Homeomorphisms and ##\mathbb{R}##

[kent davidge]

Is it hard to show that $\mathbb{R}$ with the usual topology and $\mathbb{R}$ with the discrete topology are not homeomorphic? I'm reasoning this way: in the discrete topology there are every possible subsets of $\mathbb{R}$, which includes those with just one element of the type $\{x \}$. There cannot be bijection from an open subset of the usual topology, which requires a distance greater than zero, and a set with just one element. Is this sufficient in proving that these two topological spaces are not homeomorphic?

 

[fresh_42]

Is it hard to show that $\mathbb{R}$ with the usual topology and $\mathbb{R}$ with the discrete topology are not homeomorphic? I'm reasoning this way: in the discrete topology there are every possible subsets of $\mathbb{R}$, which includes those with just one element of the type $\{x \}$. There cannot be bijection from an open subset of the usual topology, which requires a distance greater than zero, and a set with just one element. Is this sufficient in proving that these two topological spaces are not homeomorphic?

You already said it. Assume a homeomorphism $f \, : \,\mathbb{R}_E \longrightarrow \mathbb{R}_D$ from the Euclidean reals to the discrete reals. Now $\{x\} \subseteq \mathbb{R}_D$ is open. Now what is required for $f^{-1}(\{x\}) \subseteq \mathbb{R}_E\,$?

 

[kent davidge]

Now what is required for $f^{-1}(\{x\}) \subseteq \mathbb{R}_E\,$?

To be open?

 

[fresh_42]

Yes, $f$ has to be continuous, i.e. pre-images of open sets have to be open. But open sets in Euclidean $\mathbb{R}_E$ contain at least two elements or none. Either way, if $f$ is a bijection, we have only one available for $f^{-1}(\{x\})$.