Is uniform continuity well defined on Polish spaces? 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.
 A: 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
http://en.wikipedia.org/wiki/Separable_space#Cardinality
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.
A: 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?
A: 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.
A: 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).
