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Is every homeomorphism between topological spaces an order isomorphism (for orders of inclusion $\subseteq$ of sets)?

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    Of course: every **bijection** is, whether it’s a homeomorphism or not.2012-07-21
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    I wish people would comment on their downvotes. I see nothing wrong with this question.2012-07-21
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    @BrianM.Scott Maybe he meant open sets: otherwise it seems to be irrelevant whether there are topologies or not. And then the answer will only be yes if $f$ is a homeomorphism, I think, or am I missing something?2013-02-18
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    @Matt: Apparently he did, judging from the accepted answer. However, two different topological spaces can have the same partial order of open sets: consider $X$ and $X\times T$, where $T$ is the indiscrete two-point space.2013-02-18
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    @BrianM.Scott Thank you for pointing that out! In retrospect you may ignore my previous comment since your first comment answers the question whether OP wants open sets or all sets. : )2013-02-28

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Every bijection $f \colon X\to Y$ induces an order-isomorphism between $(\mathcal P(X),\subseteq)$ and $(\mathcal P(Y),\subseteq)$.

This follows easily from the following two observations:

  • $A\subseteq B$ $\Rightarrow$ $f[A]\subseteq f[B]$ for any map $f$
  • $f^{-1}[f[A]]=A$, if $f$ is a bijection.