3
$\begingroup$

We say a space $(X,\tau)$ is homogeneous if for any $x,y\in X$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x) = y$.

What is an example of a connected, homogeneous $T_2$-space $(X,\tau)$ with $|X| = 2^{\aleph_0}$ such that $(X,\tau)$ is not homeomorphic to a subspace of $\mathbb{R}^\omega$?

| cite | improve this question | | | | |
$\endgroup$
  • 3
    $\begingroup$ How about the weak topology on a separable Banach space? It's not metrizable nor even first countable, so can't be homeomorphic to any subspace of $\mathbb{R}^\omega$. $\endgroup$ – Nate Eldredge Nov 5 '18 at 15:17
2
$\begingroup$

According to this research in $\pi$-base, the sigma product of incountably many copies of $\mathbb{R}$ and the boolean product topology on $\mathbb{R}^{\omega}$ are connected, $T_2$ and homogeneous.

However, they are not first countable, hence they cannot be homeomorphic to subspaces of $\mathbb{R}^{\omega}$.

| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ Well, there is probably a mistake in $\pi$-base, since the second space I gave is the box topology in $\mathbb{R}^{\omega}$, which is not connected (Steen-Seebach, Counterexamples in Topology, p. 128-129). $\endgroup$ – Francesco Polizzi Nov 5 '18 at 15:54
  • 2
    $\begingroup$ If you actually click on the "connected" listing for that space, it says "this is wrong, please delete me". It seems that the Github code for $\pi$-base correctly says "false": github.com/pi-base/data/blob/master/spaces/S000107/properties/…. I don't know why they are out of sync. The "this is wrong" text doesn't even appear in the git history of that file. $\endgroup$ – Nate Eldredge Nov 5 '18 at 22:51
2
$\begingroup$

Or take the long line: $\omega_1\times[0,1)$ with the lexicographic order, minus the first point. Every bounded open interval is isomorphic to $(0,1)$, so it is homogeneous. It is first-countable but not second-countable, hence not embeddable into $\mathbb{R}^\omega$.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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