0
$\begingroup$

Let $(X,\tau), (Y,\sigma)$ be $T_2$-spaces such that there are injective lattice homomorphisms $f: \tau\to \sigma$ and $g:\sigma\to \tau$.

Does this imply that $(X,\tau)\cong (Y,\sigma)$?

$\endgroup$
3
$\begingroup$

No, $T_2$-spaces with this property aren't necessarily homeomorphic.

Let $\mathbb{N} = \{0,1,2,\ldots\}$ be the set of non-negative integers and let $\tau = {\cal P}(\mathbb{N})$ be the discrete topology, and define $$\sigma = {\cal P}(\mathbb{N}\setminus\{0\}) \cup \{U\subseteq \mathbb{N}: 0\in U\text{ and } \mathbb{N}\setminus U\text{ is finite}\}.$$

(Essentially $\sigma$ is the 1-point compactification of the discrete topology on ${\cal P}(\mathbb{N}\setminus \{0\})$.)

Clearly $(\mathbb{N},\tau)$ and $(\mathbb{N},\sigma)$ are not homeomorphic as only one of them is compact, but not the other. Moreover, the inclusion map $\iota: \sigma \to \tau$ is an injective lattice homomorphism, as well as the map $f:\tau \to \sigma$ defined by $A \mapsto A+1:=\{a+1: a\in A\}$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.