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Suppose $X,Y$ are connected non-empty topological spaces, $A \subset X$, $B \subset Y$ are strict subsets. Show that

$$(X \times Y) \setminus (A \times B)$$

is connected.

My attempted proof. Suppose for a contraction the space were not connected, so that

$$(X \times Y) \setminus (A \times B) = U \cup V$$

for some non-empty open $U,V \subset (X \times Y) \setminus (A \times B)$. Then there must exist some $\{x\} \times Y$ or $X \times \{y\}$ such that its intersection with both $U$ and $V$ are non-empty. Without loss of generality let this set be $\{x\} \times Y$, then since $\{x\} \times Y$ is homeomorphic to $Y$, we see that $\{x\} \times Y$ is connected, but it can be written as a disjoint union of non-empty open sets $(\{x\} \times Y) \cap U$ and $(\{x\} \times Y) \cap V$. $\square$

Feel free to share your comments. Thanks.

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Let $f:X\times Y-A\times B\rightarrow\{0,1\}$ be a continuous function. There exists $x\in X-A, y\in Y-B$, suppose $f(x,y)=0$. We have $f(x\times Y)=f(X\times y)=0$ since $x\times Y$ and $X\times y\subset :X\times Y-A\times B$. Let $(u,v)\in :X\times Y-A\times B$. You have $u$ is not in $X$ or $v$ is not in $Y$. If $u$ is not in $X$, then $u\times Y\subset :X\times Y-A\times B$ and $f(u,v)=f(u\times Y)=f(u,y)=f(X\times y)=f(x,y)=0$, if $v$ is not in $Y$ a similar argument shows that $f(u,v)-0$, so $f$ is constant and $X\times Y-A\times B$ is connected.

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  • $\begingroup$ Ok thanks I get the idea now. $\endgroup$ – Bernoulli Apr 4 '18 at 12:25
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Pick $p \in X \setminus A$, $q \in Y \setminus B$, by properness of inclusions.

For $x \in X$ define $V_x = \{x\} \times Y$ the vertical stalk at $X$, homeomorphic to $Y$, so connected.

Likewise for $y \in Y$, $H_y = X \times \{y\}$, the horizontal "stalk" at $y$, homeomorphic to $X$ and hence connected.

Let $\mathcal{V} = \{ V_x : x \in X \setminus Y\}$, a set of vertical stalks, all of which are subsets of $(X \times Y) \setminus (A \times B)$ as the $x$ coordinate of each stalk lies outside $A$, and also, $V$ already contains $H_p$, so is non-trivial.

Then the set $H_q$ intersects all $V_x$ in $\mathcal{V}$ (namely in $(x,q)$). So by a standard therem on unions of connected subsets we have that $V' = \bigcup \mathcal{V} \cup H_q$ is connected. (the stripe of "glue" $H_q$ glues the vertical stalks together).

Likewise $\mathcal{H} = \{H_y: y \in Y \setminus B\}$ is a set of horizontal stalks (all inside $(X \times Y) \setminus (A \times B)$ again), all of which intersect $V_p$ (and we already have $H_q \in \mathcal{H}$) so again

$H' = \bigcup \mathcal{H} \cup V_p$ is connected.

$H'$ and $V'$ intersect in $(p,q)$ and both are connected and one easily checks that $$(X \times Y) \setminus (A \times B) = H' \cup V'$$ and so our space is connected.

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