# Separability of subspaces of homogeneous topological spaces

Let $$\ X\$$ be a homogenous separable topological space (i.e. for every $$\ x\ y\in X\$$ there exists a homeomorphism $$\ f:X\to X\$$ such that $$\ f(x)=y,\$$ and there is a countable dense subset of $$\ X).\$$

Question  Is every subspace of $$\ X\$$ separable?

The question allows several variations by considering different stronger kinds of homogeneity. The other direction would be to consider different separability properties like Hausdorff, normal, etc. (No, forget the metric case :) ).

I would conjecture that in the simple case of homogeneity there should be an example of $$\ X,\$$ and of its subspace $$\ Y\$$ which is not separable while $$\ X\$$ is.

This fails even for topological groups. For example, the separable topological group $$\mathbb{Z}^{\mathfrak{c}}$$ contains a subgroup of uncountable cellularity (hence certainly not separable!). This is a result of my colleague Vladimir Uspenskij in the paper below:

Uspenskij, V. V., On the Suslin number of subgroups of products of countable groups, Acta Univ. Carol., Math. Phys. 36, No. 2, 85-87 (1995). ZBL0854.20064.

• Be warned that I am not an expert on topological groups, even though I have heard many talks on them! – Todd Eisworth Feb 21 '20 at 4:25
• The homogeneous compact space $\{0, 1\}^{\frak c}$ is separable but contains the non-separable $\Sigma$-product with any base point. – Anonymous Feb 21 '20 at 13:28
• That the subgroup $\mathbb{Z}^{(\mathfrak{c})}$ of finitely supported elements is not separable is immediate. – YCor Feb 21 '20 at 17:18
• Todd, thank you for your answer. (I've upvoted it, and now I wait patiently or less patiently for more answers). Only after seeing your answer I recalled the Szpilrajn-Marczewski's theorem about the separability of the product of continuum of separable spaces. A generalization of this theorem for general cardinals (and arbitrary "separability") must be known, I am sure (it's also in my old UW script on topology for 2nd-year students). – Wlod AA Feb 22 '20 at 5:02

Any compact homogeneous hereditarily separable space has size at most $$\mathfrak{c}$$ (this is a result of Ismail). Thus any compact homogeneous separable space of bigger size provides a counterexample (e.g. $$2^\mathfrak{c}$$ as in Anonymous' comment).

There are Banach spaces $$X$$ for which the dual space $$X^*$$ is separable in the weak* topology yet the unit ball $$B_{X^*}$$ of $$X^*$$ is not. A notable example is $$X=J\!L_2$$, the Johnson–Lindenstrauss space.

W.B. Johnson, and J. Lindenstrauss, Some remarks on weakly compactly generated Banach spaces, Israel J. Math. 17 (1974), 219–230. Correction ibid 32 No. 4 (1979), 382–383.

• Tomek, is the said unit ball not separable under the topology induced by the weak* topology of $X^*$? – Wlod AA Feb 22 '20 at 1:30
• I trust you. :) – Wlod AA Feb 22 '20 at 5:04
• @WlodAA, yes, the ball is not weak*-separable as $JL_2$ does not embed into $\ell_\infty$. – Tomasz Kania Feb 22 '20 at 9:13