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I have a question that relates to this question about Venn diagrams. Has anyone shown that all Venn diagrams can (theoretically) be constructed?

Fred Daniel Kline
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    Can you please clarify what you mean by 'all Ven diagrams' and 'theoretically' and 'be constructed' mean? Without precise meaning for these the question is too vague. – Ittay Weiss Jan 24 '13 at 01:56
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    Would you allow uncountable families of sets? These don't admit Venn diagrams in any reasonable sense. – Trevor Wilson Jan 24 '13 at 02:06

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There is a recursive construction for Venn diagrams (see the Wikipedia page for Venn Diagram), so by induction there is a Venn diagram for any finite $n$. For $n\approx\omega$, it is more complicated, since $2^\omega\approx\mathbb R$, but I think it can be done. But if $n\succ\omega$, then $2^n\succ\mathbb R^2$, so it can definitely not be done on a plane (because there simply aren't enough points on the plane to do the job).

Martin Sleziak
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Mario Carneiro
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