Exercise 17.3

Question

Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A 	imes B$ is closed in $X 	imes Y$.

Amit Awasthi · Accepted Answer

\begin{proof}
  We see that $X \setminus A, Y \setminus B$ are open in $X,Y$ respectively by definition of a closed set. By definition of the product topology, $(X \setminus A) 	imes Y, X 	imes (Y \setminus B)$ are open in $X 	imes Y$. We see that $(X \setminus A) 	imes Y = (X 	imes Y) \setminus (A 	imes Y), X 	imes (Y \setminus B) = (X 	imes Y) \setminus (X 	imes B)$, and so $A 	imes Y, X 	imes B$ are closed in $X 	imes Y$. Finally, $A 	imes B = (A 	imes Y) \cap (X 	imes B)$, and so is the intersection of closed sets, i.e., closed.
\end{proof}

Dan Whitman · Answer

\begin{lemma}\label{lem:closedlim:setid}
  If $X$, $Y$, $A$, and $B$ are sets then $X 	imes Y - A 	imes B = (X-A) 	imes Y \cup X 	imes (Y - B)$.
\end{lemma}
\begin{proof}
  We show this via logical equivalences:
  \begin{align*}
    (x,y) \in X 	imes Y - A 	imes B &\Leftrightarrow (x,y) \in X 	imes Y \land (x,y) 
otin A 	imes B \
                                      &\Leftrightarrow (x \in X \land y \in Y) \land \lnot(x \in A \land y \in B) \
                                      &\Leftrightarrow (x \in X \land y \in Y) \land (x 
otin A \lor y 
otin B) \
                                      &\Leftrightarrow (x \in X \land y \in Y \land x 
otin A) \lor (x \in X \land y \in Y \land y 
otin B) \
                                      &\Leftrightarrow (x \in X - A \land y \in Y) \lor (x \in X \land y \in Y - B) \
                                      &\Leftrightarrow (x,y) \in (X-A) 	imes Y \lor (x,y) \in X 	imes (Y-B) \
                                      &\Leftrightarrow (x,y) \in (X-A) 	imes Y \cup X 	imes (Y-B)
  \end{align*}
  as desired.
\end{proof}

extbf{Main Problem.}
\begin{proof}
  Since $A$ is closed we have that $X-A$ is open in $X$.
  Since also $Y$ itself is open in $Y$, we have that $(X-A) 	imes Y$ is a basis element in the product topology by definition and is therefore obviously open.
  An analogous argument shows that $X 	imes (Y-B)$ is also open in the product topology since $B$ is closed in $Y$.
  Hence their union is also open in the product topology, but by Lemma~ef{lem:closedlim:setid} we have
  \begin{gather*}
    (X-A) 	imes Y \cup X 	imes (Y-B) = X 	imes Y - A 	imes B
  \end{gather*}
  so that $X 	imes Y - A 	imes B$ is also open in the product topology.
  It then follows by definition that $A 	imes B$ is closed as desired.
\end{proof}

Exercise 17.3

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Exercise 17.3

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