The set of all sets does not exist.

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In summary, the proof uses the assumption that the set of all sets exists and shows that this leads to a contradiction, proving that the set of all sets cannot exist. This is done by first defining a type that states a set cannot belong to itself, and then using the axiom schema of specification to show that a set $B$ exists where $y$ belongs to $B$ if and only if $y$ does not belong to itself. This leads to the contradiction $B \in B \Leftrightarrow B \notin B$, showing that the initial assumption of the set of all sets existing is incorrect.
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evinda
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Hey! (Wave)

Theorem (Russell's paradox is not a paradox in axiomatic set theory)

The set of all sets does not exist.

Proof

We suppose that the set of all sets exist, let $V$. So, for each set $x$, $x \in V$.
We define the type $\phi: \text{ a set does not belong to itself, so } x \notin x$.

So, from the Axiom schema 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: $V' \in V' \leftrightarrow V' \notin V'$, contradiction.How do we know that $V'$ is a set? Also, could you explain me why, having found this contradiction, we have proven that the set of all sets do not exist? (Thinking)
 
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  • #2
The proof starts with the assumption: suppose that the set of all sets exists. At the end the proof concludes that $V' \in V' \Leftrightarrow V' \notin V'$ which is a contradiction since a statement can not be equivalent with it's negation. Because of the contradiction the assumption (this is the only thing what was assumed) is incorrect hence the set of all sets does not exist.
 
  • #3
We want to prove that the set of all sets do not exist.

We suppose that the set of all 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 schema of specification, we conclude that there is the set $B=\{ x \in V: x \notin x \}$

$$\forall y(y \in B) \Leftrightarrow y \notin y$$

How can we continue? (Thinking)
 

FAQ: The set of all sets does not exist.

What does it mean for the set of all sets to not exist?

The set of all sets refers to a hypothetical collection of all possible sets. However, this set cannot exist because it leads to contradictions in mathematical logic.

Why is the concept of the set of all sets problematic?

The concept of the set of all sets is problematic because it leads to the paradox known as the "Russell's Paradox." This paradox states that if the set of all sets exists, then it must contain itself, but if it does not exist, then it must not contain itself.

Can we prove that the set of all sets does not exist?

Yes, the inability to define the set of all sets consistently and logically has been proven by mathematicians using mathematical logic and axiomatic set theory.

What are the implications of the non-existence of the set of all sets?

The non-existence of the set of all sets has significant implications in the foundations of mathematics. It limits the ability to define universal sets and raises questions about the completeness and consistency of mathematical systems.

Is it possible for the set of all sets to exist in other systems of logic?

No, the concept of the set of all sets leads to contradictions in any consistent system of logic, making its existence impossible in any system.

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