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.
5.21 Nowhere dense sets
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. It suffices to prove the lemma for two nowhere dense sets as the result in general will follow by induction. Let $A,B \subset X$ be nowhere dense subsets. We have $\overline{A \cup B} = \overline{A} \cup \overline{B}$. Hence, if $U \subset \overline{A \cup B}$ is an open subset of $X$, then $U \setminus U \cap \overline{B}$ is an open subset of $U$ and of $X$ and contained in $\overline{A}$ and hence empty. Similarly $U \setminus U \cap \overline{A}$ is empty. Thus $U = \emptyset $ as desired. $\square$
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. Because $f(X)$ is closed in $Y$ and $f$ is a homeomorphism of $X$ onto $f(X)$, we see that the closure of $f(T)$ in $Y$ equals $f(\overline{T})$. Hence if $V \subset Y$ is open and contained in the closure of $f(T)$, then $U = f^{-1}(V)$ is open and contained in $\overline{T}$. Hence $U = \emptyset $, which in turn shows that $V = \emptyset $ as desired. $\square$
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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