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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?

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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.

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    $\begingroup$ The Lemma is correct but the family you are working on may be uncountable too. So, after clubbing too many of them you may not be able to make it a finite collection. So this Lemma is not useful here. $\endgroup$ Apr 28 '20 at 1:44
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    $\begingroup$ The finite many sets are $\{$ $x$ $\}$ $\times Y$, $X\times \{b\}$, $ \{a \} \times Y$, $X\times \{y\}$. I'm invoking the lemma on these sets Call $A_1=\{ x\} \times Y$ and ... $\endgroup$ Apr 28 '20 at 1:45
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    $\begingroup$ Yes but then you are taking union of all the $T_{xy}$ to conclude the result. I am saying that the result may not be applicable there. $\endgroup$ Apr 28 '20 at 1:49
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    $\begingroup$ You can check on this space: $\Bbb R \times \Bbb R - \Bbb Q \times \Bbb Q$. $\endgroup$ Apr 28 '20 at 1:50
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    $\begingroup$ Yup right. I just supplied a more general version of the Lemma you are using. $\endgroup$ Apr 28 '20 at 1:51

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