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This post is probably going to be marked as duplicate, but my approach to its solution is incomplete and I can't find any solution which can help me complete it.

Suppose that $S_1, S_2, S_3, \dots$ is a sequence of compact and connected sets in some metric space and $S_1 \supset S_2 \supset S_3\supset \dots$. Is $S=\bigcap S_n$ connected?

My approach: Suppose that $S=A\sqcup A^c$, where $A$ and $A^c$ are both closed. Since all $S_n$ are connected, there must exist $a_n \in S_n$ for all $n$ such that $a_n \notin S$. By compactness, we get a subsequence $(a_{n_k})_k$ of $(a_n)_n$ which converges to some $a$, which belongs to every $S_i$. Hence $a\in S$.

Is it possible to carry on with this approach? I've noticed that I haven't yet used the fact that $S$ (as well as $A$ and $A^c$) is compact. Also, https://math.stackexchange.com/a/2114566/633683 does something quite similar to the above, but I feel there could be a shorter solution...

SyntasticMonoid
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  • there must exist $a_n \in S_n$ for all $n$ such that $a_n \notin S$ is a strange sentence! – mathcounterexamples.net Sep 02 '20 at 07:58
  • Why would such "$a_n\in S_n$" exist? Note that $A\sqcup A^c$ is a decomposition of $S$, not $S_n$. I don't get how you utilize connectedness of $S_n$ here. And even if such sequence exists, how the convergence of a subsequence of $(a_n)$ brings us to a solution? – freakish Sep 02 '20 at 07:59
  • Such an $a_n \in S_n$ would exist because $S_n - S \neq \emptyset$, otherwise $S_n$ could be decomposed into $A$ and $A^c$... – SyntasticMonoid Sep 02 '20 at 08:08
  • @freakish Yes I agree, there seems to be a dead end after finding that convergent subsequence of $(a_n)$ :( But the solution in https://math.stackexchange.com/a/2114566/633683 motivated me to carry on with my approach. – SyntasticMonoid Sep 02 '20 at 08:13
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    This is Corollary 6.1.19 in R. Engelking, General Topology. The $S_n$ have just to be compact and connected, not necessarily be subsets of the plane. – Ulli Sep 02 '20 at 17:31
  • OK thanks, I'll edit it – SyntasticMonoid Sep 03 '20 at 03:40

1 Answers1

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Suppose $\bigcap_n S_n$ is disconnected. That is, there exists non-empty, disjoint open sets $A,B$ such that

$$\bigcap_n S_n = A \cup B.$$

Then as metric spaces are normal, there exists disjoint open set $U,V$ containing $A,B$ respectively, this is because $A,B$ are complements of one another and they are thus both closed as a result of both being open. Then define

$$W_n := S_n \setminus (U \cup V).$$

Then $\bigcap_n W_n = \emptyset$ forcing $W_k = \emptyset$ for some $k \in \Bbb{Z}^+$. Then we have that (for this $k$)

$$S_k \subset U \cup V.$$

But as $S_k \cap A, S_k \cap B \neq \emptyset$, then $S_k$ intersects $U$ and $V$ a contradiction to it being connected, it can only lie in either $U$ or $V$. Therefore $\bigcap_n S_n$ is connected.

MyMathYourMath
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  • "The intersection of $W_n$ is empty" cannot imply $W_k$ is empty for some $k$. Consider $(0,1/n)$. – Helen Oct 04 '24 at 16:53