There Exists No Universal Set

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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.

$\blacksquare$