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Suppose $A$ be an index set, and $A$ is uncountable. $\mathcal{F} = \{F_a \}_{a \in A}$ is a family of compact subsets in $\mathbb{R},$ that is, $F_a$ is a compact subset in $\mathbb{R}$ for any $a \in A$. Suppose for any countable subset $I \subset A$, there is an interval of length $1$ contained in the intersection of $\{F_i \}_{i_I}$, i.e. , the interval is in the $\cap_{i \in I} F_i$. My question is, is there an interval of length $1$ contained in the intersection of $\{F_a \}_{a\in A}$ ( $\cap_{a \in A} F_a$ ) ?

bof
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Haipeng Chen
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2 Answers2

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Let $\mathcal{F}=\{F_a\}_{a\in A}$ be a family of (arbitrary) subsets of $\mathbb R$ such that, for each countable set $D\subseteq A,$ the intersection $\bigcap_{a\in D}F_a$ contains a closed interval of length $1,$ and assume for a contradiction that $\bigcap_{a\in A}F_a$ contains no closed interval of length $1.$

Let $\{I_n:n\in\mathbb N\}$ be the set of all intervals of length $\ge1$ with rational endpoints. For each $n\in\mathbb N$ choose an index $b_n\in A$ such that $I_n\not\subseteq F_{b_n}.$ Let $B=\{b_n:n\in\mathbb N\}$ and let $F_B=\bigcap_{a\in B}F_a.$

The set $F_B$ contains no rational interval of length $\ge1,$ and no interval at all of length $\gt1.$ Since $B$ is countable, the set $F_B$ contains one or more closed intervals of length exactly $1,$ but these intervals must be pairwise disjoint, so there are at most countably many of them.

Let $\{J_n:n\in\mathbb N\}$ be the set of all closed intervals of length $1$ contained in $F_B.$ For each $n\in\mathbb N,$ choose an index $c_n\in A$ such that $J_n\not\subseteq F_{c_n}.$ Let $C=\{c_n:n\in\mathbb N\}.$ Then $F_{B\cup C}=\bigcap_{a\in B\cup C}F_a$ contains no closed interval of length $1.$ Since $B\cup C$ is countable, we have arrived at a contradiction.

bof
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  • I have 2 questions. Why for each $n \in \mathbb{N}$, there exist an index $b_n \in A$ such that $I_n$ is not contained in $F_{b_n}$? Why $F_B$ contains no rational interval of length $\geq 1$? – Haipeng Chen Sep 13 '16 at 09:10
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    I assumed that the intersection of all the $F_a$ contains no interval of length $1.$ Since $I_n$ is an interval of length $1,$ it follows that $I_n$ is not contained in the big intersection. Therefore, there is some index $a$ such that $I_n$ is not contained in $F_a.$ – bof Sep 13 '16 at 09:15
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    If $I$ is a rational interval of length $\ge1,$ then $I=I_n$ for some $n.$ Thus $I\not\subseteq F_{b_n}$ by the choice of $b_n.$ Since $I\not\subseteq F_{b_n}$ and $F_B\subseteq F_{b_n}$ it follows that $I\not\subseteq F_B.$ – bof Sep 13 '16 at 09:19
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Given F_a, there exist some interval L of length 1 contained in F_a. Given F_b, there exist some interval of length 1 contained in the intersection of F_a and F_b. Since L is such an interval, you can suppose that F_b also contains L. The result follows.

Lázaro Albuquerque
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