There Exists No Universal Set/Proof 3

From ProofWiki
Jump to navigation Jump to search


There exists no set which is an absolutely universal set.

That is:

$\map \neg {\exists \, \UU: \forall T: T \in \UU}$

where $T$ is any arbitrary object at all.

That is, a set that contains everything cannot exist.


Let $\SS$ be the set of all sets.

Then $\SS$ must be an element of itself:

$\SS \owns \SS$

Thus we have an infinite descending sequence of membership:

$\SS \owns \SS \owns \SS \owns \cdots$

But by No Infinitely Descending Membership Chains, no such sequence exists, a contradiction.


Axiom of Foundation

This proof depends on the Axiom of Foundation, by way of No Infinitely Descending Membership Chains.

Most mathematicians accept the Axiom of Foundation, but theories that reject it, or negate it, have found applications in Computer Science and Linguistics.