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Prove that if $X$ is Hausdorff and $\mathfrak{C}$ is a nonempty chain of compact and connected subsets of $X$, then $\bigcap \mathfrak{C}$ is compact and connected.

Here are the definitions which were provided:

A family $\mathfrak{F}$ of sets is a chain if for all $A,B \in \mathfrak{F}$ either $A \subset B$, or $B \subset A$. (If $\mathfrak{C}$ is a nonempty chain of nonempty compact sets, then $\bigcap \mathfrak{C} \neq \emptyset$.)

A space is Hausdorff (or $T_2$) if for all distinct $a, b \in X$ there are $U, V \in \mathcal{T}$ so that $a \in U$, $b \in V$ and $U \cap V = \emptyset$. Here, $\mathcal{T}$ is the topology on $X$.

My confusion starts in what seems to me like a contradiction between definitions. We are assuming that $X$ is Hausdorff, so there is a separation of $X$ so there are disjoint $U, V \in \mathcal{T}$ with $a \in U$ and $b \in V$. But we are also given that there is a chain of compact and connected sets $\mathfrak{C}$, which by definition, $U \subset V$ or $V \subset U$. But they are supposed to be disjoint according to the Hausdorff property.

Can somebody please explain this to me and assist me in solving this problem with a proper proof? Many thanks in advance for your time and patience, it is greatly appreciated.

user642796
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Jamil_V
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  • The fact that distinct points are contained in disjoint open sets has nothing to do with the chain of closed, connected sets that you’re given by hypothesis. For instance, $\Bbb R$ is Hausdorff, and ${[-x,x]:x>0}$ is a chain of closed, connected subsets of $\Bbb R$. There is nothing in the hypothesis to suggest that the open sets of $X$ form a chain. – Brian M. Scott Apr 16 '15 at 20:21
  • The fact that $U,V$ are disjoint open non-empty does not imply that $X$ is disconnected unless $X=U\cup V$, so you need not confuse T$_2$ (points could be separated by disjoint neighborhoods) with $X$ being disconnected ($X$ could be partitioned into disjoint nonempty open sets). – Mirko Apr 16 '15 at 21:48

2 Answers2

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To start, notice that the intersection of any chain of nonempty compact sets in a Hausdorff space must be nonempty (by the finite intersection property for closed sets). Let $D = \cap \mathfrak{C}$. Then $D$ is nonempty by the preceding statement.

To prove that $D$ is compact, choose some $C_0 \in \mathfrak{C}$. Since $X$ is Hausdorff, all compact subsets of $X$ are closed in $X$, and thus all of the sets in $\mathfrak{C}$ are closed in $X$. It follows that $D=\cap \mathfrak{C}$ is closed in $X$ (since intersections of closed sets are closed). Thus $\cap \mathfrak{C}$ is closed in $C_0$, so since closed subsets of compact spaces are compact, it follows that $D$ is compact.

To prove that $D$ is connected, argue by contradiction. Suppose that $D$ is not connected, so we can find closed subsets $A,B \subset X$ such that $D \subset A \cup B$ and $A \cap D \neq \emptyset$ and $B \cap D \neq \emptyset$ and $A \cap B \cap D = \emptyset$. Notice that for any $C \in \mathfrak{C}$, the set $A \cap B \cap C$ is compact because it is a closed subset of the compact set $C$. Therefore, there exists some $C_1 \in \mathfrak{C}$ such that $A \cap B \cap C_1 = \emptyset$, otherwise $\{ A \cap B \cap C\; | \; C \in \mathfrak{C} \}$ would form a chain of nonempty compact sets, and thus $\bigcap_{C \in \mathfrak{C}} A \cap B \cap C= A \cap B \cap D$ would be nonempty (which we know is false). Also notice that $A \cap C_1$ and $B \cap C_1$ are both nonempty, since they contain the nonempty sets $A \cap D$ and $B \cap D$, respectively. Thus $A \cap C_1$ and $B \cap C_1$ form a separation of the set $C_1$ which contradicts the fact that $C_1$ is connected.

shalin
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  • You want to prove that $D$ is compact, so you can't use its compactness. But it's much easier: every member of $\mathfrak{C}$ is closed (the space is Hausdorff), so their intersection $D$ is closed; then $D$ is a closed subset of a compact set, hence compact. In the proof of connectedness, you state that $U\cap V=\emptyset$, but all you can say is that $U\cap V\cap D=\emptyset$. – egreg Apr 16 '15 at 21:46
  • @egreg: Yes, my post was circular (I was very careless), but I have edited it, and I think that it works now. Can someone verify if it is correct now? – shalin Apr 17 '15 at 04:12
  • Seems good now! – egreg Apr 17 '15 at 08:48
  • I confirm, quite nice proof, clean concise and clear. – am70 Feb 28 '20 at 17:20
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It seems the following.

Fix any element $K\in\mathfrak{C}$. Then the set $C=\bigcap \mathfrak{C}$ is compact as an intersection $\bigcap\{ K\cap L:L\in\mathfrak{C}\}$ of a family $ \{ K\cap L:L\in\mathfrak{C}\}$ of closed subsets of a compact set $K$.

Assume that the set $C$ is not connected. This means that $C$ can be represented as a union $C_1\cup C_2$ of two its disjoint non-empty clopen (that is, closed and open) subsets. Since $C_1$ and $C_2$ are disjoint compact subsets of a Hausdorff space $X$, it is well known and easy to prove that they can be separated by disjoint open neighborhoods, that is there exist two disjoint open subsets $U_1$ and $U_2$ of the space $X$ such that $C_1\subset U_1$ and $C_2\subset U_2$. Put $U=U_1\cup U_2$.

Now we are going to find a set $L\in\frak C$ such that $L\subset U$. If $K\subset U$, we are done. In the opposite case consider a family ${\frak C}’=\{(K\cap L)\setminus U:L\in\frak C\}$. Since $\bigcap{\frak C}=C\subset U$, $\bigcap \mathfrak{C}'=\varnothing$. Since $\frak C’$ is a family of compact subsets of the compact space $K$, it contains an empty element $(K\cap L)\setminus U$. Since the family $\frak C $ is a chain, this implies that there exists a set $L\in\frak C$ such that $L\subset U$. Then $(L\cap U_1)\cup (L\cap U_2)$ is a partition of the set $L$ into two its disjoint clopen subsets. Each of these sets is non-empty. Indeed, since $\frak C$ is a chain, $L\supset \bigcap \mathfrak{C}=C$. Then $L\cap U_i\supset C\cap U_i\ne\varnothing$. So the set $L$ is not connected, a contradiction.

Alex Ravsky
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