Is every Hausdorff space metric?

Question

My question is very simple.

I know every metric space is Hausdorff, but the converse is true? anyone knows some counterexample?

Thanks

Every metric space has a countable basis of neighborhoods around any of its points (namely balls of radius $1/n$). — Berci, Jul 16 '15 at 22:31
Which means look for a counterexample that has so many neighbourhoods around a point that there's no countable basis there. That motivates looking at uncountable ordinals. — Robert Israel, Jul 16 '15 at 22:45
Or: A perfect Hausdorff space that is not metrizable., or Example of Hausdorff and Second Countable Space that is Not Metrizable, or other examples that you would find if bothered to search. — , Jul 16 '15 at 23:48

score 2 · Answer 1 · answered Jul 16 '15 at 22:55

Consider the space $X = \{0,1\}^\mathbb R$ of all functions from $\mathbb R$ to $\{0,1\}$ (actually any uncountable set would do in place of $\mathbb R$). The topology is the product topology, i.e. a base is the cylinder sets, the sets obtained by specifying the values of the function at finitely many points. This is Hausdorff, indeed for any $x \ne y \in X$ there is some $t \in \mathbb R$ such that $x(t) \ne y(t)$. It is not metrizable because no point can have a countable base of neighbourhoods: the intersection of a countable collection of cylinder sets containing $x$ will still contain lots of points besides $x$, because it only restricts the values at countably many members of $\mathbb R$.

score 2 · Answer 2 · answered Jul 17 '15 at 00:28

2

The ordinal $\omega_1+1$ with the order topology is Hausdorff but not metrizable.

answered Jul 17 '15 at 00:28

Noah Schweber

245,398

Is every Hausdorff space metric?

2 Answers2