Section 5.21 (03HM): Nowhere dense sets

5.21 Nowhere dense sets

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

Given a subset $T \subset X$ the interior of $T$ is the largest open subset of $X$ contained in $T$.
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. Denote $\overline{T}_ i$ the closure of $T \cap U_ i$ in $U_ i$. We have $\overline{T} \cap U_ i = \overline{T}_ i$. Taking the interior commutes with intersection with opens, thus

\[ (\text{interior of }\overline{T}) \cap U_ i = \text{interior of }(\overline{T} \cap U_ i) = \text{interior in }U_ i\text{ of }\overline{T}_ i \]

By assumption the last of these is empty. Hence $T$ is nowhere dense in $X$. $\square$

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 $T$.) $\square$

Comments (1)

Comment #8799 by Maxime CAILLEUX on February 12, 2024 at 07:40

For 5.21.2 a proof could be :

Naturally it suffices to show it for the union of two nowhere dense sets as the result will follow by induction. So, let $A,B\subset X$ be nowhere dense subsets. As for every pair of subsets of $X$ , $\overline{A\cup B}=\overline{A}\cup\overline{B}$ so if $\mathcal{U}\subset\overline{A\cup B}$ is an open subset of $X$ , then, as $\mathcal{U}=\left(\mathcal{U}\cap\overline{A}\right)\bigcup\left(\mathcal{U}\cap(X-\overline{A})\right)$ but $\mathcal{U}\cap(X-\overline{A})=\emptyset$ as $B$ is nowhere dense and $\mathcal{U}\cap(X-\overline{A})\subset\overline{B}$ , we have that $\mathcal{U}\subset\overline{A}$ and since $A$ is nowhere dense, $\mathcal{U}=\emptyset$ .

For 5.21.5 a proof could be :

As $f(X)$ is closed in $Y$ , $\overline{f(T)}\subset f(X)$ and as $f:X\rightarrow f(X)$ is an homeomorphism, $f(\overline{T})$ is closed in $f(X)$ and since it contains $f(T)$ , $\overline{f(T)}\subset f(\overline{T})$ so $(\text{interior of }\overline{f(T)})\subset f(\overline{T})$ which then gives $f^{-1}((\text{interior of }\overline{f(T)}))\subset f^{-1}(f(\overline{T}))=\overline{T}$ . As $T$ is nowhere dense, $f^{-1}((\text{interior of }\overline{f(T)}))=\emptyset$ so $(\text{interior of }\overline{f(T)})=\emptyset$ as $(\text{interior of }\overline{f(T)})\subset f(X)$ .

The Stacks project

5.21 Nowhere dense sets

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