# saying if two topologies are homeomorphic

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

• Are the definitions of the two topologies written correctly? Sep 28, 2023 at 17:10
• No my bad, let me fix Sep 28, 2023 at 17:11
• They are still not written correctly, the topology needs to be a subset of the powerset of $[0,1]$ Sep 28, 2023 at 17:59
• @Carlyle to me this definition seems clear why do you think is poorly written? Sep 28, 2023 at 18:23
• @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 Sep 28, 2023 at 18:59

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.
• 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 Sep 28, 2023 at 17:42