# Is there a Tychonoff space with $G_\delta$ diagonal that isn't submetrizable?

Recently, I'm interested in the topological spaces with $G_\delta$ diagonal. Could someone give me some examples such that the given topology space is a Tychonoff space with a $G_\delta$ diagonal but not submetrizable? The more, the better.

Submetrizable = if we can choose a coarser topology on the space $X$ and thus make it a metrizable space.

One more question: How could I know the topological space is not submetrizable? Such as $X$ is not $T_2$ or has not $G_\delta$ diagonal, or has not regular $G_\delta$ diagonal, or has not zeroset diagonal. Is there some other tools that I could use to judge that the space is not submetrizable?

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A Space with G-delta Diagonal that is not Submetrizable at Dan Ma's Topology Blog. – Martin Sleziak Apr 26 '13 at 15:55

You probably want to look at Aleksander V. Arhangel’skii & Raushan Z. Buzyakova, The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597, which is available here in PDF. Example 2.9 is a separable Tikhonov Moore space that is not submetrizable but has a $G_\delta$-diagonal. (In fact it has a rank $3$ diagonal.) Example 2.17, due to Mike Reed, is a non-separable Tikhonov Moore space that is not submetrizable but has a $G_\delta$-diagonal.
Another example is the Mrówka space $\Psi$, which is Tikhonov, separable, pseudocompact, not countably compact, and not submetrizable but does have a $G_\delta$-diagonal (even a rank $2$ diagonal).
Thanks! One more question: How could I know the topological space is not submetrizable? Such as $X$ is not $T_2$ or has not $G_\delta$ diagonal, or has not regular $G_\delta$ diagonal, or has not zeroset diagonal. Is there some other tools that I could use to judge that the space is not submetrizable? – Paul Dec 28 '11 at 12:16
corollary 2.3 of that paper has a necessary and sufficient condition for star-Lindelöf spaces (weakly normal and rank for the diagonal 5), which thus also applies to separable spaces. $\Psi$ is not submetrizable because a submetrizable pseudocompact space is metrizable (also in the paper). – Henno Brandsma Dec 28 '11 at 14:07