Dense and Nowhere Dense Sets in a Topological Space

Dense Sets in a Topological Space

Definition: Let $(X, \tau)$ be a topological space. The set $A \subseteq X$ is said to be Dense in $X$ if the intersection of every nonempty open set with $A$ is nonempty, that is, $A \cap U \neq \emptyset$ for all $U \in \tau \setminus \{ \emptyset \}$.

Given any topological space $(X, \tau)$ it is important to note that $X$ is dense in $X$ because every $U \in \tau$ is such that $U \subseteq X$, and so $X \cap U = U \neq \emptyset$ for all $U \in \tau \setminus \{ \emptyset \}$.

For another example, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals. Then the set of rational numbers $\mathbb{Q} \subset \mathbb{R}$ is dense in $\mathbb{R}$. If not, then there exists an $U \in \tau \setminus \{ \emptyset \}$ such that $\mathbb{Q} \cap U = \emptyset$.

Since $U \in \tau$ we have that $(a, b) \subseteq U$ for some open interval $(a, b)$ with $a, b \in \mathbb{R}$ and $a < b$. Suppose that $\mathbb{Q} \setminus U = \emptyset$. Then we must also have that:

(1)

\begin{align} \quad \mathbb{Q} \cap U = \mathbb{Q} \cap (a, b) = \emptyset \end{align}

The intersection above implies that there exists no rational numbers in the interval $(a, b)$, i.e., there exists no $q \in \mathbb{Q}$ such that $a < q < b$. But this is a contradiction since for all $a, b \in \mathbb{R}$ with $a < b$ there ALWAYS exists a rational number $q \in \mathbb{Q}$ such that $a < q < b$, i.e., $q \in (a, b)$. So $\mathbb{Q} \cap (a, b) \neq \emptyset$ for all $U \in \tau \setminus \{ \emptyset \}$. Thus, $\mathbb{Q}$ is dense in $\mathbb{R}$.

We will now look at a very important theorem which will give us a way to determine whether a set $A \subseteq X$ is dense in $X$ or not.

Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is dense in $X$ if and only if $\bar{A} = X$.

• Proof: $\Rightarrow$ Suppose that $A$ is dense in $X$. Then for all $U \in \tau \setminus \{ \emptyset \}$ we have that $A \cap U = \emptyset$. Clearly $\bar{A} \subseteq X$ so we only need to show that $X \subseteq \bar{A}$.

Nowhere Dense Sets in a Topological Space

Definition: Let $(X, \tau)$ be a topological space. A set $A \subseteq X$ is said to be Nowhere Dense in $X$ if the interior of the closure of $A$ is empty, that is, $\mathrm{int} (\bar{A}) = \emptyset$.

For example, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usually topology of open intervals on $\mathbb{R}$, and consider the set of integers $\mathbb{Z}$. The closure of $\mathbb{Z}$, $\bar{\mathbb{Z}}$ is the smallest closed set containing $\mathbb{Z}$. The smallest closed set containing $\mathbb{Z}$ is $\mathbb{Z}$ since $\mathbb{Z}^c$ is open as $\mathbb{Z}^c$ is an arbitrary union of open sets:

(2)

\begin{align} \quad \mathbb{Z}^c = ... (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (1, 2) \cup ... \end{align}

So what is the interior of $\bar{\mathbb{Z}} = \mathbb{Z}$? It is the largest open set contained in $\bar{\mathbb{Z}} = \mathbb{Z}$. All open sets of $\mathbb{R}$ with respect to this topology $\tau$ are either the empty set, an open interval, a union of open intervals, or the whole set (the union of all open intervals). But no open intervals are contained in $\mathbb{Z}$ and so:

(3)

\begin{align} \quad \mathrm{int} (\bar{\mathbb{Z}}) = \emptyset \end{align}

Therefore $\mathbb{Z}$ is a nowhere dense set in $\mathbb{R}$ with respect to the usual topology $\tau$ on $\mathbb{R}$.