Why Munkres §26 Exercise 11 is nontrivial?

Question

This is probably a silly question, but I have a trivial (most likely wrong) reading of Munkres §26 Exercise 11:

Let $X$ be a compact Hausdorff space. Let $\mathcal{A}$ be a collection of closed connected subsets of $X$ that is simply ordered by proper inclusion. Show that $$Y = \bigcap_{A \in \mathcal{A}} A$$ is connected. [Hint: If $C \cup D$ is a separation of $Y$ , choose disjoint open sets $U$ and $V$ of $X$ containing $C$ and $D$ respectively and show that $$\bigcap_{A \in \mathcal{A}} (A − (U \cup V ))$$ is nonempty.]

My question is: We are given a collection $\mathcal{A}$ of sets simply ordered by proper inclusion. So let $A_0$ be the least element, such that for all $A \in \mathcal{A}$, $A_0 \subseteq A$. So why not $\bigcap_{A \in \mathcal{A}} = A_0$?

But if this is true, $Y$ is trivially connected and Munkres' Hint is pointless, which does not seem to be the case.

So what am I missing here?

Thanks!

Answer 1

The collection $\mathcal{A}$ may be infinite, in which case it need not have a smallest element. For instance, taking $X=[-1,1]$, you could have $\mathcal{A}=\{[-r,r]:0<r\leq 1\}$. This collection is simply ordered, but it has no smallest element.

Answer 2

Your "trivial proof" doesn't use compactness, does it? So what's wrong with the following counterexample:

Let $X=\mathbb R^2$ and let $\mathcal A=\{A_n:n\in\mathbb N\}$ where $$A_n=\mathbb R^2\setminus (-1,1)\times(-n,n) .$$

Each $A_n$ is a connected closed set, and the collection $\mathcal A$ is simply ordered by inclusion, yet the intersection $Y=\bigcap_{n\in\mathbb N}A_n$ is the disconnected set $\{(x,y):x\le-1\text{ or }x\ge1\}.$ How can that be??

The collection $\mathcal A$ has no least element.