-1
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Ittay Weiss Jan 24 '13 at 1:56
  • 1
    $\begingroup$ Would you allow uncountable families of sets? These don't admit Venn diagrams in any reasonable sense. $\endgroup$ – Trevor Wilson Jan 24 '13 at 2:06
1
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.