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I know this question has asked to death here on MSE but I have not found a satisfactory solution. A solution found online is extremely elegant but I do not quite understand it!

Given nested sequence of closed nonempty connected subsets of a compact metric space $X$. Prove that $\bigcap_{i=1} X_i$ is nonempty and connected.

It is a standard fact that arbitrary intersection of compact set is nonempty and compact.

I wish to understand the bit of the proof that this is also connected

Proof: https://math.berkeley.edu/sites/default/files/pages/f10solutions.pdf

Suppose that $\bigcap_{i=1} X_i$ is not connected. Let $A$ and $B$ be two disjoint nonempty closed sets so that $\bigcap_{i=1} X_i = A \cup B$. Find disjoint open sets $U$ and $V$ so that $A \subset U$ and $B \subset V$.

Put $F_i = X_i − (U \cup V )$. Then $\{Fi\}$ is a nested sequence of compact sets, whose intersection is empty. Thus $F_i = ∅$ for some $i$. That is, $X_i \subset U \cup V$.

However, $X_i$ intersects both $U$ and $V$ , since $X_i \cap A\neq ∅$ and $X_i \cap B \neq ∅$. This contradicts the assumption that $X_i$ is connected.

Can someone please elaborate on some of the details of the proof?

1) Why bother finding open sets containing $A,B$ and how do we know that they even exist? This move seems sort of unnatural.

2) What is so special about $U \cup V$ especially, why does $F_i = X_i − (U \cup V )$ imply $\{F_i\}$ has empty intersection? Why does intersection being empty imply the existence of $F_i = \varnothing$?

3) Can someone please justify $F_i = \varnothing \Leftrightarrow X_i \subset U \cup V$.

4) Can someone please justify the contradiction?

I know this is the pretty much the entire proof, but I have seriously tried to crack it and failed to understand it. Perhaps the proof is too difficult.

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1) We have some fiddley-ness at the start to get $U, V$ disjoint. It's a bit long, so see Disjoint compact sets in a Hausdorff space can be separated

This is possible because the space is Hausdorff (given that it is metrizable). There's a nice phrase by Willard, who says that this is an example of compact sets behaving like points - as $A, B$ are closed in a compact space, they are also compact, and because the space is Hausdorff, there are disjoint open sets containing $A$ and $B$. Neat!

The reason we want such sets will become apparent as we wade on.

2) $U\cup V$ is special because it covers $\bigcap_{i=1}X_i$, and because it's open. So defining $F_i\equiv X_i\setminus (U\cup V)$ is a closed subset of $X$, hence compact, and doesn't contain the intersection $\bigcap_{i=1}X_i$; finally note the $F_i$ also form a nested sequence (draw a diagram!). So if all $F_i$ were non-empty, their intersection would be non-empty - but that's not true! As $$\bigcap_{i=1}F_i=\bigcap_{i=1}(X_i\setminus (U\cup V))=(\bigcap_{i=1}X_i)\setminus (U\cup V)=\emptyset$$ because we already knew $U\cup V$ covers that intersection. So, there is some specific $F_k$ that is empty!

3) Nearly there now. As this $F_k$ is defined as $X_k\setminus (U\cup V)$, then if $F_k=\emptyset$, that implies every element of $X_k$ was also in $U\cup V$. i.e. $X_k\subset (U\cup V)$.

4) So we have that there is a pair of disjoint open sets covering this connected $X_k$. But surely they provide a separation of $X_k$? Just taking the complements of the open sets $U, V$ (namely our original $A$ and $B$...) and then intersecting with $X_k$ to get a separation. The only way this wouldn't be true is if $X_k$ had trivial intersection with one of those complements, i.e. if $X_k$ was entirely contained by one of the open sets. But this can't be true, because we know $$X_k\cap A\supseteq (\bigcap_{i=1}X_i)\cap A\neq\emptyset$$ The same holds for $B$, and so $X_k$ really has a separation. This is a contradiction because it was supposed to be connected!

Phew. I hope that makes sense, please ask for clarifications if anything is too terse. As often seems to be the case, a short looking proof is actually an example of economy of writing, rather than a simple idea. That's not to say this isn't a nice proof, it's just not one you can 'see' in a flash without having spent some time thinking topologically.

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    $\begingroup$ Hi thank you for this well thought out and complete answer. One point of confusion, if $F_k = \varnothing$, doesn;t this imply that $X_k = U \cup V$ instead of just merely being a subset of $U \cup V$? $\endgroup$ – Carlos - the Mongoose - Danger Jan 29 '16 at 2:17
  • $\begingroup$ Nah, not necessarily. Note that $\bigcap_{i=1}X_i\subseteq U\cup V$, so $U\cup V$ could certainly be bigger than this. The question I linked you too works through this; you might end up with open sets slightly bigger than the closed ones (though you probably don't need to, I feel like Hausdorff could take care of that). Either way the proof works - the crucial fact is that $U\cup V$ covers this $X_k$. $\endgroup$ – Alexander Heyes Jan 29 '16 at 2:26
  • $\begingroup$ Thanks yeah sorry I mixed up the notation. I wish the original proof used different indices $\endgroup$ – Carlos - the Mongoose - Danger Jan 29 '16 at 2:31
  • $\begingroup$ @AlexanderHeyes How does $F_i$ being closed imply it is compact. $\endgroup$ – DLB Mar 10 at 23:41

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