Problem Let $A$ be a proper subset of $X$ , and let $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, Show that $(X\times Y)\smallsetminus(A\times B)$ is connected.
I wanna prove it by contradiction. I know the alternative proof which is
here.
Thanks!
Best Answer
Suppose $Z:=(X\times Y)\setminus(A\times B)$ is disconnected by disjoint open subsets $U,V\subseteq Z$.
For $x\notin A$, $y\notin B$, let $C_{x,y}:=(\{x\}\times Y)\cup(X\times\{y\})$, the connected cross formed from the pair $(x,y)$.
Suppose $(x,y)\in U$. Then, $x\notin A$ or $y\notin B$. For the former, since $\{x\}\times Y$ is connected, it must be entirely in $U$, so there is a pair $(x,y')\in U$, $y'\notin B$. For the latter, $X\times\{y\}$ is connected, so similarly we can assume there is $x'\notin A$. In any case there is a pair $(x',y')\in U$, $x'\notin A$, $y'\notin B$. But $C_{x',y'}$ is connected, so it must be entirely in $U$.
Any pair $(x,y)$ with either $x\notin A$ or $y\notin B$ must belong to $U$. For the former, the connected set $\{x\}\times Y$ intersects $C_{x',y'}$ in $(x,y')$, so their union is connected and thus belongs to $U$. Similarly for the latter case. Thus any such $(x,y)\in U$, so $U\supseteq Z$, leaving $V=\emptyset$.