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.

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 $T$.)
$\square$

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #8799 by Maxime CAILLEUX on

Comment #9287 by Stacks project on