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Let $X$ be a Polish space with the compatible metric being $d_1$. So $(X,d_1)$ is a separable complete metric space, and the topology is generated by $d_1$. Can there be a metric $d_2$ such that $(X,d_2)$ is also a separable complete metric space, but $d_2$ generates a different topology?

Many thanks in advance!

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    $\begingroup$ 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 $\endgroup$
    – Alessandro Codenotti
    Commented Apr 14 at 10:42
  • $\begingroup$ @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... $\endgroup$
    – J.R.
    Commented Apr 14 at 10:49
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    $\begingroup$ 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 $\endgroup$
    – Alessandro Codenotti
    Commented Apr 14 at 10:55
  • $\begingroup$ @AlessandroCodenotti Yep, makes sense, thanks a lot! $\endgroup$
    – J.R.
    Commented Apr 14 at 11:02

1 Answer 1

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Every Polish space is countable or of cardinality $\mathfrak{c}$.

Consequently, every countable Polish space has a homeomorphic copy with underlying set the natural numbers, and every uncountable one has a copy with $\mathbb{R}$ as its underlying set. In the latter case the bijection can be made a Borel isomorphism (see Theorem 15.6 in Kechris' Classical Descriptive Set Theory).

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