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Let $(B_i)_{i\in I}$ be an indexed family of closed, connected sets in a compact space X. Suppose $I$ is ordered, sucht that $i < j \implies B_i \supset B_j$.

Is $B = \bigcap_i B_i$ necessarily connected?

I can prove it, if I assume $X$ to be Hausdorff as well: If $B$ is not connected, then there are two disjoint, closed, nonempty sets $C$, $D$ in $B$, such that $C \cup D = B$. Now these sets are also closed in $X$, hence by normality there exist open disjoint neighborhoods $U$, $V$ of $C$ and $D$, respectively.

Then for all $i$: $B_i \cap U^c \cap V^c \ne \emptyset$, since $B$ is contained in $B_i$ and $B_i$ is connected. Thus we must also have

$$ B \cap U^c \cap V^c = \bigcap_i B_i \cap U^c \cap V^c \ne \emptyset $$

by compactness and the fact that the $B_i$ satisfy the finite intersection property. This is a contradiction to the choice of $U$ and $V$.

I can neither see a counterexample for the general case, nor a proof. Any hints would be greatly appreciated!


S. L.

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You statement is true in Hausdorff spaces. – mez Apr 18 '13 at 18:51
What is the meaning of "by normality"? – hengxin Feb 13 '14 at 14:04
@hengxin Every compact Hausdorff space is normal. see… – amirbd89 Jul 21 '15 at 20:35
up vote 15 down vote accepted

I think the following is a counterexample: take $Y=[-1,1]\times\{a,b\}$, where $[-1,1]$ has the standard topology, and $\{a,b\}$ the discrete topology, and let $X=Y/\sim$, $y\sim y'$ if and only if there exist $x\in[0,1]$ such that $y=(x,a)$ and $y'=(x,b)$, or $y=(x,b)$ and $y'=(x,a)$. That is, identify all points except $0$. Give $X$ the quotient topology

This is the interval $[-1,1]$ with "a doubled origin", a common proving ground because the space is $T_1$ but not $T_2$, but any two points other than the doubled origins can be separated by open neighborhoods. (So, in a sense, it is "almost" Hausdorff; the Hausdorff property only fails for one choice of points, and there are lots of other points around).

Since $Y$ is compact and the quotient map is continuous and onto, $X$ is compact.

For every positive integer $n$, let $\mathcal{B}_n\subseteq Y$ be the set $[-\frac{1}{n},\frac{1}{n}]\times\{a,b\}$, and let $B_n$ be the image of $\mathcal{B_n}$ in $X$; that is, $B_n$ is the interval from $-\frac{1}{n}$ to $\frac{1}{n}$, including both origins.

$B_n$ is closed, since $X-B_n = [-1,-\frac{1}{n})\cup(\frac{1}{n},1]$ is a union of two open sets. It is also connected, because $B_n$ is a union of two connected subsets (the two copies of the interval $[-\frac{1}{n},\frac{1}{n}]$ obtained by removing one of the two $0$s) and the two subsets intersect.

What is $\cap_{n=1}^{\infty}B_n$? It's a set whose only two elements are the doubled origin points. But this subset of $X$ is not connected, because $X$ is $T_1$, so there exist open neighborhoods $U$ and $V$ such that $(0,a)\in U-V$ and $(0,b)\in V-U$. So $B\subseteq U\cup V$, $U\cap B\neq\emptyset \neq V\cap B$, and $B\cap U\cap V = \emptyset$.

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Thanks, that seems like the canonical choice for a counterexample requiring a non-Hausdorff space! I've actually seen it many times, so I'm a bit annoyed I haven't thought of it myself. Well, luckily there are better mathematicians around to help. ;) – Sam Mar 27 '11 at 4:21

The line with two origins should work as a counterexample. Take the intersection of nested closed intervals containing both origins. The intersection is the two origins, which is two points with the discrete topology, so is disconnected.

Edit: If you want the ambient space to be compact, let it be a closed, bounded interval containing the origins. (Thanks to Arturo for pointing this out).

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Well, if you either restrict the line to some finite closed interval (containing the origin), or compactify the line (either one or two points). Otherwise, your space is not compact. – Arturo Magidin Mar 27 '11 at 4:19

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