
Lemma 5.25.1. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that $f(X)$ is dense in $Y$ and that $Y$ is Hausdorff. Then the cardinality of $Y$ is at most the cardinality of $P(P(X))$ where $P$ is the power set operation.

Proof. Let $S = f(X) \subset Y$. Let $\mathcal{D}$ be the set of all closed domains of $Y$, i.e., subsets $D \subset Y$ which equal the closure of its interior. Note that the closure of an open subset of $Y$ is a closed domain. For $y \in Y$ consider the set

$I_ y = \{ T \subset S \mid \text{ there exists } D \in \mathcal{D}\text{ with }T = S \cap D\text{ and }y \in D\} .$

Since $S$ is dense in $Y$ for every closed domain $D$ we see that $S \cap D$ is dense in $D$. Hence, if $D \cap S = D' \cap S$ for $D, D' \in \mathcal{D}$, then $D = D'$. Thus $I_ y = I_{y'}$ implies that $y = y'$ because the Hausdorff condition assures us that we can find a closed domain containing $y$ but not $y'$. The result follows. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).