The Stacks project

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. 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 (2)

Comment #8799 by Maxime CAILLEUX on

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 be nowhere dense subsets. As for every pair of subsets of , so if is an open subset of , then, as but as is nowhere dense and , we have that and since is nowhere dense, .

For 5.21.5 a proof could be :

As is closed in , and as is an homeomorphism, is closed in and since it contains , so which then gives . As is nowhere dense, so as .


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