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.

  • 3
    $\begingroup$ 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

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 $\{2\} \in \tau_1 \setminus \tau_2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.