Exercise 17.19

Proof. Consider any $x\in \mathrm{Int}A$ so that there is a neighborhood of $x$ that is entirely contained in $A$. Then, for any $y\in U$, we have that $y\in A$ and hence $y\notin X-A$. This shows that $U$ does not intersect $X-A$, which suffices to show that $x$ is not in the closure of $X-A$ by Theorem 17.5 part (a). Thus $x$ is not in the boundary of $A$ since $\mathrm{Bd}A=\overline{A}\cap \overline{(X-A)}$. This of course shows that $\mathrm{Int}A$ and $\mathrm{Bd}A$ are disjoint since $x$ was arbitrary.

To show that $\overline{A}=\mathrm{Int}A\cup \mathrm{Bd}A$, first consider any $x\in \overline{A}$. If $x\in \mathrm{Int}A$ then clearly $x\in \mathrm{Int}A\cup \mathrm{Bd}A$, so assume that $x\notin \mathrm{Int}A$. Consider any neighborhood $U$ of $X$. Then it has to be that $U$ is not a subset of $A$ since otherwise, $x$ would be in the union of open subsets of $A$ and hence in the interior. It then follows that there is a point $y\in U$ where $y\notin A$ and therefore $y\in X-A$. This shows that $U$ intersects $X-A$ so that $x$ is in the closure of $X-A$ since $U$ was an arbitrary neighborhood. Since also $x\in \overline{A}$, we have that $x\in \overline{A}\cap \overline{(X-A)}=\mathrm{Bd}A$. Hence clearly $x\in \mathrm{Int}A\cup \mathrm{Bd}A$ so that $\overline{A}\subset \mathrm{Int}A\cup \mathrm{Bd}A$ since $x$ was arbitrary.

Now consider any $x\in \mathrm{Int}A\cup \mathrm{Bd}A$. If $x\in \mathrm{Int}A$ then also $x\in \overline{A}$ since we have that $\mathrm{Int}A\subset A\subset \overline{A}$. On the other hand, if $x\in \mathrm{Bd}A=\overline{A}\cap \overline{(X-A)}$ then of course $x\in \overline{A}$. This shows that $\mathrm{Int}A\cup \mathrm{Bd}A\subset \overline{A}$ in either case since $x$ was arbitrary. Since both directions have been shown, it follows that $\overline{A}=\mathrm{Int}A\cup \mathrm{Bd}A$ as desired. □

Proof. $\text{(⇒)}$ First suppose that $\mathrm{Bd}A=\varnothing$. Then by part (a) we have that $\overline{A}=\mathrm{Int}A\cup \mathrm{Bd}A=\mathrm{Int}A\cup \varnothing =\mathrm{Int}A$. Hence $A\subset \overline{A}=\mathrm{Int}A$ so that $A=\mathrm{Int}A$ since it is also always the case that $\mathrm{Int}A\subset A$. This shows that $A$ is open since $\mathrm{Int}A$ is always open. We also have $\overline{A}=\mathrm{Int}A\subset A$ so that $A=\overline{A}$ since it is always also the case that $A\subset \overline{A}$. This of course shows that $A$ is also closed since $\overline{A}$ is always closed.

$\text{(⇐)}$ Now suppose that $A$ is both open and closed. It then follows that $\overline{A}=A=\mathrm{Int}A$. So consider any $x\in \overline{A}$ so that also $x\in \mathrm{Int}A$. Then there is a neighborhood $U$ of $x$ contained entirely in $A$. Thus, for any point $y\in U$, we have that $y\in A$ so that $y\notin X-A$, which shows that $U$ does not intersect $X-A$. Since $U$ is a neighborhood of $x$, this shows that $x\notin \overline{X-A}$ by Theorem 17.5 part (a). Then, since $x$ was an arbitrary element of $\overline{A}$, it follows that $\overline{A}$ and $\overline{X-A}$ are disjoint so that $\mathrm{Bd}A=\overline{A}\cap \overline{(X-A)}=\varnothing$ as desired. □

Proof. $\text{(⇒)}$ First suppose that $U$ is open and consider any $x\in \mathrm{Bd}U$. Then we have that $x\in \overline{U}$ and $x\in \overline{X-U}$ since $\mathrm{Bd}U=\overline{U}\cap \overline{(X-U)}$ by definition. Suppose for the moment that $x\in U$ so that $U$ itself is a neighborhood of $x$ since it is open. For any $y\in U$ we have that $y\notin X-U$, and hence $U$ does not intersect $X-U$. This shows that $x$ is not in $\overline{X-U}$ by Theorem 17.5 part (a), which is a contradiction since we know it is. Thus it must be that $x\notin U$ so that $x\in \overline{U}-U$. This of course shows that $\mathrm{Bd}U\subset \overline{U}-U$ since $x$ was arbitrary.

Now consider any $x\in \overline{U}-U$ so that clearly $x\in \overline{U}$. Since also $x\notin U$, it follows that $x\in X-U$ so that of course $x\in \overline{X-U}$ as well. Hence $x\in \overline{U}\cap \overline{(X-U)}=\mathrm{Bd}U$, which shows that $\overline{U}-U\subset \mathrm{Bd}U$ since $x$ was arbitrary. This suffices to show that $\mathrm{Bd}U=\overline{U}-U$ as desired.

$\text{(⇐)}$ Now suppose that $\mathrm{Bd}U=\overline{U}-U$ and consider any $x\in U$. Then we have that $x\notin \overline{U}-U=\mathrm{Bd}U=\overline{U}\cap \overline{(X-U)}$. Since we know that $x\in \overline{U}$ (since $U\subset \overline{U}$), it must be that $x\notin \overline{X-U}$. Thus, by Theorem 17.5 part (a), there is a neighborhood of $V$ of $x$ that does not intersect $X-U$. This means that for any point $y\in V$, we have that $y\notin X-U$. Since of course $y\in X$, it follows that $y$ must be in $U$. This shows that $V\subset U$ since $y$ was arbitrary. Hence $V$ is a neighborhood of $x$ that is entirely contained in $U$ so that $x$ is in the union of open sets contained in $U$, hence $x\in \mathrm{Int}U$. Since $x$ was an arbitrary element of $U$, this shows that $U\subset \mathrm{Int}U$. As it is always the case that $\mathrm{Int}U\subset U$ as well, we have that $U=\mathrm{Int}U$ so that $U$ is open since $\mathrm{Int}U$ is always open. □

Proof. As a counterexample consider the set $U=\mathit{ℝnz}$ in the finite complement topology on $ℝ$. Clearly $U$ is open as its complement $ℝ-U=\left\{0\right\}$ is finite. It is also obvious that $U$ is an infinite set.

Now consider any real number $x$ and any neighborhood $V$ of $x$. It cannot be that $ℝ-V$ is all of $ℝ$ since then $V$ would be empty, and we know that $x\in V$. So it must be that $ℝ-V$ is finite since $V$ is open, which means that there are only a finite number of real numbers that are not in $V$. However, since $U$ is infinite, there must be an element of $U$ that is in $V$ (in fact there are an infinite number of such elements). Hence $V$ intersects $U$ so that $x\in \overline{U}$ by Theorem 17.5 part (a). Since $x\in ℝ$ was arbitrary, it must be that $\overline{U}$ is all of $ℝ$.

Clearly, $ℝ$ is open (since the set is always open in any topology on that set) so that $\mathrm{Int}(\overline{U})=\mathrm{Int}ℝ=ℝ$. Then, since $0\in ℝ=\mathrm{Int}(\overline{U})$ but $0\notin U$, we have that $U\ne \mathrm{Int}(\overline{U})$. □