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I hope this is not too basic or obvious a question.

Let $d_1$ and $d_2$ be metrics on the same set $X$, with $d_1$ being separable and $d_2$ not being separable. Is it possible that $d_1$ and $d_2$ generate the same Borel $\sigma$-algebra? If so, does the answer change if we require $d_1$ to be Polish?

Thanks.

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It is possible that $(X,d_{1})$ has the same Borel sets as $(X,d_{2})$ when $d_{1}$ is separable and $d_{2}$ is not by assuming the Martin's axiom and the negation of the continuum hypothesis by the answer https://mathoverflow.net/a/155527/22277 by Andreas Blass to my question. A set $X$ is said to be a $Q$-set if every subset of $X$ is an $F_{\sigma}$-set in $X$. If Martin's Axiom holds then every subset $X$ of $\mathbb{R}$ of cardinality less than the continuum is a $Q$-set. In particular, if MA and the negation of the continuum hypothesis holds, then if $L\subseteq\mathbb{R}$ has cardinality $\aleph_{1}$, then every subset of $L$ is a Borel set. Therefore, the metric space $L$ has the same Borel sets as the Borel sets where $L$ is given some discrete metric.

$\textbf{Added 8/17/2015}$

I claim that there is no Polish metric $\rho$ and non-separable metric $d$ on a set $X$ so that $(X,\rho)$ and $(X,d)$ have the same Borel sets.

Suppose that $(X,d)$ is a non-separable metric space. A metric space satisfies the countable chain condition if and only if it is separable. Therefore $(X,d)$ does not satisfy the countable chain condition. Therefore, let $\mathcal{U}$ be an uncountable pairwise disjoint collection of open subsets of $X$. Let $x_{U}\in U$ for each $U\in\mathcal{U}$. Let $A=\{x_{U}|U\in\mathcal{U}\}$. If $B\subseteq A$, then $B=\overline{B}\cap\bigcup\mathcal{U}$ which is a Borel set in $X$ being the intersection of a closed set with an open set. Thus $A$ is an uncountable set where every subset of $A$ is Borel in $X$.

On the other hand, I claim that there does not exist a polish space $(X,\rho)$ and an uncountable subset $A\subseteq X$ so that every subset of $A$ is a Borel set in $X$. If $(X,\rho)$ is a Polish space and $A$ is an uncountable Borel subset of $X$, then $A$ has cardinality continuum since every Borel subset of a Polish space is either countable or has cardinality continuum. Therefore, there are at least $2^{\mathfrak{c}}$ subsets of $A$, but there are only $\mathfrak{c}$ Borel subsets of $A$. Therefore some subset of $A$ is not Borel.

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  • $\begingroup$ Thank you very much for this. Now I think the fact that every metric space satisfying ccc is separable relies on the axiom of choice. Do you know if it is possible to prove that no Polish metric is measurably isomorphic to a non-separable metric just using ZF+DC? $\endgroup$ Sep 22, 2015 at 0:55

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