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Let $(X,\tau)$ be a topological space. We assign to $(X,\tau)$ an equivalence relation $\simeq_{(X,\tau)}$ in the following way:

$x\simeq_{(X,\tau)} y$ if and only if there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x) = y$.

(Reflexivity, symmetry, and transitivity of this relation are easy to see.)

Given a non-empty set $X$ and an equivalence relation $\simeq$, is there a topology $\tau$ on $X$ such that $\simeq$ equals $\simeq_{(X,\tau)}$?

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I think the following works in quite a lot of situations:

  • The equivalence relation $\cong$ induces a partition $\mathcal{P}$.
  • Under AC we can well-order $\mathcal{P}$, and for every $P \in \mathcal{P}$, define $U_{P} = \bigcup_{Q < P} Q$.
  • The $U_{P}$ form a topology $\tau$, and for all $x,y \in P$, we have $x \cong_{\tau} y$.

NB: It is clear that if $\mathcal{P}$ is countable, then this answers your question. (And we don't need AC.)

Edit: I changed the definition of $U_{P}$ to $\bigcup_{Q < P} Q$, as per Joseph van Name's comment below.

Problem: For the trivial equivalence relation $\cong$ on $\mathbb{R}$, this gives the long-line topology $\tau$, and $x \cong_{\tau} y$, for all $x,y \in \mathbb{R}$.

Edit: The problem above is not a problem, because a well-ordering on $\mathbb{R}$ does not induce the long-line topology. See Eric Wofsey's comment below.

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    $\begingroup$ This is a good answer. I suppose that you meant $U_{P}=\bigcup_{Q<P}Q$ instead of $U_{P}=\bigcup_{Q\leq P}Q$ to cover the limit ordinal case. Also, one can turn this example into a $T_{0}$-example simply by partially ordering $X$ by letting $x<y$ iff $x\in U_{P},y\in Y_{Q}$ and $P<Q$ and giving $X$ the specialization ordering. Of course, for finite spaces, $T_{0}$ is the highest well known separation axiom that always gives you an example since every finite $T_{1}$-space has the discrete topology. $\endgroup$
    – Joseph Van Name
    Commented Feb 23, 2015 at 16:06
  • $\begingroup$ @JosephVanName — My set theory is not that good, so I'm fine with changing it to $U_{P} = \bigcup_{Q<P} Q$. In the meantime I haven't found a solution to the problem I sketched in my answer… So I'm not sure what the answer to OP's question will be in general. $\endgroup$
    – jmc
    Commented Feb 23, 2015 at 19:03
  • $\begingroup$ This construction always works; countability has nothing to do with it. The trivial equivalence relation on $\mathbb{R}$ will just give the Alexandrov topology on some well-ordering of $\mathbb{R}$, which is rigid since all well-ordered sets are rigid. $\endgroup$
    – Eric Wofsey
    Commented Feb 24, 2015 at 1:37
  • $\begingroup$ @EricWofsey — I see, I confused the usual ordering on $\mathbb{R}$ with a well-ordering (stupid mistake!). I did not know that all well-ordered sets are rigid; but if that is true, then my answer is actually a full answer to OP's question. Do you have a reference? Feel free to copy my answer, and add this new information, so that you can get rep for it. Yours should be the accepted answer. $\endgroup$
    – jmc
    Commented Feb 24, 2015 at 7:56
  • $\begingroup$ Thanks for your comments, and in particular your construction @jmc! $\endgroup$
    – Dominic van der Zypen
    Commented Feb 24, 2015 at 8:01

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