There Exists No Universal Set
Theorem
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.
Proof 1
Aiming for a contradiction, suppose such a $\UU$ exists.
Using the Axiom of Specification, we can create the set:
- $R = \set {x \in \UU: x \notin x}$
But from Russell's Paradox, this set cannot exist.
Thus:
- $R \notin \UU$
and so $\UU$ cannot contain everything.
$\blacksquare$
Proof 2
Aiming for a contradiction, suppose such a $\UU$ exists.
By No Injection from Power Set to Set, $\powerset \UU$ has no injection to $\UU$.
Let $f : \powerset \UU \to \UU$ be the identity mapping.
By Identity Mapping is Injection, it is an injection, which leads to a contradiction.
$\blacksquare$
Proof 3
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.
$\blacksquare$
Proof 4
Aiming for a contradiction, suppose such a $\UU$ exists.
Using the Axiom of Specification, we can create the set of all ordinals:
- $\set {x \in \UU: x \text{ is an ordinal} }$
But from Burali-Forti Paradox, this set cannot exist, which is a contradiction.
$\blacksquare$