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An isomorphism between two measurable spaces $(X_1,\mathcal{B}_1), (X_2,\mathcal{B}_2)$ is a measurable bijection $f:X_1\rightarrow X_2$ whose inverse is also measurable.

QUESTION. Can there be an isomorphism between an uncountable Polish space and a non-Hausdorff topological space, each endowed with its respective Borel $\sigma$-algebra?

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    $\begingroup$ Yes. Just consider the identity of $\mathbf{R}\cup\{\infty\}$. In the first case it's endowed with its topology of 1-point compactification. In the second it's endowed with the topology making $\{\infty\}$ dense (closed subsets are the whole subsets and closed subsets of $\mathbf{R}$). Borel subsets are the same for both topologies. $\endgroup$
    – YCor
    Jun 24, 2020 at 21:27
  • $\begingroup$ @YCor: thank you. $\endgroup$
    – T. Amdeberhan
    Jun 26, 2020 at 15:27

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Here is a particularly simple way to see why the answer is yes, as YCor already pointed out in a comment. Uncountability is a red herring.

Ignore the uncountability condition for now. Consider the set $\{0,1\}$ once with the discrete topology and once with the non-Hausdorff topology whose only nontrivial open set is $\{1\}$, the Sierpiński space. They are clearly Borel isomorphic.

Now take your favorite uncountable Polish space $X$. Taking the disjoint union $X\sqcup\{0,1\}$ of both two point spaces will again give you two Borel isomorphic spaces, only one of which is Polish.

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    $\begingroup$ thank you for your answer. $\endgroup$
    – T. Amdeberhan
    Jun 26, 2020 at 15:28

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