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Let $X$ be a Hausdorff space and $\{C_s:s\in S\}$ be a family of non-empty connected compact subsets of $X$. Suppose that for every $s_1,s_2\in S$ there exists $s_3\in S$ such that $C_{s_3}\subseteq C_{s_1}\cap C_{s_2}$. Show that $C=\displaystyle\bigcap_{s\in S}C_s$ is compact, connected and non-empty.

The condition $C_{s_3}\subseteq C_{s_1}\cap C_{s_2}$ implies $\{C_s:s\in S\}$ has the finite intersection property, because every $C_s$ is non-empty. Also, every $C_s$ is closed because it's compact and $X$ is Hausdorff. Fix $s_0\in S$. Then the family $\{C_s\cap C_{s_0}:s\in S\}$ is formed by closed subsets of $C_{s_0}$ with the finite intersection propery. Because $C_{s_0}$ is compact then $C=\displaystyle\bigcap_{s\in S}C_s\neq\emptyset$.

Also, $C$ is a closed subspace of $C_{s_0}$, hence $C$ is compact.

It remains to show $C$ is connected. This I can't do yet. Would anyone give me a hint?


marked as duplicate by Eric Wofsey, user91500, user223391, BLAZE, Fabian Jan 20 '16 at 9:01

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  • $\begingroup$ Have you tried proof by contradiction? $\endgroup$ – graydad Jan 20 '16 at 5:02
  • $\begingroup$ @EricWofsey in that question $X$ is also compact, though $\endgroup$ – JonSK Jan 20 '16 at 5:19
  • $\begingroup$ @JonSK: Just pick some $s\in S$ and restrict your attention to $C_s$. $\endgroup$ – Eric Wofsey Jan 20 '16 at 5:20
  • $\begingroup$ It's not obvious to me how the existence of $s_3$ such that $C_{s_3}\subset C_{s_1} \cap C_{s_2}$ implies that for all $s_1, s_2$ either $C_{s_1} \subset C_{s_2}$ or $C_{s_2} \subset C_{s_1}$ as in the question you referenced as the duplicate. $\endgroup$ – David Kleiman Jan 20 '16 at 16:27

Suppose $U,V$ are open sets such that $C = U\sqcup V$ and $U\cap V = \emptyset$.

If $C_{s}\cap V = \emptyset$ for some $s$, then $C \subset U$ and $V$ is empty and $C$ is connected.

This just leaves the case where $C_s \cap U \ne \emptyset$ and $C_s \cap V \ne \emptyset$ for all $s\in S$.

Suppose that $C_{s_1},C_{s_2}$ are such that $C_{s_1}\cap C_{s_2} \setminus U \setminus V = \emptyset$. Then there exists $s_3$ such that $C_{s_3} \cap U^c \cap V^c = \emptyset$ which implies that $C_{s_3}$ is disconnected, which is a contradiction. This implies that for every $s_1,s_2\in S$ you can find $s_3$ such that $(C_{s_3} \cap U^c \cap V^c) \subset (C_{s_1}\cap C_{s_2} \cap U^c \cap V^c)$. Thus $C_s\cap U^c\cap V^c$ is nonempty and has the finite intersection property. This gives the contradiction that

$$ \bigcap_{s\in S} C_s \cap U^c \cap V^c = C \cap C^c \ne \emptyset. $$

Thus $C$ is connected.

  • $\begingroup$ But we have $C\subseteq C_s$, and most likely $C\neq C_s$. I don't think you can say $C_s$ is contained in $U\cup V$. $\endgroup$ – JonSK Jan 20 '16 at 5:56
  • $\begingroup$ Sorry. You were right. That was definitely not true. I came up with something new. $\endgroup$ – David Kleiman Jan 20 '16 at 8:20

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