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Apr 14 at 13:24 vote accept J.R.
Apr 14 at 12:44 answer added KP Hart timeline score: 2
Apr 14 at 11:02 comment added J.R. @AlessandroCodenotti Yep, makes sense, thanks a lot!
Apr 14 at 10:55 comment added Alessandro Codenotti Sure, take any bijection $f\colon\Bbb R^d\to\Bbb R$ and define a new metric on $\Bbb R^d$ by $d'(x,y)=|f(x)-f(y)|$. Then $(\Bbb R^d,d')$ is homeomorphic to $\Bbb R$ (with the usual topology). The point is that the Polish space you're starting with it's irrelevant since you're forgetting its topology, what you are really asking is whether there are two different Polish topologies on an uncountable set, or equivalently whether any two uncountable Polish spaces are homeomorphic
Apr 14 at 10:49 comment added J.R. @AlessandroCodenotti Thank you for the quick response! May I ask an even simpler question, if $X=\mathbb R ^d$, then I know that equipped with the usual Euclidean metric, $\mathbb R ^d$ is a separable metric space. Is there another metric $d$ on $\mathbb R ^d$ that makes $\mathbb R ^d$ separable complete, but $d$ generates a different topology (so different open sets)? Apologies if my question is a bit lacking, I usually work within probability theory, so trying to grasp the topology side of it requires some effort...
Apr 14 at 10:42 comment added Alessandro Codenotti If you don't have any requirements on the relationship between $d_1$ and $d_2$ the answer is obviously yes, let $(X,d_1)$ and $(Y,d_Y)$ be two uncountable Polish spaces which are not homeomorphic, let $f\colon X\to Y$ be any bijection and use it to define a new metric $d_2$ on $X$ by $d_2(x,x')=d_Y(f(x),f(x'))$. If you want the topology generated by $d_2$ to be finer than the one generated by $d_1$ the answer is still yes, but less obviously so
Apr 14 at 10:34 history asked J.R. CC BY-SA 4.0