Are metrizable subspaces of separable spaces separable?

Certainly subspaces of separable metrizable spaces are separable but subspaces of separable spaces need not be separable in general.


3 Answers 3


The answer is no, if we interpret separable as "has a countable dense subset" (Fedor Petrov's answer appears to have interpreted it as possessing a countable base). Consider the Moore plane, or Niemytzki tangent disc topology, as it is called on page 100 of Steen & Seebach's Counterexamples in Topology. This is a topology on the closed upper half-plane. A neighbourhood base for points $(x,y)$ with $y > 0$ is consists of open discs of radius $< y$. A neighbourhood base for points $(x,0)$ consists of sets of the form $\{ (x,0) \} \cup B_r(x,r)$ for $0 < r < \infty$, where $B_r(x,r)$ is the open disc of radius $r$ with centre $(x,r)$.

The rational points contained in the open upper half-plane form a countable dense set, so this is a separable space. The subspace topology on the horizontal axis is discrete. Now, the discrete topology is always metrizable, and as the real axis has continuum cardinality and every set is equal to its closure, it is not separable. So we have a non-separable metrizable subspace of a separable space.

  • $\begingroup$ @bof I should emphasize that I do not take that interpretation of the word "separable" myself, or in my answer, and I am well aware that second countability is hereditary for subspaces. Maybe your comment should be directed to Fedor Petrov? $\endgroup$ Jun 24, 2018 at 3:47
  • 1
    $\begingroup$ @bof thank you, I indeed wrote something, meaningless $\endgroup$ Jun 24, 2018 at 6:15

It looks so. A non-separable metric space $X$ contains a more than countable family of disjoint open sets (namely, balls of sufficiently small radius). If now $X\subset Y$ is a metrizable topological subspace and $\{A_\alpha\}$ is a more than countable family of (non-empty) open subspaces of $X$, for all $\alpha$ choose a point $x_\alpha\in A_{\alpha}$, open $U_\alpha$ in $Y$ such that $U_\alpha\cap X=A_{\alpha}$ and an open set $V_\alpha\subset U_\alpha$ containing the point $x_\alpha$ from a countable base in $Y$. The sets $V_\alpha\cap X$ are all disjoint, thus all distinct --- a contradiction.

  • $\begingroup$ This is just an unusually hard proof of the partial case of the obvious claim that having a countable base is hereditary property. $\endgroup$ Jun 24, 2018 at 6:17

Let me support @RobertFurber and contradict Fedor (I enjoyed greatly many of your posts, Fedor, but possibly the general topology is not yet your cup of tea). Indeed,

let $\ \Phi\ $ be the family of all non-trivial ultrafilters in $\ \mathbb Z\ $ (assume, how reasonably!, that $\mathbb Z\cap\Phi=\emptyset).\ $ Let

$$ X\ :=\ \mathbb Z\cup\Phi $$

Let a topology in $\ X\ $ be defined via bases of neighborhoods of points, where $\ \{z\}\ $ is the only element of neighborhood base of $\ z\ $ for every $\ z\in \mathbb Z,\ $ and family

$$ U_\phi\ :=\ \{\{\phi\}\cup F\ :\ F\in\phi\} $$

is the base of neighborhoods of $\ \phi\ $ for every ultrafilter $\ \phi\in\Phi\ \in X $

Space $\ X\ $ is Hausdorff because of the ultra property (maximality) of the ultrafilters (it's just a beautiful space :) ).

We see that

a.   space $\ X\ $ is separable (countable $\ \mathbb Z\ $ is dense in $\ X$);

b.   the subspace $\ \Phi\ $ is discrete hence metrizable, while it is not separable since it is the only dense subset of itself, and it is not countable--in fact, it has cardinality equal to

$$ 2^{2^{^{\aleph_0}}} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.