# Spaces without maximal homogeneous subspaces

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.

• 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. May 3, 2018 at 12:49
• 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. May 3, 2018 at 13:50

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. $\square$

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

• (So, if I didn't oversimplified Taras' construction, $X_k:=\mathbb{R}^k$ works as well). Jun 14, 2018 at 15:35
• 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. Jun 14, 2018 at 17:44
• 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 $X$). Jun 14, 2018 at 18:27
• Thank you Taras! yes, now I see it -- for a moment I thought it could be made even simpler, but it can't Jun 14, 2018 at 20:47