Cauchy completion (which is again a uniform space), and assume $Z$ is compact (i.e. $X$ is precompact). Let $K$ be a compact space, and $f:X\to K$ a continuous function. Is $f$ necessarily uniformly continuous ?

If the answer is yes, then it can be extended to a map $g: Z\to K$ by the usual construction. If the answer is no, does this last bit still hold: even if $f$ is not uniformly continuous, can it always be extended to the completion when the codomain is compact ? (here the completion is not assumed to be compact)

My thoughts on how to answer these questions:

for the first one, I've been stuck for quite a while and have not made much progress. I've tried using the fact that for an open set $U\subset K$, there is $V$ open with $\overline{V}\subset U$, but this hasn't led me anywhere. Similarly, finding that a neighbourhood of the diagonal contains another such neighbourhood of the form $\displaystyle\bigcup_{i=1}^n U_i\times U_i$, where each $U_i$ open hasn't given me anything.

For the second one, I think the answer is yes, but only because I think that the map can be extended, and since $Z$ is compact, then $g$ would be uniformly continuous, and thus its restriction to $X$ too. So I've tried actually answering the third question (i.e. not proving that $f$ is uniformly continuous, and not any assumptions on $Z$; simply proving that it can be extended), but I run into some complications when I try to prove that if $x_i \to x, y_j \to x$, and $f(x_i), f(y_j)$ both converge in $K$, then it's to the same limit, and haven't managed to show this (which would help)

• Since these are two distinct questions, it would be better to ask them separately. – mathematics2x2life Mar 23 '18 at 21:03
• The second question has a simple counterexample $\sqrt{x}$ as a function $(0,1)\to[0,1]$ with the usual metrics – Dap Mar 24 '18 at 9:58
• @Dap : this will be uniformly continuous, and this follows from the fact that it has a continuous extension $[0,1]\to [0,1]$ – Max Mar 24 '18 at 10:18
• @Max: oh, sorry, I was thinking of something else. $\sin 1/x$ then – Dap Mar 24 '18 at 10:19
• @mathematics2x2life : you are probably right but I thought since the two questions are related I could ask them both at once – Max Mar 24 '18 at 10:19

For 1, by regularity any point is contained in some open $A$ with $\overline{A}\subset B$ with $B$ open and $B\times B\subset U.$ This means there is a finite open cover $\mathcal A=\{A_1,\dots,A_n\}$ where each $A_i$ comes with a corresponding open $B_i$ satisfying $\overline{A_i}\subset B_i$ and $B_i\times B_i\subset U.$ For each $i$ we have a two-element open cover $\mathcal A_i=\{K\setminus \overline{A_i}, B_i\},$ and the original open cover $\mathcal A.$ Let $\mathcal V$ be an open cover finer than all these covers, such as the family of sets of the form $X\cap X_1\cap \dots \cap X_n$ with $X\in \mathcal A$ and $X_i\in\mathcal A_i$ for each $i.$ Let $V=\bigcup_{X\in\mathcal V} X \times X.$ It's an open neighborhood of the diagonal. If $(x,y),(y,z)\in V$ then $x,y\in A_i$ for some $i,$ so $y,z\in B_i$ and $(x,z)\in B_i\times B_i\subset U$ as required.
• thank you ! I'll accept your answer because you already answered the rest in the comments with $sin(1/x)$ which can't be continuously extended to $[0,1]$ (moreover as pointed out I should have asked separate questions) – Max Mar 24 '18 at 10:24