User:WillF63

I enjoy mathematics and mathematical physics.

 The axiom of unrestricted comprehension (AoUC). $$\exist x \forall y [y\in x \iff \Psi(y)]$$ for any wff $$\Psi$$ in the language of set theory.

Theorem. AoUC leads to a contradiction.

Proof. Suppose AoUC:

$$\exist x \forall y [y\in x \iff \Psi(y)]$$ for any wff $$\Psi$$. $$\lnot (x\in x)$$ is a wff.

$$\implies \exists\mathcal{R}\forall x[x\in \mathcal{R} \iff \lnot (x\in x)]$$

$$\implies [\mathcal{R}\in \mathcal{R} \iff \lnot (\mathcal{R}\in \mathcal{R}).$$

There is a contradiction. $$\square$$

Remark. $$\mathcal{R}$$ is the Russell set.