Let $A$ be a proper subset of $X$ and $B$ a proper subset of $Y$. If $X,Y$ are connected. Show that
$X\times Y\backslash (A\times B)$ is connected.
Lemma: Let $X$ be a space and $A_1,A_2…,A_n$ a finite sequence of connected subsets in $X$. If $A_j\cap A_{j+1}\neq \varnothing$ for each $j=1,2…,n-1$ then $A_1\cup A_2 \cup…A_n$ is connected.
My attempt:
Fix $a\in X\backslash A$ and fix $b\in Y\backslash B$..
Observe, $\{$ $a$ $\} \times Y$ and $X\times \{$ b $\}$ are connected. Let $x\in X\backslash A$ and$y\in Y\backslash B$. Put
$T_{xy}=(\{x \}\times Y)$ $\cup (X\times \{b\}) \cup (\{a\} \times Y$) $\cup$ $(X \times \{y \})$
Then, $T_{xy}$ is connected because of the above lemma. Put $T=\bigcup_{x\in X\backslash A}\bigcup_{y\in Y\backslash B} T_{xy}$ . Then, $T$ is connected because each $T_{xy}$ contains $(a,b)$ and each $T_{xy}$ is connected
Is this correct?
Best Answer
Yes, your approach is correct. But the lemma doesn't apply here since you don't have a finite (Or even it may not be enumerable) collection of connected subsets.
The lemma which applies here would be : Let, $X$ be a topological space and $\{A_{\alpha}\}$ be a collection of connected subsets of $X$ such that $A_{\alpha} \cap A_{\beta} \neq \emptyset \forall \alpha, \beta$ then $\cup_{\alpha} A_{\alpha}$ is connected.