For a topological space $X$,
$\mathcal{H}(X)$ denotes the set of all autohomeomorphism of $X$. Note that, with the composition of functions, $(\mathcal{H}(X),\circ)$ is a group with identity $\mbox{Id}_{X}$.
If $x\in X$, let $\mathcal{O}(x,X)=\{h(x):h\in\mathcal{H}(X)\}$, we call this set the orbit of $x$ (under homeomorphism). Note that $x=\mbox{Id}_{X}(x)\in \mathcal{O}(x,X)$.
Note that for each $x\in X$, $\mathcal{O}(x,X)$ is fixed under autohomeomorphisms of $X$ (that is, for each $h\in\mathcal{H}(X)$, $h[\mathcal{O}(x,X)]=\mathcal{O}(x,X)$)
A topological space $X$ is called countable dense homogeneous if
- $X$ is separable, and
- for any pair $D, E$ of countable dense subsets there exists $h\in\mathcal{H}(X)$ such that $h[D]=E$.
A topological space $X$ is homogeneous if for every $x, y \in X$ there is $h\in \mathcal{H}(X)$ such that $h(x) = y$.
The motivation for this question is to demonstrate the following claim
- Every connected countable dense homogeneous space is homogeneous.
For this, we follow the following preliminary steps
- For every $x\in X$, $\overline{\mathcal{O}(x,X)}$ is a clopen subset of $X$.
- If $x,y\in X$ are such that $\overline{\mathcal{O}(x,X)}\cap \overline{\mathcal{O}(y,X)}\not=\emptyset$, then $\overline{\mathcal{O}(x,X)}=\overline{\mathcal{O}(y,X)}$.
- Let $x\in X$, then $\overline{\mathcal{O}(x,X)}$ is a countable dense homogeneous space.
- Let $x\in X$, then $\overline{\mathcal{O}(x,X)}=\mathcal{O}(x,X)$.
Finally, let $X$ be a connected countable dense homogeneous space and let $x,y\in X$. Note that $\mathcal{O}(x, X)$ and $\mathcal{O}(y, X)$ are clopen, also are non-empty sets. Then $\mathcal{O}(x, X)=\mathcal{O}(y, X)=X$. Therefore there is $h\in\mathcal{H}(X)$ such that $h(x)=y$.
Unfortunately I just couldn't prove step 4, someone has a suggestion on how to prove that, or some other new suggestion to directly prove the claim.
For the case of a first countable space we have that: