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Let $\mathscr K$ be an uncountable set such that every $K\in\mathscr K$ is a compact subset of $[0,1]$ with positive Lebesgue measure. Does it then follow that there exists an uncountable $\mathscr A\subseteq\mathscr K$ with $\bigcap\,\mathscr A\not=\emptyset$ ?

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    $\begingroup$ We may assume without loss of generality that the measures are as close to $1$ as we like, because by the Lebesgue density theorem, every positive measure set has density as close to $1$ as we like on some small intervals, and since there are only countably many intervals with rational endpoints, we can restrict to an uncountable subfamily that all has big measure on that interval. (This seems relevant, but I'm not sure how to use it...) $\endgroup$ Sep 28, 2013 at 20:40
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    $\begingroup$ In light of the answers, I've added the tags set-theory and continuum-hypothesis. $\endgroup$ Sep 29, 2013 at 13:53

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Assuming the continuum hypothesis, the answer is negative.

First of all, we can index ${\cal K}$ by $[0,1]$, let $K_x$ be the set in ${\cal K}$ with index $x$. Now well-order $[0,1]$ in a way corresponding to the first uncountable ordinal. For any $x$, there are only countably many points preceding $x$ in the well ordering. We can therefore choose $K_x$ so that it does not contain any point that precedes $x$. Now let $A$ be any uncountable set and consider $\bigcap_{x\in A} K_x$. For any $y$, there is an $x\in A$ which comes after $y$. Hence this set must be empty.

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    $\begingroup$ It took me a while to remember that if $D\subseteq[0,1]$ is at most countable and $\lambda$ is the Lebesgue measure, then $1=\lambda([0,1]\setminus D)=\sup\{\lambda(K): K\subseteq [0,1]\setminus D\ ,\ K\ \mbox{compact}\}$. So choosing $K_{x}$ with $\lambda(K_{x})>0$ is possible. For those who struggle like me, I thought I'd make the point. $\endgroup$ Sep 29, 2013 at 9:17
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    $\begingroup$ @NoelVaillant: We can also do it explicitly: let $D = \{x_1, x_2, \dots\}$, and for $\epsilon > 0$, let $K := [0,1] \setminus \bigcup_n (x_n - \epsilon 2^{-n}, x_n + \epsilon 2^{-n})$. Then $K$ is compact, disjoint from $D$, and has measure at least $1-\epsilon$. $\endgroup$ Sep 29, 2013 at 13:56
  • $\begingroup$ @NateEldredge Thank you, yes that works too. $\endgroup$ Sep 29, 2013 at 15:36
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Assuming $MA_{\aleph_1}$, the answer is positive.

Let $P$ be the collection of all positive measure finite intersections of elements of $\mathcal{K}$, ordered by inclusion. Then $P$ is a ccc uncountable partial order so (by $MA_{\aleph_1}$) it contains an uncountable centered subset $Q \subseteq P$. If we let $\mathcal{A}$ be the collection of all elements of $\mathcal{K}$ that contain some element of $Q$, it follows that $\mathcal{A}$ has the finite intersection property and therefore $\bigcap \mathcal{A} \neq \emptyset$.

Edit: A subset $Q$ of a poset $P$ is called centered if any finite $F \subseteq Q$ has a lower bound in $P$. It was proved by Velickovic and Todorcevic in "Martin's axiom an partitions" (1987) that $MA_{\aleph_1}$ is equivalent to the statement that every ccc uncountable partial order contains an uncountable centered subset.

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    $\begingroup$ So how to reconcile the two answers? $\endgroup$ Sep 29, 2013 at 14:43
  • $\begingroup$ @LevBorisov it is a theorem of ZFC that $\mathtt{MA}({\bf c})$ fails. So if CH holds, then $\mathtt{MA}(\aleph_{1})$ will fail. If $\lnot$CH holds, then $\mathtt{MA}(\aleph_{1})$ is consistent with ZFC+$\lnot$CH. Just reading on wiki about Martin's axioms. $\endgroup$ Sep 29, 2013 at 16:00
  • $\begingroup$ @ Ramiro de la Vega. Reading the wikipedia definition en.wikipedia.org/wiki/… , I am trying to understand how your reasoning fits in it. What is a "centered" subset, and what is wp's $D$ in your setting? Why is $Q$ uncountable? $\endgroup$
    – TaQ
    Sep 30, 2013 at 22:37
  • $\begingroup$ @TaQ: I edited my answer to clarify these points. $\endgroup$ Sep 30, 2013 at 23:24
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    $\begingroup$ @TaQ: I wouldn´t call $MA_{\aleph_1}$ "exotic"; after all it is just a "mild" generalization of Baire´s category theorem. It is also the first axiom one usually looks at as an alternative to $CH$, since $\lnot CH$ alone is just too weak. $\endgroup$ Oct 1, 2013 at 11:50

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