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I meet these two exercises:

  • Q1: let $A$ be a proper subset of $X$, and $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X\times Y)\setminus(A\times B)$ is connected.

  • Q2: Let $Y\subset X$. Assume that $X$ and $Y$ be connected. Show that if $A$ and $B$ form a separation of $X\setminus Y$, then $Y\cup A$ and $Y\cup B$ are connected.

My attempt for Q2

I think to prove it by contradiction, assume $Y\cup A$ and $Y\cup B$ are not connected then for $P$ and $Q$ disjoint $Y\cup A=P\coprod Q$ and for $M$ and $N$ disjoint $Y\cup B=M\coprod N$ $$(Y\cup A)\cup (Y\cup B)=(P\coprod Q)\cup (M\coprod N)$$ The left side will give $X$,and the right side can be written as a disjoint union , this contradicts the fact that $X$ is connected, so $Y\cup A$ and $Y\cup B$ must be connected.

I need help for Q1.

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To address Q2 by way of contradiction, you want to assume that either $Y\cup A$ or $Y\cup B$ is not connected as this is the negation of the condition that both sets are connected. – wckronholm Jun 11 '11 at 17:24
@aloycetemba, your proof of Q2 is great! I tried to prove it much with no success. I tried to show that $Y \cup A$ is connected and then a similar argument establish the same for $Y\cup B$ but I couldn't find such an argument. your one is very impressive in fact! thanx – Maths Lover Aug 21 '14 at 12:44
up vote -1 down vote accepted

In question 2 you have a logical mistake in the assumption: if you prove it by contradiction then you have to assume that one of them is disconnected, not necessarily both. Also, you need to mention that $P,Q$ are open (or closed).
As for Q1, prove by contradiction: If $(X\times Y)\setminus(A\times B)$ is disconnected then $$(X\times Y)\setminus(A\times B)=F_1\times F_2 ⊔ G_1\times G_2$$ $F_1,G_1$ are closed in $X$, $F_2,G_2$ are closed in $Y$. Since $A$ is proper, take any $x\notin A$. Then for any $y\in Y$ you have $(x,y)\in(X\times Y)\setminus(A\times B)$, hence $(x,y)\in F_1\times F_2 ⊔ G_1\times G_2$. It follows that $Y=F_2 ⊔G_2$ (Fill in all the details)

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Open sets in the product topology are not just products of open sets, but arbitrary unions of products of open sets, so your answer to Q1 is incorrect - you can't characterize a separation as a union of product sets. – Thomas Andrews Jun 11 '11 at 19:34
Thomas is right. This "solution" exposes a common misconception about the product topology. Open sets of the form $F_1\times F_2$ form a basis for the product topology, but don't represent arbitrary open sets. (Think of an open disk in the plane. It's not a product of two sets.) – Grumpy Parsnip Jun 12 '11 at 1:24
@Jim Conant: I'm sorry, was in a bit of a harry, and wrote a complete nonsense. I tried to vote to delete, but the answer was accepted, so I can't delete it. – Dennis Gulko Jun 15 '11 at 6:49

As to Q1, for every $x$ in $X \setminus A$, $S_{x} = \{x\} \times Y$ is connected, and for every $y$ in $Y \setminus B$, $T_{y} = X \times \{y\}$ is connected, as these sets are homeomorphic to the connected spaces $Y$ resp. $X$. Also, every $S_{x}$ intersects $T_{y}$, so for fixed $x \in X \setminus A$, $U_{x} = S_{x} \cup \cup_{y \in Y \setminus B} T_{y}$ is connected as well.

Now $( X \times Y) \setminus (A \times B)$ is the union of these sets $U_x$, which all intersect as well, making it connected. This uses standard theorems on unions of connected sets.

As to Q2, it suffices to show (by symmetry) that $Y \cup A$ is connected. Suppose not, then it can be written as a disjoint clopen union (non-trivially) of say $U$ and $V$. By connectedness of $Y$ one of them, say $U$ must miss $Y$, and then one checks that $U$ is clopen in $A$. As $A$ is clopen in $X \setminus Y$, $U$ would then be clopen in $X \setminus Y$ and thus in $X$, contradicting its connectedness. Some details omitted.

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