I have the following question:
Show that two discrete spaces are homeomorphic iff they have the same cardinality:
I have tried the following:
Let $f: (X, \mathcal{T}_{discrete}) \to (Y, \mathcal{T}_{discrete})$
$(\Rightarrow )$
As we have that the topology is the discrete then in particular each $\{x\} \in X$ is an open set, as we have that $f$ in a homeomorphism then $\{f(x)\} \in Y$ is an open set, this allows to conclude that both spaces have the same cardinality.
$(\Leftarrow )$
Let suppose that both spaces have the same cardinality, since both spaces are discrete then we have that each $\{x\} \in X$ is an open set and analogous each $\{y\} \in Y$ is an open set.
If we define the function $f$ as
$$f: (X, \mathcal{T}_{discrete}) \to (Y, \mathcal{T}_{discrete}) \text{ as } x_i \to y_i i \in |X|$$
then we have that is a bijective map, and for the affirmation that I made before $f$ and $f^{-1}$ are continuous. So we have that both spaces are homeomorphic.
Is correct what I have done?, thank you
Best Answer
If both sides have discrete topology, then any map is continuous. So as long as you have a bijection it is a homeomorphism. Thus this is true iff the underlying sets have same cardinality.