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Let $\mathcal T$ and $\mathcal S$ be two topologies on a set $X$ and $(X,\mathcal T)$ and $(X,\mathcal S)$ be homeomoric and compact and Hausdorff.

Is $\mathcal S$ equal to $\mathcal T$?

I know the answer is No (one can relabel the elements and get a homeomorphic one). But I'm looking for a good counterexample.

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    $\begingroup$ Take $X=[0,1]$ with the usual topology $\mathcal T$. Now, simply relabel the elements by any bijection that isn't a homeomorphism $(X,\mathcal T)\to(X,\mathcal T)$, for example the one interchanging $0$ and $1$. $\endgroup$ – Dejan Govc Sep 2 '14 at 8:34
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Let $X = \mathbb{N}$ and let

$\tau_1 = \mathcal{P}(\mathbb{N} \setminus \{1\}) \cup \{U\subseteq \mathbb{N}: 1\in U\textrm{ and } \mathbb{N}\setminus U \textrm{ is finite } \}$ and

$\tau_2 = \mathcal{P}(\mathbb{N} \setminus \{2\}) \cup \{U\subseteq \mathbb{N}: 2\in U\textrm{ and } \mathbb{N}\setminus U \textrm{ is finite } \}$.

(Note that $\mathcal{P}(.)$ denotes the power set.)

Then $(\mathbb{N},\tau_1)$ and $(\mathbb{N},\tau_2)$ are homeomorphic, and both topological spaces are compact and Hausdorff, but $\tau_1 \neq\tau_2$ because $\{2\} \in \tau_1 \setminus \tau_2$.

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