3
$\begingroup$

Suppose $X\neq \emptyset$ is a set. Let $\tau_1, \tau_2$ be Hausdorff topologies on $X$ with the property that the partially ordered sets $(\tau_1,\subseteq)$ and $(\tau_2,\subseteq)$ are order-isomorphic.

Does this imply that $(X,\tau_1)\cong (X,\tau_2)$ as topological spaces?


If Hausdorffness is dropped, then there are easy counterexamples, for instance $X = \omega$ and $\tau_1 = \{\emptyset,\{0\},\omega\}, \tau_2 = \{\emptyset, \omega\setminus\{0\},\omega\}$.

$\endgroup$
3
$\begingroup$

If $h:(\tau_1, \subseteq) \to (\tau_2, \subseteq)$ is an order isomorphism.

Define $M(\tau_1) = \{O \in \tau_1: O \neq X \land \forall O' \in \tau_1: O \subseteq O' \implies O' = X\}$, the "submaximal" elements.

Because the definition is purely order theoretical, $h[M(\tau_1)] = M(\tau_2)$.

Then all $O \in M(\tau_1)$ are of the form $X\setminus \{p\}$ for some $p \in X$ when $X$ is a $T_1$ space: if $|X \setminus O| \ge 2$ pick $x$ in this complement, and $O \subseteq X\setminus \{x\} \subseteq X$, where the middle set is open by $T_1$-ness, this shows $O \notin M(\tau_1)$. So $\forall O \in M(\tau_1): |X \setminus O| =1$. Also, all sets of the form $X\setminus \{p\}$ are in $M(\tau_1)$.

So the above implies that if both topologies are $T_1$, we can define $f: X \to X$ by $\{f(x)\} = X\setminus h(M\setminus \{x\})$.

One can now check that $f$ is the required homeomorphism, using that $h$ is an order isomorphism.

$\endgroup$
3
$\begingroup$

There is more than an implication of a homeomorphism but actually an induced homeomorphism. It is so for all $T_1$-spaces.

Let $\ (X\ \tau)\ $ be a $T_1$-space. Then points of $\ X\ $ are characterized by the topology (treated as a Birkhoff lattice) as maximal elements of $\ \tau\ $ which are strictly contained in $\ X$.

$\endgroup$

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.