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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. –  Brian M. Scott Jul 21 '12 at 9:55
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I wish people would comment on their downvotes. I see nothing wrong with this question. –  Ben Millwood Jul 21 '12 at 16:47
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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? –  Matt N. Feb 18 '13 at 13:56
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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. –  Brian M. Scott Feb 18 '13 at 14:10
    
@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. : ) –  Matt N. Feb 28 '13 at 14:26

1 Answer 1

up vote 10 down vote accepted

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.
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