Exercise 2.1.9

Question

Show that if $S$ is a subset of a topological space $X$, then
\begin{enumerate}[label=(\alph*)]
\item $\overline{X \setminus S} = X \setminus \operatorname{int}(S)$,
\item $\operatorname{int}(X \setminus S) = X \setminus \bar{S}$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item \begin{proof} Using \href{./exercise 2-1-7}{Exercise 7} and \href{./exercise 2-1-8}{Exercise 8}, we can equivalently rewrite the theorem assertion as:
$$\bigcap \left\{ V \subset X \ \left|\ \begin{array}{ll} V 	ext{ is closed}\ S \subset V \end{array}ight.ight\} = X \setminus \bigcup \left\{ U \subset X \ \left|\ \begin{array}{ll} U 	ext{ is open}\ U \subset S \end{array}ight.ight\}.$$

Using De-Morgan's laws, we can rewrite the right-hand side as 
$$\bigcap \left\{ V \subset X \ \left|\ \begin{array}{ll} V 	ext{ is closed}\ S \subset V \end{array}ight.ight\} = \bigcap \left\{ X\setminus U \subset X \ \left|\ \begin{array}{ll} U 	ext{ is open}\ U \subset S \end{array}ight.ight\}.$$

Obviously, if $U$ is open, then $F \leftrightarrow X \setminus U$ is closed. Similarly, if $U$ is contained in $S$, then $F \leftrightarrow X \setminus U$ must contain $S$. Thus, by change of variables, we see that both sides must indeed be equal.

%$X \setminus \operatorname{int}(S)$ is a complement of an open set and is thus closed. Furthermore, it contains $X \setminus S$. Since $\overline{X \setminus S}$ is a closed hull of $X \setminus S$ (see \href{./exercise 2-1-8}{Exercise 8}), we have $\overline{X \setminus S} \subset X \setminus \operatorname{int}(S)$.
ewline
%For the other direction, let $x \in X \setminus \operatorname{int}(S)$. We have two possible cases: either $x$ is not boundary to $S$, or it is. In the former case, clearly $x \in X \setminus S \subset \overline{X \setminus S}$. In the latter case, $x$ is also a boundary point of $X\setminus S$ (cf. page 62) and thus $x \in \overline{X \setminus S}$.

\end{proof}

\item Follows directly from (a) by resetting $S:= X \setminus S$.
\end{enumerate}

Exercise 2.1.9

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Exercise 2.1.9

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