**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. $\square$