A homogeneous space $(X,\tau)$ is a topological space such that for all $x,y\in X$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x)=y$. As a previous question implies, the union of an ascending chain of homogeneous spaces need not be homogeneous (example, see below).

What is an example of a Hausdorff space $(X,\tau)$ that does not contain a homogeneous subspace that is maximal with respect to $\subseteq$?

Note. The supremum (union) of an ascending chain of homogeneous spaces need not be homogeneous: Endow $\mathbb{N}$ with the discrete topology and consider the disjoint union $X = (\mathbb{N}\times\{0\}) \cup (\mathbb{Q}\times\{1\})$, where $\mathbb{Q}$ carries the Euclidean topology. Since $X$ is countable, it is the the union of an ascending chain of finite sets, all of which carry the discrete topology and are homogeneous, but $X$ is not.

  • 2
    $\begingroup$ I think there are easier examples of nonhomogeneous spaces arising as the union of a chain of homogeneous spaces. For example, consider a single convergent sequence with its limit point. $\endgroup$ May 3, 2018 at 12:49
  • $\begingroup$ Thanks @JoelDavidHamkins for pointing this out. I think all these examples point to there being a space without maximal homogeneous subspaces - but I can't find one. $\endgroup$ May 3, 2018 at 13:50

2 Answers 2


Theorem. The topological sum $X=\bigoplus_{n\in\omega}\ell_2(\aleph_n)$ of Hilbert spaces of density $\aleph_n$ does not contain maximal homogeneous subspaces.

Proof. Let $H$ be a non-empty homogeneous subspace in $X$. Then for some $k\in\omega$ the intersection $H\cap\ell_2(\aleph_k)$ is not empty and hence $H\cap\ell_2(\aleph_k)$ is a closed-and-open subspace of density $\le\aleph_k$ in $H$. By the homogeneity of $H$, each point $x\in H$ has a closed-and-open neighborhood of density $\le\aleph_k$.

Claim. For every $n>k$ the intersection $H\cap \ell_2(\aleph_n)$ is nowhere dense in $\ell_2(\aleph_n)$.

Proof. To derive a contradiction, assume that for some $n>k$ the set $H\cap\ell_2(\aleph_n)$ is not nowhere dense in $\ell_2(\aleph_n)$. Then for some non-empty open set $W\subset\ell_2(\aleph_n)$ the intersection $H\cap W$ is dense in $W$. Since each point of $H$ has an open neighborhood of density $\le\aleph_k$, we can replace $W$ by a smaller open subset of $W$ and assume that $W\cap H$ has density $\le\aleph_k$. Then $W\subset\overline{H\cap W}$ also has denisty $\le\aleph_k$, which is not true as non-empty open sets in $\ell_2(\aleph_n)$ have density $=\aleph_n>\aleph_k$. This contradiction shows that $H\cap\ell_2(\aleph_n)$ is nowhere dense in $\ell_2(\aleph_n)$ for all $n>k$.

By Claim, for $n=k+1$ the set $H\cap\ell_2(\aleph_n)$ is nowhere dense in $\ell_2(\aleph_n)$. So, we can find a non-empty open set $U\subset \ell_2(\aleph_n)$ whose closure does not intersect the closure of $H$ in $X$. Since $U$ contains a topological copy of $\ell_2(\aleph_k)$, we can find a subspace $D'\subset U$, homeomorphic to the closed-and-open subspace $D:=H\cap\ell_2(\aleph_k)$ of $H$.

We claim that the subspace $H':=H\cup D'$ of $X$ is homogeneous. Fix a homeomorphism $h:D\to D'$ and extend $h$ to a homeomorphism $\bar h:H'\to H'$ letting $\bar h|D=h|D$, $\bar h|D'=h^{-1}|D'$ and $\bar h|H\setminus D=id$.

Given any points $x,y\in H'$ we should find a homeomorphism $f':H'\to H'$ such that $f'(x)=y$.

