non-countable subset of $\mathbb 2^{\mathbb Z}$ with finite pairwise intersection.

Asked Aug 28 '15 at 23:20

Active Aug 28 '15 at 23:25

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Does a non countable subset of the power-set of $\mathbb Z$ exist so that the intersection of any two elements is a finite set?

If we ask for the sets to be pairwise disjoint then the answer is a clear no. I'm pretty stumped, something which may be helpful is that the number of finite subsets of $\mathbb Z$ is countable.

edited Aug 28 '15 at 23:25

Asaf Karagila

393,674

asked Aug 28 '15 at 23:20

Asinomás

105,651

(it's not quite a duplicate, but you can just take a bijection of $\Bbb{Z}$ and $\Bbb{Q}$ to get the result from the other answer) – Peter Woolfitt Sep 05 '15 at 20:44
Yeah, I got that part, thanks. – Asinomás Sep 06 '15 at 05:24

non-countable subset of $\mathbb 2^{\mathbb Z}$ with finite pairwise intersection.

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