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let $ \tau := \{A\subseteq [0,1] : 0\in A\} \cup \{[0,1]\} $

And $\sigma := \{A\subseteq [0,1] : 0\notin A\} \cup \{[0,1]\} $

Prove that $( x,\tau)$, $(x,\sigma)$ with $x=[0,1]$ Are homeomorphic.

I already proved that the two family of sets are topologies, therefore I need to find out if there exists an $f:( x,\tau)\rightarrow(x,\sigma)$ that's continues, bjective and such that $f^{-1}$ is also continuous

I guess that the two topological spaces aren't homeomorphic, but I don't know how to prove that and also how to prove the case in Which they're homeomorphic, do I need to find out the exact analytical function usually?

The only thing I can use to prove it is the denition of continuous function ( the retro image of an open set is also open) but I don't know how to do it, if the Open set it's ∅ or X self this propety uholds but I don't know how to do about other open sets in sigma

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  • $\begingroup$ Are the definitions of the two topologies written correctly? $\endgroup$ Sep 28, 2023 at 17:10
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    $\begingroup$ No my bad, let me fix $\endgroup$ Sep 28, 2023 at 17:11
  • $\begingroup$ They are still not written correctly, the topology needs to be a subset of the powerset of $[0,1]$ $\endgroup$
    – Carlyle
    Sep 28, 2023 at 17:59
  • $\begingroup$ @Carlyle to me this definition seems clear why do you think is poorly written? $\endgroup$ Sep 28, 2023 at 18:23
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    $\begingroup$ @TurquoiseTilt They are not subsets of the powerset, so they are not Topologies. I don't think its poorly written I think OP just made a mistake in copying them down, probably just forgot the "\" before "{" and "}". I have suggested the edit, but OP still needs to confirm that that is what they meant $\endgroup$
    – Carlyle
    Sep 28, 2023 at 18:59

1 Answer 1

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Looking at the definition of $\sigma$ we can see that all the singleton $\{x\}$ with $x\in (0,1]$ are open sets so for each of them in order to have a continuous function we need $f^{-1}(\{x\})$ to be open so $0\in f^{-1}(\{x\}), \forall x \in (0,1]$. This means that $f(0) = (0,1]$ and is not even a function. Thus every continuous "function" that are not constant could not be a function, but constant function aren't bijection.

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  • $\begingroup$ Do you proved that there are more elements with the same retroinage 0? $\endgroup$ Sep 28, 2023 at 17:30
  • $\begingroup$ Well the fact is, if you want $f$ to be continuous $0$ must be in the preimage but this lead to some well definition problem of the function $f$ itself $\endgroup$ Sep 28, 2023 at 17:42

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