# Schema of separation and set of all sets

The schema of separation states (this is probably simplified) states that if $P$ is a property and $X$ is a set, then there exists a set $Y = \{ x \in X : P(x)\}$.

The notes I'm reading say that from this we can conclude that the set of all sets doesn't exist, by applying the schema of separation to the property $x \notin x$.

I'm not seeing how applying the schema of separation to that property shows it. Can someone clarify this?

Let $V$ be the collection of all sets, and assume that $V$ was a set, let $W=\{x\in V\mid x\notin x\}$, then by the separation schema $W$ is a set.

Now ask yourself, $W\in W$ or $W\notin W$?

1. If $W\in W$ then $W$ satisfies $W\notin W$, which is a contradiction to the fact that $W\in W$.
2. If $W\notin W$ then $W$ does not satisfy $W\notin W$, which is again a contradiction!

Therefore $W$ cannot be a set, but this can only mean that $V$ cannot be a set to begin with.

This is also known as Russell's paradox.

• Was the separation schema created to resolve Russell's paradox? May 14, 2013 at 5:16
• Yes, yes it was. May 14, 2013 at 5:17

This is Russell's paradox. Starting with the universal set $U$ containing all sets, consider the set

$$S = \{ x \in U : x \notin x \}$$

So now consider the question: Is $S \in S$? Well, if $S \in S$, then by its own construction ($S$ is the set of all sets that don't contain themselves) it has to be the case that $S \notin S$, which is impossible. Otherwise, if $S \notin S$, then by its own construction it must be the case that $S \in S$, a contradiction. Therefore, $S \in S$ iff $S \notin S$, giving a contradiction. The only assumption we made (besides the comprehension schema) was that the universal set $U$ exists, and therefore there cannot be a set of all sets.

Hope this helps!