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Here is Prob. 9, Sec. 23, in the book Topology by James R. Munkres, 2nd edition:

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 $$ ( X \times Y ) - ( A \times B ) $$ is connected.

My Attempt:

It can be shown that $$ (X \times Y) \setminus (A \times B) = [ X \times (Y \setminus B) ] \cup [ (X \setminus A ) \times Y ]. $$ Let us choose a point $u \in X \setminus A$ and a point $v \in Y \setminus B$.

The space $u \times Y$, being homeomorphic with the connected space $Y$, is connected, and so is the space $X \times y$ for any point $y \in Y$ and hence for any point $y \in Y \setminus B$.

Now, let us take any point $y \in Y \setminus B$. As the spaces $u \times Y$ and $X \times y$ are connected and as these have the point $u \times y$ in common, so the union $ ( u \times Y ) \cup (X \times y)$ is also connected, by virtue of Theorem 23.3 in Munkres.

Moreover, the space $X \times v$, being homeomorphic with the connected space $X$, is connected, and so is the space $x \times Y$ for any point $x \in X$ and hence for any point $x \in X \setminus A$.

Now, let us take any point $x \in X \setminus A$. As the spaces $x \times Y$ and $X \times v$ are connected and as they have the point $x \times v$ in common, so their union $( x \times Y) \cup ( X \times v)$ is also connected.

Now, as, for each point $x \in X \setminus A$ and for each point $y \in Y \setminus B$, the spaces $ ( u \times Y ) \cup (X \times y)$ and $( x \times Y) \cup ( X \times v)$ are connected and as they have the point $u \times v$ in common, so the union $$ [ ( u \times Y ) \cup (X \times y) ] \cup [ ( x \times Y) \cup ( X \times v) ] $$ is also connected.

Finally, as, for each point $x \in X \setminus A$ and for each point $y \in Y \setminus B$, the spaces $$ [ ( u \times Y ) \cup (X \times y) ] \cup [ ( x \times Y) \cup ( X \times v) ] $$ are connected and as these have the point $u \times v$ in common, so the union $$ \bigcup_{ x \in X\setminus A, \ y \in Y \setminus B} \left[ \ [ ( u \times Y ) \cup (X \times y) ] \cup [ ( x \times Y) \cup ( X \times v) ] \ \right] $$ is also connected, and this set coincides with $( X \times Y) \setminus ( A \times B)$.

Am I right?

Is this proof correct? If so, then is my presentation good enough too? Have I introduced any redundancies in my reasoning?

If this proof (or portions thereof) is not correct, then where lie the problems?

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  • $\begingroup$ I think the proof works. $\endgroup$ – Andres Mejia Dec 10 '17 at 18:53
  • $\begingroup$ @AndresMejia thank you for your comment. Can you please read through my post once again? $\endgroup$ – Saaqib Mahmood Dec 12 '17 at 4:35
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I would like to start with a generalization of theorem 23.3 that makes things more easy.


Generalization of theorem 23.3: If $\left\{ A_{\alpha}\right\} $ is a collection of subspaces of $X$ such that every $A_{\alpha}$ is connected and such that for every pair $\left\langle \alpha,\beta\right\rangle $ the intersection $A_{\alpha}\cap A_{\beta}$ is not empty, then the union of that collection is connected.

Proof: let $Y$ denote the union and suppose that $Y=C\cup D$ is a separation of $Y$. Since $A_{\alpha}$ is connected it must be a subset of $C$ or $D$. WLOG we may assume that $A_{\alpha_{0}}\subseteq C$ for some $\alpha_{0}$. Then the condition that $A_{\alpha_{0}}\cap A_{\alpha}\neq\varnothing$ leads to the conclusion that also $A_{\alpha}\subseteq C$ for every $\alpha$. So we find that $Y\subseteq C$ or $Y\subseteq D$, which contradicts that $C\cup D$ is a separation of $Y$. This shows that no separation of $Y$ exists.


Note that in the generalization it is not demanded that the collection has a common point but only that every pair has a common point.

Now I go on with a solution of problem 9 based on that generalization.


For $x\in X$ and $y\in Y$ define $C_{x,y}:=\left(\left\{ x\right\} \times Y\right)\cup\left(X\times\left\{ y\right\} \right)\subseteq X\times Y$. So if $X$ and $Y$ are connected then $C_{x,y}$ is the union of two connected sets with non-empty intersection, hence $C_{x,y}$ is connected. Further we have $C_{x,y}\cap C_{x',y'}\neq\varnothing$ because it contains $\left\langle x,y'\right\rangle $ and $\left\langle x',y\right\rangle $ as elements. Now observe that: $$\left(X\times Y\right)-\left(A\times B\right)=\bigcup_{x\in X-A,y\in Y-B}C_{x,y}$$This especially because $A$ is proper subset of $X$ and $B$ is a proper subset of $Y$. Then applying the generalization of 23.3 we conclude that $\left(X\times Y\right)-\left(A\times B\right)$ is connected.


I haven't checked your proof yet, but soon I will do that and give you an account of that.


edit:

I have checked your proof. It is completely okay and does not contain essential redundancies.

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  • $\begingroup$ thank you very much for your answer; it's very easy to follow. And, thank you also for your comment on my proof. $\endgroup$ – Saaqib Mahmood Mar 18 '18 at 7:15
  • $\begingroup$ Glad to hear that. You are welcome. Actually I do not understand why Munkres did not gave this himself. It is definitely more handsome and not more difficult. I can only find one reason: he just overlooked. $\endgroup$ – drhab Mar 18 '18 at 7:28
  • $\begingroup$ yes, that may well be the case. He might include it in a 3rd edition of his book, if one ever comes out. Or, perhaps somebody can bring this (and similar other possibilities) to his attention. $\endgroup$ – Saaqib Mahmood Mar 18 '18 at 8:36

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