# Restriction of non-metrizable topology to dense subset is non-metrizable

Let $$(X,\tau)$$ be a non-metrizable topological space which is not first-countable and let $$\emptyset \neq Y\subset X$$ be a proper dense subset. Is it possible for $$(Y,\tau_Y)$$ (where $$\tau_Y$$ is the relativisation of $$\tau$$ to $$Y$$) be metrizable?

• Since second countability wouldn't work I edited your question to require first countability. – AIM_BLB Jan 14 at 10:11
• You're right, that does indeed make sense. – AnnieTheKatsu Jan 14 at 10:12
• Did you mean to ask whether it's possible for $Y$ to be metrizable? As you've phrased it, the answer to your question seems obvious: of course it's possible for $Y$ to be non-metrizable, e.g., if we take $Y = X$. – Will Brian Jan 14 at 13:52
• @WillBrian Indeed, I had made an English hickup. – AnnieTheKatsu Jan 14 at 13:53
• Another class of examples is to let $Y$ be any non-compact metrizable space, and $X = \beta Y$ its Stone-Čech compactification. – Nate Eldredge Jan 14 at 18:03

Yes: the order topology on $$\omega_1+1$$ (the first uncountable successor ordinal) is an example. It is not first countable, because it's "top" point $$\omega_1$$ has no countable neighborhood base. But the set of all isolated points of this space is dense in it, and the relative topology on this set is discrete (hence metrizable).
On the other hand, let me point out that if $$x \in X$$ has no countable neighborhood base, and if $$X$$ is $$T_3$$, then $$x$$ will still fail to have a countable neighborhood base in any dense $$Y \subseteq X$$ with $$x \in Y$$. Therefore there are plenty of spaces $$X$$ for which the answer to your question would be negative. For example, no dense subset of $$[0,1]^{\kappa}$$ is metrizable (for uncountable $$\kappa$$), because every point of this space witnesses the fact that the space is not first-countable, and this will continue to be true in any dense subspace.
• Also, for a slightly smaller example, the order topology on $\omega_1$ is not metrizable. – François G. Dorais Jan 14 at 21:26
• @FrançoisG.Dorais: The OP asks that the original space not be first countable. While $\omega_1$ is not metrizable, it is first countable. – Will Brian Jan 14 at 22:40