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We recall that a topological space $(X,\tau)$ is submetrizable, if there is a coarser metrizable topology $\tau'$ that $\tau\supseteq\tau'$.

one of the properties of these topological spaces is that each point of them is a $G_\delta$-point.

there are a lot of interesting topological spaces which are not metrizable.For example the surgenfery line denoted by $(\mathbb{R}_l)$ is a classical example of a normal lindelof nonmetrizable space in which every point of it is a $G\delta$-point. notice that this space is submetrizable.

another classical example is the moore plane denoted by $(\Gamma)$ which is non normal,non lindelof, separable space in which every point of it is a $G_\delta$-point.notice that this space is also a submetrizable space.

With the above summary, I can pose my Question.

Q. Is there an example of a topological space $(X,\tau)$ such that every point of it, is a $G_\delta$, but this space is not submetrizable?

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The Alexandroff´s double arrow space (i.e. $[0,1] \times \{0,1\}$ with the lexicographic order) is first countable (so any point is a $G_\delta$), but is compact and non-metrizable (hence it is not submetrizable).

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  • $\begingroup$ Thank You Dear Ramiro. Your example is very beautiful. When I posed this question I found another example. It suffices to consider the first uncountable ordinal(i.e.$\omega_1$). then the space $[0,\omega_1)$ works for this situation. Anyway,Let me Know that if your example is not metrizable because this space is not second countable. thank you very much again. $\endgroup$
    – Ali Reza
    Commented May 10, 2012 at 14:03

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