Let $I$ be a non-empty set and $(A_i)_{i\in I}$ a family of sets.

Is it true that there exists a subset $J\subset I$ such that $\bigcap_{j\in J}A_j=\bigcap_{i\in I}A_i$ and, for any $j_0\in J$, $\bigcap_{j\in J-\{j_0\}}A_j\neq\bigcap_{j\in J}A_j$?

If $I=\mathbb{N}$, the answer is yes (if I am not mistaken): $J$ can be constructed by starting with $\mathbb{N}$ and, at the $n$-th step, removing $n$ if that does not affect the intersection.

What if $I$ is uncountable? I guess the answer is still "yes" and tried to prove it by generalizing the above approach using transfinite induction, but I failed.

The answer "yes" or "no" and a sketch of a proof (resp. a counterexample) would be nice.

  • $\begingroup$ Since I expect the proof to use the axiom of choice in some form, I've tagged this (non-elementary) set theory. $\endgroup$ – Stefan May 26 '12 at 10:44
  • $\begingroup$ Stefan, not all uses of axiom of choice are non-elementary (e.g. $|A|+|B|=\max\{|A|,|B|\}$ for infinite sets). I also have to ask, why did you expect the axiom of choice to appear here? $\endgroup$ – Asaf Karagila May 26 '12 at 11:22
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    $\begingroup$ @Asaf: 1. Axiom of choice is explicitly listed in the tag description of (set-theory). 2. because I wrongly thought there was a proof using transfinite induction. I'm just answering your question, not disagreeing with your retagging. $\endgroup$ – Stefan May 26 '12 at 11:31

The answer is no, even in the case $I=\mathbb N$. to see this, consider the collection $A_i=[i,\infty)\subset \mathbb R$. Then $\bigcap\limits_{i\in I}A_i=\emptyset$ and this remains true if we intersect over any infinite subset $J\subseteq I$, yet is false if we intersect over a finite subset. Thus there is no minimal subset $J$ such that $\bigcap\limits_{i\in I}A_i=\bigcap\limits_{j\in J}A_j$.

  • $\begingroup$ I think you meant $A_n=[n,\infty)$. $\endgroup$ – Asaf Karagila May 26 '12 at 11:03
  • $\begingroup$ @AsafKaragila Thanks, fixed. $\endgroup$ – Alex Becker May 26 '12 at 11:08
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    $\begingroup$ Surprise! Thank you very much. $\endgroup$ – Stefan May 26 '12 at 11:13

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