# Generalization of cantors intersection theorem

Let $A_1\supset A_2\supset\cdots$ be a sequence of connected compact subsets of $\mathbb{R}^2$.

Is it true that their intersection $A=\bigcap_{i=1}^{\infty}A_i$ is connected also?

Suppose it is not connected then there exists non empty $U\subset A$ which is both open and closed..

So, there exists $M$ open and $N$ closed in $\mathbb{R}^2$ such that $U=\bigcap_{i=1}^{\infty}(M\cap A_i)$ and $U=\bigcap_{i=1}^{\infty}(N\cap A_i)$

So, we have $U=\bigcap_{i=1}^{\infty}(M\cap A_i)=\bigcap_{i=1}^{\infty}(N\cap A_i)$

I do not know where to go from here..

The theorem is proved in several places on this site, but since you want only a hint, I’ll see what I can do. Let $V=A\setminus U$; you to make sure that $U$ is not just non-empty, but also a proper subset of $A$, so that $V\ne\varnothing$.

• Show that $U$ and $V$ are disjoint closed sets in $\Bbb R^2$.
• Show that there are disjoint open sets $G$ and $H$ in $\Bbb R^2$ such that $U\subseteq G$ and $V\subseteq H$.
• Show that there is an $n\in\Bbb Z^+$ such that $A_n\subseteq G\cup H$, and get a contradiction by showing that $A_n$ is ... ?

Added: For the third point, if there is no such $n$, then for each $n\in\Bbb Z^+$ there is a point $x_n\in A_n\setminus(G\cup H)$. Get a contradiction by showing that $\langle x_n:n\in\Bbb Z^+\rangle$ must have a subsequence converging to a point of ... ?

• I think second point is easy... Because $\mathbb{R}^2$ is normal given two disjoint closed sets there exists disjoint open sets containing them – user312648 Mar 3 '16 at 7:32
• @cello: Yes, that’s right. For the third point you’ll need to use compactness. – Brian M. Scott Mar 3 '16 at 7:34
• Yes Yes.. I think i got it.. just filling the gaps.. Once i am done i will write it down.. please see that and let me know if it is correct.. – user312648 Mar 3 '16 at 7:35
• @cello: You’re welcome. Don’t worry about it; it’s happened to all of us. Sometimes one’s mind just doesn’t want to go in the right direction! – Brian M. Scott Mar 3 '16 at 8:26
• Ok Ok... I got it... :) – user312648 Mar 3 '16 at 9:18

Let $A_1\supset A_2\supset\cdots$ be a sequence of connected compact subsets of $\mathbb{R}^2$.

Is it true that their intersection $A=\bigcap_{i=1}^{\infty}A_i$ is connected also?

Suppose it is not connected then there exists non empty $U\subset A$ which is both open and closed..

So, we have $U=M\cap A=N\cap A$ for $M$ open and $N$ closed in $\mathbb{R}^2$...

As $A$ is closed in $\mathbb{R}^2$, $N\cap A=U$ is closed in $\mathbb{R}^2$..

We have $V=A\setminus U=A\cap U^c=A\cap (M^c\cup A^c)=A\cap M^c$ which is closed in $\mathbb{R}^2$ as $A$ is closed and $M^c$ is closed...

Clearly $U\cap V=\emptyset$...

As $\mathbb{R}^2$ is normal, there exists two disjoint open sets $G$ and $H$ in $\mathbb{R}^2$ such that $U\subset G$ and $V\subset H$..

We claim that $A_n\subset G\cup H$ for some $n$... If this is true then we have a decomposition for $A_n$ contradicting the connectedness of $A_n$..

So, we try to prove that $A_n\subset G\cup H$... Suppsoe we have $A_n\not\subset G\cup H$ for each $n$... Then we have $x_n\in A_n\setminus(G\cup H)$...

Clearly, $x_n\in A_1$ for all $n$.. As $A_1$ is compact, this has a convergent subsequence converging to $x$.. So, $x\in A_1$ as $A_1$ is closed..

Clearly, $x_n\in A_2$ for all $n\geq 2$... convergence of subsequence assures that $x\in A_2$..

For similar reasons, we have $x\in A_n$ for all $n$. So, $x\in A$.

Mow, $x_n\in \mathbb{R}^2\setminus\{G\cup H\}$ for all $n$ and as $\mathbb{R}^2\setminus\{G\cup H\}$ is closed we see that subsequence limit $x\in \mathbb{R}^2\setminus\{G\cup H\}$..

But then we have $A=U\cup V\subset G\cup H$.. SO, there can be no element in $A$ that is not in union $G\cup H$.. so, we have a contradiction and $A_n\subseteq G\cup H$ for some $n$...

Done!

• You’re almost there: to finish, it you need to explain why having $A_n\subseteq G\cup H$ is impossible (namely, because $G\cap A_n$ and $H\cap A_n$ are then disjoint, non-empty subsets of $A_n$ that are relatively open in $A_n$, contradicting the connectedness of $A_n$. – Brian M. Scott Mar 3 '16 at 21:27
• Yes Yes.. I forgot to write that.. Thanks.. @BrianM.Scott – user312648 Mar 4 '16 at 0:45