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I am looking at the proof of the theorem: The set of all the sets does not exist.

Proof:

We suppose that the set of all the sets, let $V$, exists.So, for each set $x, x \in V$.

We define the type $\phi: \text{ a set does not belong to itself, so }, x \notin x$.

From the axiom shema of specification, the set $\{ x \in V: x \notin x \}$ exists.

Since $V$ is the set of all sets,

$$\{ x \in V: x \notin x \} \subset V$$

So, $V'=\{ x: x \notin x \}$ is a set.

Therefore: $$V' \in V' \leftrightarrow V' \notin V', \text{ Contradiction.}$$

Could you explain me why we define $V'$ and how we concluded that it is a set?

Andrés E. Caicedo
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evinda
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2 Answers2

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The proof shows something stronger than "the set of all sets does not exist", but we can get that as an easy consequence. So easy, that we really forget that we prove something stronger.

The axiom schema of specification states that if $A$ is a set, and $\phi(x)$ is a formula, then $\{x\in A\mid\phi(x)\}$ is also a set. This means that we begin by assuming that $A$ is a set, then there is a subset which includes exactly those elements satisfying $\phi$. (I have omitted the parameters here for sake of readability.)

What the proof you cite actually show is this:

If $A$ is a set, then there exists a set $B$ such that $B\subseteq A$ and $B\notin A$.

The proof is exactly this, take $B=\{x\in A\mid x\notin x\}$. Since $A$ is a set, $B$ is also a set, and it is clear why $B\subseteq A$. If $B\in A$, then either $B\in B$ in which case we have a contradiction, or $B\notin B$ in which case we have a contradiction. Therefore $B\notin A$, and the proof is complete.

But what about the set of all sets? Now it's an easy consequence, since no set contains all sets, there is no set of all sets.

Asaf Karagila
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  • Is this superfluous at the proof of the theorem: $$$$ Since $V$ is the set of all sets,

    $${ x \in V: x \notin x } \subset V$$ $$$$

    ? and is this a right proof of the theorem: We suppose that the set of all the sets, let $V$, exists.So, for each set $x, x \in V$.

    We define the type $\phi: \text{ a set does not belong to itself, so }, x \notin x$.

    From the axiom shema of specification, the set $V'={ x \in V: x \notin x }$ exists.

    Therefore: $$V' \in V' \leftrightarrow V' \notin V', \text{ Contradiction.}$$

     – evinda Oct 05 '14 at 17:33
  • I'm not sure that I understand your comment. – Asaf Karagila Oct 05 '14 at 18:09
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We used the axiom of comprehension to define $V'$: $V' = \{x \in V : x \notin x\}$. We define $V'$ to obtain a contradiction. The crux of the matter is that assuming $V$ is a set, so is $V'$, and so $V' \in V$ and the contradictory argument works.

Yuval Filmus
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  • So you mean that the proof of the theorem is this:

    We suppose that the set of all the sets, let $V$, exists.So, for each set $x, x \in V$.

    We define the type $\phi: \text{ a set does not belong to itself, so }, x \notin x$.

    From the axiom shema of specification, the set ${ x \in V: x \notin x }$ exists.

    Therefore: $$V' \in V' \leftrightarrow V' \notin V', \text{ Contradiction.}$$

    ? and we don't have to say this:

    "Since $V$ is the set of all sets,

    $${ x \in V: x \notin x } \subset V$$

    So, $V'={ x: x \notin x }$ is a set."

    ?

     – evinda Oct 05 '14 at 16:56
  • Formally speaking, you define $V$ as the set of $x$ such that $x=x$, and if it exists, then it follows that $V'\in V$. – Yuval Filmus Oct 05 '14 at 18:51