# A Product of Connected Spaces Minus A Product of Proper Subsets of Both

Let $A$ be a proper subset of $X$, and let $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, what can we say about the set $$(X \times Y) - (A \times B)$$ being connected or not?

• What are your thoughts? Do you think it should be connected or disconnected? Have you thought of any examples? – Clayton Feb 12 '13 at 10:50
• Hint: how to disconnect $\mathbb{R}^2$ deleting a line? – Sigur Feb 12 '13 at 10:52
• Well, though I still can't locate the accept button in my display settings, I have been upvoting answers. I've even explicitly thanked a member for their contribution. – Saaqib Mahmood Feb 12 '13 at 12:04

It’s connected. Since $A \neq X$ there exists $x_0 \in X \setminus A$. Let $V=\{x_0\}\times Y$ and $y_0 \in Y$. Then consider $\langle x_1,y_1\rangle\in(X\times Y)\setminus(A\times B)$. If $y_1\notin B$, let $H=X\times\{y_1\}$. Then $H$ and $V$ are connected subsets of $(X\times Y)\setminus(A\times B)$ and intersect at $\langle x_0,y_1\rangle$, so $H\cup V$ is a connected subset of $(X\times Y)\setminus(A\times B)$ containing $\langle x_0,y_0\rangle$ and $\langle x_1,y_1\rangle$. If $y_1\in B$, then $x_1\notin A$; in this case let $V'=\{x_1\}\times Y$, pick any $y\in Y\setminus B$, and let $H=X\times\{y\}$. Then $V,V'$, and $H$ are connected, $V$ and $H$ intersect at $\langle x_0,y\rangle$, and $H$ and $V'$ intersect at $\langle x_1,y\rangle$, so $V\cup H\cup V'$ is a connected subset of $(X\times Y)\setminus(A\times B)$ containing $\langle x_0,y_0\rangle$ and $\langle x_1,y_1\rangle$.
• @SaaqibMahmuud Suppose for $x,y \in X$ there exists a connected subspace $C(x,y)$ such that $x,y \in C$. Then for fixed $x$, $X = \cup_{y \in X} C(x,y)$, which is a union of connected sets, all of whom intersect in the same point $x$, so their union, i.e. $X$, is connected too. The same fact is also used in Brian's proof (it holds for any number of sets, not just finitely many). Other proof: in a supposed decomposition, pick $x$ and $y$ in different parts and conclude $C(x,y)$ would be disconnected... – Henno Brandsma Feb 12 '13 at 14:39