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Let F be a non-empty family of subcontinua of a continuum X such that for any finite subfamily $F_{1},F_{2},...,F_{n}$ in F there is $C\in F$ such that

$C \subset F_{1} \cap F_{2} \cap... \cap F_{n}$

I want to prove that

$\cap \{C: C \in F\}$ is a continuum.

Does anyone know how to do it ?

I was thinking of using the following theorem:

Theorem: If X is a continuum and $A_{1}, A_{2},...$ are nested subcontinuum $A_{n+1} \subset A_{n}, n \in \mathbf{N}$ then $A= A_{1}\cap A_{2} \cap...$ is a continuum of X.

But I don't know if this is enough.

Any help is appreciated.

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  • $\begingroup$ What do you mean by subcontinua? My intuitive interpretation would make the theorem you mention false, so… $\endgroup$ Commented May 15, 2022 at 0:27
  • $\begingroup$ I mean subcontinuum a subset of the continuum X that is a non-empty, connected and compact set $\endgroup$ Commented May 15, 2022 at 0:55
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    $\begingroup$ Since the family $F$ may not be countable, your proposal won't work. You should try to just attack the definitions directly. For example, here's how to show that that intersection is nonempty (recall that a continuum is a connected compact metric space). Suppose $\bigcap F=\emptyset$. Each $C\in F$ has $X\setminus C$ open in $X$, so $\{X\setminus C: C\in F\}$ is an open cover of $X$. Since $X$ is compact, this cover must have a finite subcover, but that contradicts your finite intersection hypothesis on the family $F$. $\endgroup$ Commented May 16, 2022 at 1:08
  • $\begingroup$ You're right, thanks this is very helpful $\endgroup$ Commented May 16, 2022 at 1:45

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