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*}

5.21 Nowhere dense sets

Definition 5.21.1. Let $X$ be a topological space.

  1. Given a subset $T \subset X$ the interior of $T$ is the largest open subset of $X$ contained in $T$.

  2. A subset $T \subset X$ is called nowhere dense if the closure of $T$ has empty interior.

Lemma 5.21.2. Let $X$ be a topological space. The union of a finite number of nowhere dense sets is a nowhere dense set.

Proof. Omitted. $\square$

Lemma 5.21.3. Let $X$ be a topological space. Let $U \subset X$ be an open. Let $T \subset U$ be a subset. If $T$ is nowhere dense in $U$, then $T$ is nowhere dense in $X$.

Proof. Assume $T$ is nowhere dense in $U$. Suppose that $x \in X$ is an interior point of the closure $\overline{T}$ of $T$ in $X$. Say $x \in V \subset \overline{T}$ with $V \subset X$ open in $X$. Note that $\overline{T} \cap U$ is the closure of $T$ in $U$. Hence the interior of $\overline{T} \cap U$ being empty implies $V \cap U = \emptyset $. Thus $x$ cannot be in the closure of $U$, a fortiori cannot be in the closure of $T$, a contradiction. $\square$

Lemma 5.21.4. Let $X$ be a topological space. Let $X = \bigcup U_ i$ be an open covering. Let $T \subset X$ be a subset. If $T \cap U_ i$ is nowhere dense in $U_ i$ for all $i$, then $T$ is nowhere dense in $X$.

Proof. Omitted. (Hint: closure commutes with intersecting with opens.) $\square$

Lemma 5.21.5. Let $f : X \to Y$ be a continuous map of topological spaces. Let $T \subset X$ be a subset. If $f$ is a homeomorphism of $X$ onto a closed subset of $Y$ and $T$ is nowhere dense in $X$, then also $f(T)$ is nowhere dense in $Y$.

Proof. Omitted. $\square$

Lemma 5.21.6. Let $f : X \to Y$ be a continuous map of topological spaces. Let $T \subset Y$ be a subset. If $f$ is open and $T$ is a closed nowhere dense subset of $Y$, then also $f^{-1}(T)$ is a closed nowhere dense subset of $X$. If $f$ is surjective and open, then $T$ is closed nowhere dense if and only if $f^{-1}(T)$ is closed nowhere dense.

Proof. Omitted. (Hint: In the first case the interior of $f^{-1}(T)$ maps into the interior of $T$, and in the second case the interior of $f^{-1}(T)$ maps onto the interior of $T$.) $\square$

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