Let $A$ be a proper subset of $X$, and let $B$ is a proper subset of $Y$. If $X$ and $Y$ are connected, show that
$$(X×Y)-(A×B)$$
is connected.
I know there's already a question answering this problem, but as I was working on this problem, I did it from another way and wanted to make sure is right.
Proof Idea:
Suppose by contradiction that $H=(X×Y)-(A×B)$ is not connected. Therefore There exist a separation of H from two opens $U,V$.
$$(X×Y)-(A×B)=U\cup V = U_1\times V_1 \cup U_2 \times V_2$$
Where $U_1,U_2$ are open in $X$ and $V_1,V_2$ are open in Y. This way
$U_1\times V_1 \cap U_2 \times V_2 = (U_1 \cap U_2) \times (V_1 \cap V_2) = \emptyset$. So there are two cases. I will work only on the first case, the second is similar.
- $U_1 \cap U_2 = \emptyset$, note that $U_1, U_2 \notin A$, as consequence $U_1 \cap U_2 \cap A = \emptyset$. Consider in particular $U_c=U_1 \cup U_2$. Observe that $U_c \cup A = X$, is a separation of $X$, but $X$ is connected. $\Rightarrow\!\Leftarrow$.
Let me know if there's any mistake, will help me prepare for my exam.
Best Answer
An open set need not be of the form $U_1 \times V_1$, you can only say that $U = \cup_i (U_i \times V_i) - (A \times B)$, $U_i, V_i, i \in I$ are open in $X$, rep. $Y$. This doesn't help you for the proof, really.
A more direct appraoch is to write this difference $(X\times Y)- (A \times B)$ as a union of sets like $\{x\} \times Y, X \times \{y\}$, which are connected subspaces, and in such a way that there are intersections to ensure the connectedness of the union.