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Dense and Nowhere Dense Sets in Topological Spaces Examples 1
Recall from the Dense and Nowhere Dense Sets in a Topological Space page that if $(X, \tau)$ is a topological space than a set $A \subseteq X$ is said to be dense in $X$ if the intersection of $A$ with any $U \in \tau$ is nonempty, that is, for all $U \in \tau$:
(1)Furthermore, we said that $A$ is nowhere dense in $X$ if the interior of the closure of $A$ is empty, i.e.,:
(2)We will now look at some example exercises regarding dense and nowhere dense sets in topological spaces.
Example 1
Consider the topological space $(\mathbb{R}^2, \tau)$ where $\tau$ is the usual topology formed by open disks. Consider the subset $A = \{ (x, y) \in \mathbb{R}^2 : x \in \mathbb{N}, y \in \mathbb{R} \} \subseteq \mathbb{R}^2$. Is this set dense? Is this set nowhere dense?
We claim that the set $A$ is NOT dense. To show this, we must find an open neighbourhood $U$ of $\mathbb{R}^2$ whose intersection with $A$ is empty.
Consider the following open disk in $\mathbb{R}^2$:
(3)This is a disk centered at the point $\left ( \frac{1}{2}, 0 \right \}$ and with radius $\frac{1}{2}$ that does not intersect $A$ as seen in the diagram below:
Therefore $A \cap U = \emptyset$ so $A$ is not dense.
So, let's determine whether $A$ is nowhere dense. Note that $A$ itself is a closed set since $A^c$ is open. Therefore $\bar{A} = A$. Furthermore, $\mathrm{int}(\bar{A}) = \emptyset$, so $A$ is nowhere dense.
Example 2
Consider the topological space $(\mathbb{R}^2, \tau)$ where $\tau$ is the usual topology formed by open disks. Consider the subset $A = \{ (x, y) \in \mathbb{R}^2 : xy \neq 0 \} \subseteq \mathbb{R}^2$. Is this set dense? Is this set nowhere dense?
Note that $A$ is the set of points $(x, y) \in \mathbb{R}^2$ which satisfy $xy \neq 0$. Note that $xy \neq 0$ if and only if $x \neq 0$ and $y \neq 0$. If $x = 0$ then the point $(x, y) = (0, y)$ lies on the $y$axis, and if $y = 0$ then the point $(x, y) = (x, 0)$ lies on the $x$axis. Therefore $A$ is equal to $\mathbb{R}^2$ with the $x$axis and $y$axis removed.
We claim that this set is dense. To show this, let $U$ be any open disk in $\mathbb{R}^2$. If this disk does not intersect the $x$ or $y$ axes then $U$ is fully contained in $A$ and so $A \cap U \neq \emptyset$. If not, then this open disk intersects either axis at either one point or an infinite number of points. In either case, the open disk will still contain elements of $A$ and so $A \cap U \neq \emptyset$ for every open disk $U$ in $\mathbb{R}^2$.
Since these open disks form a basis of the topology on $\mathbb{R}^2$ we see that the intersection of $A$ with any open set in $\mathbb{R}^2$ is nonempty and set $A$ is a dense set.
Furthermore, this set is clearly not nowhere dense. Note that $A$ is an open set and that $\bar{A} = \mathbb{R}^2$. Furthermore, $\mathrm{int}(\bar{A}) = \mathbb{R}^2 \neq \emptyset$.