# There is no set of all sets

From Maths

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## Contents

## Statement

We claim there is no set of all sets^{[1]}
Claim:

- [ilmath]\forall X\exists Y[Y\notin X][/ilmath]
- For any set [ilmath]X[/ilmath] there exists a set [ilmath]Y[/ilmath] such that [ilmath]Y\notin X[/ilmath]

## Significance

If we take [ilmath]X[/ilmath] to be the "set of all sets" we'd reach Russell's paradox, this statement shows half of it. The remainder of the paradox shows that [ilmath]Y\notin X[/ilmath] is absurd too.

## Proof

- Let [ilmath]X[/ilmath] be given
- Define [ilmath]Y:\eq\{x\in X\ \vert\ x\notin x\} [/ilmath] - which exists by the axiom schema of separation, accordingly [ilmath]Y\subseteq X[/ilmath]
^{[Note 1]}- Suppose that [ilmath]Y\in X[/ilmath]
- We now have two cases (as [ilmath]Y\in Y[/ilmath] is conceivable (as is [ilmath]Y\notin Y[/ilmath]) as [ilmath]Y[/ilmath] is some subset of [ilmath]X[/ilmath]):
- [ilmath]Y\in Y[/ilmath]
- But [ilmath]\forall A\in Y[A\notin A][/ilmath] is the defining property of [ilmath]Y[/ilmath]
- So then [ilmath]Y\notin Y[/ilmath] (the case where [ilmath]A:\eq Y[/ilmath]) - a contradiction! So we cannot have this!

- But [ilmath]\forall A\in Y[A\notin A][/ilmath] is the defining property of [ilmath]Y[/ilmath]
- [ilmath]Y\notin Y[/ilmath]
- As [ilmath]Y\in X[/ilmath] by hypothesis, if [ilmath]Y\notin Y[/ilmath] then [ilmath]Y\in Y[/ilmath] by definition of what [ilmath]Y[/ilmath] contains from [ilmath]X![/ilmath]
- A contradiction again! So we cannot have this!

- As [ilmath]Y\in X[/ilmath] by hypothesis, if [ilmath]Y\notin Y[/ilmath] then [ilmath]Y\in Y[/ilmath] by definition of what [ilmath]Y[/ilmath] contains from [ilmath]X![/ilmath]

- [ilmath]Y\in Y[/ilmath]
- Thus [ilmath]Y\notin X[/ilmath]

- We now have two cases (as [ilmath]Y\in Y[/ilmath] is conceivable (as is [ilmath]Y\notin Y[/ilmath]) as [ilmath]Y[/ilmath] is some subset of [ilmath]X[/ilmath]):

- Suppose that [ilmath]Y\in X[/ilmath]

- Define [ilmath]Y:\eq\{x\in X\ \vert\ x\notin x\} [/ilmath] - which exists by the axiom schema of separation, accordingly [ilmath]Y\subseteq X[/ilmath]

**Note: ** this is half of Russell's paradox, the other half is showing that [ilmath]Y\notin X[/ilmath] is absurd too!.

## Notes

- ↑ TODO:
*Any set constructed by separation is a subset of the bounding set*would be a great page to have

## References