If $x,y\in H$, then the homogeneity of $H$ yields a homeomorphism $f:H\to H$ such that $f(x)=y$. Extend $f$ to a homeomorphism $f':H'\to H'$ letting $f'|H=f$ and $f'|D'=id|D'$.

If $x\in H$ and $y\in D'$, then $\bar h(y)\in D$ and by the preceding case there exists a homeomorphism $f:H'\to H'$ such that $f(x)=\bar h(y)$. Then $f':=\bar h^{-1}\circ f$ is a homeomorphism of $H'$ such that $f'(x)=y$.

If $x\in D'$ and $y\in H$, then $\bar h(x)\in D$ and by the first case there exists a homeomorphism $f:H'\to H'$ such that $f(\bar h(x))=y$. Then $f':=f\circ \bar h$ is a homeomorphism of $H'$ such that $f'(x)=y$.

If $x,y\in D'$, then $\bar h(x),\bar h(y)\in D'$ and by the first
case there exists a homeomorphism $f:H'\to H'$ such that $f(\bar h(x))=\bar h(y)$. Then $f':=\bar h^{-1}\circ f\circ \bar h$ is a homeomorphism of $H'$ such that $f'(x)=y$.

This completes the proof of the homogeneity of the space $H'$. Then the homoheneous space $H$ is a proper subspace of the homogeneous space $H'$ and hence $H$ is not maximal homogeneous. $\square$

  • $\begingroup$ very clever construction $\endgroup$ Jun 14, 2018 at 13:17
  • $\begingroup$ @PietroMajer Thanks! I also liked it when discovered. $\endgroup$ Jun 14, 2018 at 17:38

A comment on Taras Banakh's beautiful answer (with his corrections). If I understand it correctly, the key ingredients of the construction are:

1) $X$ is a regular space, countable union of disjoint open sets $X_k$, for $k\in\omega$;

2) Any nonempty open subset of $X_{k+1}$ contains a copy of $ X_k$ as a subspace.

As a consequence of 1) and 2), any open subset of $X$ that meets infinitely many $X_k$, contains a copy of $X$ itself as a subspace (it has room enough to include a topological sum of copies $X_k$, for all $k$); in fact even with its closure.

3) Each subset of $X_{k+1}$ , which is homeomorphic to a subset of $X_{k}$ is not dense in $X_{k+1}$

In this situation if $H\subset X$ is a homogeneous subspace of $X$, then $ X_k\setminus \overline H\neq\emptyset$ for all large $k $. But then $X$ also contains a shifted copy $H'$ of $H$, separated by neighbourhoods, so that it also contains their topological sum, which is a strictly larger homogeneous space.

  • $\begingroup$ (So, if I didn't oversimplified Taras' construction, $X_k:=\mathbb{R}^k$ works as well). $\endgroup$ Jun 14, 2018 at 15:35
  • $\begingroup$ I am afraid that this is an oversimplification. It is not clear to me why the topological sum of $\mathbb R^n$ should not contain maximal homogeneous subspaces. For example, something dense and zero-dimensional? The trick with the sum of Hilbert space of the incresing density is that each homogeneous subspace is nowhere dense in large piece. So there is a place to enlarge this homogeneous space by another its piece. But in separable space there can happen that no place for such enlargement is available. $\endgroup$ Jun 14, 2018 at 17:44
  • $\begingroup$ In 1) "regular" can be removed, the whole condition 3) can be removed; instead of 4) one should require that each subset of $X_{n+1}$, which is homeomorphic to a subset of $X_n$ is not dense in $X_{n+1}$. Then the same argument works (under the condition of regularity of $X$). $\endgroup$ Jun 14, 2018 at 18:27
  • $\begingroup$ Thank you Taras! yes, now I see it -- for a moment I thought it could be made even simpler, but it can't $\endgroup$ Jun 14, 2018 at 20:47

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.