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.