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Let $\displaystyle X$ be a Polish space. Let $\displaystyle d,d'$ be two complete compatible metrics. Is $\displaystyle 1:(X,d)\to(X,d')$ uniformly continuous.

What I really want to know is: can we unambiguously speak of $\displaystyle f:X\to K$ from one Polish space to a compact Polish space as being uniformly continuous? Clearly it won't depend on the compatible metric on $\displaystyle K$, and clearly if the above question is answered in the positive, then the answer will be „yes“.

So far I have done the following. Suppose $\displaystyle 1:(X,d)\to(X,d')$ were not uniformly cts. Then there would be an $\displaystyle \epsilon\in\mathbb{R}^{+}$ for which $\displaystyle (\forall{\delta\in\mathbb{R}^{+}})(\exists{x\in X})\ \ \mathcal{B}^{d}_{<\delta}(x)\nsubseteq\mathcal{B}^{d'}_{<\epsilon}(x)$. For each $\displaystyle \delta\in\mathbb{R}^{+}$, let $\displaystyle S_{\delta}:=\{x\mid \mathcal{B}^{d}_{<\delta}(x)\nsubseteq\mathcal{B}^{d'}_{<\epsilon}(x)\}$. Then for all $\displaystyle \delta<\delta'$ in $\displaystyle \mathbb{R}^{+}$, we would have $\displaystyle S_{\delta}\neq\emptyset$ and $\displaystyle S_{\delta}\subseteq S_{\delta'}$. We also see that $\displaystyle \overline{S_{\delta}}\subseteq\bigcup_{\delta'>\delta}S_{\delta'}$.

So were $\displaystyle \bigcap_{n\in\omega}\overline{S_{\delta_{n}}}\neq\emptyset$ where $\displaystyle (\delta_{n})_{n\in\omega}\searrow 0$ in $\displaystyle \mathbb{R}$, then it would be so that $\displaystyle \bigcap_{\delta\in\mathbb{R}}S_{\delta}\neq\emptyset$. But then there would be an $\displaystyle x$ and a sequence $\displaystyle (x_{n}\in\mathcal{B}^{d}_{<\delta_{n}}(x)\setminus\mathcal{B}^{d'}_{<\epsilon}(x))_{n\in\omega}$ converging to $\displaystyle x$ in the space $\displaystyle (X,d)$ and not converging to $\displaystyle x$ in $\displaystyle (X,d')$, which would contradict that $\displaystyle d,d'$ are compatible for $\displaystyle X$. Thus it would be that the countable intersection $\displaystyle \bigcap_{n\in\omega}\overline{S_{\delta_{n}}}=\emptyset$.

If I can show, this must not be the case, then I would be done.

However (1) there might be a completely different way to proving this; (2) the question might have a negative answer.

Any ideas would be great.

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What does it mean for two metrics to be compatible? – Qiaochu Yuan Dec 7 '10 at 6:28
I think he means that they generate the same topology. After all one of the requirements we make of a Polish space is that it be completely metrizable, but not that any particular metric be singled out as "canonical". Also... isn't $X = \mathbb{R}$, $d(x,y) = |x-y|$, and $d'(x,y) = \left\lvert x^{\frac{1}{3}} - y^{\frac{1}{3}}\right\rvert$ a counterexample? – kahen Dec 7 '10 at 6:34

4 Answers 4

This is an answer to the question in one of your answers: is the Stone-Cech Compactification of a Polish space necessarily Polish?

The answer is a resounding no: not even close, in some sense. Let $X$ be a countably infinite set equipped with the discrete metric (and thus the discrete topology). Its Stone-Cech compactification is the space of ultrafilters on $X$, which has cardinality $2^{2^{|X|}}$ (added: e.g., see here). Evidently this space is separable: $X$ itself is a countable dense subspace. Now, if you consult

you will see that a first countable separable space can have at most continuum cardinality. (Disclosure: I wrote this part of the article.) So the Stone-Cech compactification is not first countable, so is not even close to being metrizable.

I'm not an expert on Polish spaces, but I think you just do not have the right structure here to define a reasonable notion of uniform continuity.

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Hmm, yes that is. Also $X=[0,\infty)$ with $d(x,y)=|y-x|$ $d'(x,y)=|y^{2}-x^{2}|$ would work too. Thanks for that.

Okay then, scrapping that too general a question\ldots can my other concern be addressed, i.e. can we unambiguously call a map from a Polish space to a compact Polish space uniformly continuous without referring to the metrics?

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You mean something like definining $f:X\to Y$ to be uniformly continuous between Polish spaces $X$ and $Y$ iff $f$ is uniformly continuous from $(X,d)$ to $(Y,d')$ for all complete compatible metrics $d$ and $d'$ on $X$ and $Y$ respectively? If $X$ were compact, this would be all continuous maps, but if $X$ is not compact and $Y$ isn't too small my guess is that there aren't many of these. (I don't have much reason for my guess.) – Jonas Meyer Dec 7 '10 at 7:10

Yeah i worked it out on the train. It's not the case. Consider $X=[0,\infty)$ and $d:(x,y)\mapsto|y-x|$, $d':(x,y)\mapsto|\sqrt{y}-\sqrt{x}|$ and consider the map $f:X\to[0,1]$ a periodic continuous function like $sin$. Then $f:(X,d)\to[0,1]$ is uniformly cts, but $f:(X,d')\to[0,1]$ is not.

Ultimately I'm looking for this: let $X$ be a polish space, then does it embed continuously to a unique polish compactification, universal in the sense that every continuous map from $X$ to another compact polish space lifts uniquely to a map from the compactification. Doesn't seem likely. I can't see that the Stone-Cech compactification is polish (it's at least separable compact hausdorff, but that actually does not imply second countable).

What I can do, is take $(X,d)$ where $d$ is a compatible complete metric, and embed this via a uniformly continuous map to a compact polish space, universal in the sense that every uniformly cts map from $(X,d)$ to another compact polish space lifts uniquely to a map from the compactification.

What that compactification is might depend on $d$. Though it would be nice if it didn't, so that we could talk about „the“ compactification of a polish space.

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thanks for the (counter-)example. Yes, there's a few basic results (i) if $X$ is compact dann es sei metrisable $\Leftrightarrow$ es sei hausdorff and second countable; (ii) compact metrisable implies polish (obviously!); (iii) any uncountable polish space has cardinality $2^{\aleph_{0}}$.

So your claim runs through, provided $Ult(\omega)$ the space of ultrafilters over $\omega$, does indeed have cardinality $>2^{\aleph_{0}}$. However I'll have to run a few calculations to check this is the case.

Nevertheless, it is certain that it is consistent to assume that compact hausdorff separable does not imply polish ($\Leftrightarrow$ second countable in this context).

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@Raj: I added a reference for the fact that any infinite set $X$ carries $2^{2^{|X|}}$ ultrafilters. Sorry for not including justification before: this is a standard fact, but it is certainly not obvious. – Pete L. Clark Dec 8 '10 at 15:19

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