I have a question:
Does a second countable $T_2$ space $X$ always have a $G_\delta$-diagonal?
(If $X$ is regular, then it is metrizable, and it obviously has a $G_\delta$-diagonal.)
Thanks for your help.
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2I think that the answer may not be known; the authors of this recent paper [PDF] did not know. On page $25$ they do give an example of a second countable Hausdorff space that does not have a regular $G_\delta$-diagonal, though it does have a $G_\delta$-diagonal. – Brian M. Scott Apr 01 '15 at 06:30
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This question now has a negative answer at MathOverflow: Taras Banakh shows that the quotient of $\Bbb R^\omega\setminus{0}^\omega$ (with the topology inherited from the product topology on $\Bbb R^\omega$) by the equivlence relation $x\sim y$ iff $\Bbb R x=\Bbb R y$ is a counterexample. – Brian M. Scott Apr 09 '22 at 03:36