The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$


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