Exercise 17.4

Question

Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U-A$ is open in $X$, and $A-U$ is closed in $X$.

Dan Whitman · Accepted Answer

\begin{lemma}\label{lem:closedlim:setmid}
  If $A$, $B$, and $C$ are sets then $A - (B - C) = (A-B) \cup (A \cap C)$.
\end{lemma}
\begin{proof}
  We show this by a sequence of logical equivalences:
  \begin{align*}
    x \in A - (B - C) &\Leftrightarrow x \in A \land x 
otin B - C \
                      &\Leftrightarrow x \in A \land \lnot(x \in B \land x 
otin C) \
                      &\Leftrightarrow x \in A \land (x 
otin B \lor x \in C) \
                      &\Leftrightarrow (x \in A \land x 
otin B) \lor (x \in A \land x \in C) \
                      &\Leftrightarrow x \in A - B \lor x \in A \cap C \
                      &\Leftrightarrow x \in (A-B) \cup (A \cap C)
  \end{align*}
  as desired.
\end{proof}

\begin{corollary}\label{cor:closedlim:cc}
  If $A \subset X$ and $B = X - A$, then $A = X - B$.
\end{corollary}
\begin{proof}
  By Lemma~ef{lem:closedlim:setmid}, we have that
  \begin{gather*}
    X - B = X - (X - A) = (X - X) \cup (X \cap A) = \varnothing \cup (X \cap A) = X \cap A = A
  \end{gather*}
  since $A \subset X$.
\end{proof}

extbf{Main Problem.}
\begin{proof}
  First, since $A$ is closed in $X$, we have that $B = X - A$ is open in $X$, and it follows from Corollary~ef{cor:closedlim:cc} that $A = X - B$.
  Then we have that
  \begin{gather*}
    U - A = U - (X - B) = (U - X) \cup (U \cap B)
  \end{gather*}
  by Lemma~ef{lem:closedlim:setmid}.
  Since $U \subset X$, it follows that $U - X = \varnothing$, and hence
  \begin{gather*}
    U - A = \varnothing \cup (U \cap B) = U \cap B \,.
  \end{gather*}
  Then, since both $U$ and $B$ are open, their intersection is as well and therefore $U - A$ is open.

Next, we have by Lemma~ef{lem:closedlim:setmid}
  \begin{gather*}
    X - (A - U) = (X - A) \cup (X \cap U) = B \cup (X \cap U) = B \cup U.
  \end{gather*}
  since $U \subset X$ so that $X \cap U = U$.
  Since both $B$ and $U$ are open, clearly their union is as well and hence $X - (A - U)$ is open.
  This of course means that $A - U$ is closed by definition.
\end{proof}

Exercise 17.4

Answers

Comments

Exercise 17.4

Answers

Comments

Add answer