If \(A \subset X\), we define the boundary of \(A\) by the equation
\[ \operatorname{Bd} A = \overline{A} \cap \overline{(X - A)}. \]
Show that \(\operatorname{Int} A\) and \(\operatorname{Bd} A\) are disjoint, and \(\overline{A} = \operatorname{Int} A \cup \operatorname{Bd} A\).
Show that \(\operatorname{Bd} A = \varnothing\) if and only if \(A\) is both open and closed.
Show that \(U\) is open if and only if \(\operatorname{Bd} U = \overline{U} - U\).
If \(U\) is open, is it true that \(U = \operatorname{Int}(\overline{U})\)? Justify your answer.
Solution
Throughout, \(\overline{X-A}=X-\operatorname{Int}A\), since \(\operatorname{Int}A\) is the largest open set contained in \(A\) and \(\overline{X-A}\) is the smallest closed set containing \(X-A\).
1. (a) \(\operatorname{Int}A\cap\operatorname{Bd}A=\varnothing\) and \(\overline A=\operatorname{Int}A\cup\operatorname{Bd}A\).
\(\operatorname{Bd}A\subseteq\overline{X-A}=X-\operatorname{Int}A\), so \(\operatorname{Bd}A\) misses \(\operatorname{Int}A\). Since \(\operatorname{Int}A\subseteq\overline A\), \[ \overline A=\operatorname{Int}A\cup\bigl(\overline A\cap(X-\operatorname{Int}A)\bigr)=\operatorname{Int}A\cup\bigl(\overline A\cap\overline{X-A}\bigr)=\operatorname{Int}A\cup\operatorname{Bd}A. \]
2. (b) \(\operatorname{Bd}A=\varnothing\) if and only if \(A\) is both open and closed.
If \(\operatorname{Bd}A=\varnothing\), step 1 gives \(\overline A=\operatorname{Int}A\), so \(A\subseteq\overline A=\operatorname{Int}A\subseteq A\). Hence \(A=\operatorname{Int}A\) is open and \(A=\overline A\) is closed. Conversely, if \(A\) is open and closed, then \(\overline A=A\) and \(\overline{X-A}=X-A\), so \(\operatorname{Bd}A=A\cap(X-A)=\varnothing\).
3. (c) \(U\) is open if and only if \(\operatorname{Bd}U=\overline U-U\).
Since \(\overline{X-U}\supseteq X-U\) and \(\overline U\cap(X-U)=\overline U-U\), \[ \operatorname{Bd}U=(\overline U-U)\cup\bigl(\overline U\cap(\overline{X-U}-(X-U))\bigr). \] The set \(\overline{X-U}-(X-U)\) is contained in \(U\subseteq\overline U\), so the second term equals \(\overline{X-U}-(X-U)\), which is disjoint from \(\overline U-U\). Hence \(\operatorname{Bd}U=\overline U-U\) if and only if \(\overline{X-U}=X-U\), that is, if and only if \(X-U\) is closed, that is, if and only if \(U\) is open.
4. (d) An open set \(U\) need not equal \(\operatorname{Int}(\overline U)\).
In \(\RR\), the open set \(U=\RR-\{0\}\) has \(\overline U=\RR\), so \(\operatorname{Int}(\overline U)=\RR\ne U\).
5. Q.E.D.
Steps 1, 2, 3, and 4 answer (a), (b), (c), and (d).