# There is no set of all sets

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## Statement

We claim there is no set of all sets 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]):
1. [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!
2. [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!
• Thus [ilmath]Y\notin X[/ilmath]

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