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First of all, I apologixe if there is already an equivalent question with a satisfying answer on the site, but I couldn't find any.


The axiom schema of specification goes roughly like this:

$$ \forall A \ \exists B \ \forall x ( x \in B \Leftrightarrow x \in A \land P(x) ). $$

This doesn't end up in Russell's paradox, but there's still something that confuses me; to show it, I'll go through a reasoning similar to that of Russell's paradox.

First, this is an axiom schema, and as far as I understand we aren't arbitrarily excluding the statement $P(x):x \not \in x$.

So in the schema there is one axiom going like this:

$$ \forall A \ \exists B \ \forall x ( x \in B \Leftrightarrow x \in A \land x \not\in x ). $$

For this to be true $\forall x$, it must be true, for example, for $x=B$:

$$ \forall A \ \exists B( B \in B \Leftrightarrow B \in A \land B \not\in B ). $$

If we try to assume $B \in B$, we immediately encounter an absurd, since the axiom infers $B\in B \Rightarrow B \in A \land B \not \in B$.

Therefore we infer $B \not \in B $, and at first we don't encounter any issue, because we don't know if $B \in A$ necessarily.

But doesn't this imply that if a set $B$ doesn't contain itself, we cannot find any set $A$ such that $B \in A$?

To put this in perspective, what confuses me is the fact that... most sets don't contain themselves.

For example, take an element of Von Neumann's Integers:

$$ 2 = \{ \emptyset, \{ \emptyset \} \} $$

which, as far as I can see, doesn't contain itself. But (as we do when defining the other integers) surely we can find a set that contains it!

Shouldn't this contradict the axiom of specification relative to $P(x):x \not \in x$? What am I missing? Where am I going wrong in my reasoning?

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  • $\begingroup$ Your statement $x\notin x$ is implied by the axiom of foundation, not by specification. $\endgroup$ – Arthur Oct 23 '16 at 11:35
  • $\begingroup$ How do you jump to "we cannot find any set $A$ such that $B\in A?$ The set $A$ was given to start with, and the set $B$ was defined in terms of $A$, and it turns out that $B\notin A.$ What this proves is that, given a set $A,$ we can find a set $B$ such that $B\notin A.$ (Under normal circumstances, i.e. assuming the axiom of foundation, it will turn out that $B=A.$) $\endgroup$ – bof Oct 23 '16 at 11:40
  • $\begingroup$ The paradox, if you want to call it that, is that there is no universal set: no matter what set $A$ you take, you find a set $B$ which is not in $A.$ If your intuition tells you that there must be a universal set, you may consider this fact paradoxical. $\endgroup$ – bof Oct 23 '16 at 11:45
  • $\begingroup$ @bof You're right, my leap in logic was in not realizing how $B$ is dependent from $A$. In a way, I was thinking as if the statement was actually $ \forall A \ \forall B ( B \in B \Leftrightarrow B \in A \land B \not \in B) $, which doesn't make any sense. $\endgroup$ – EducatedGuest Oct 23 '16 at 12:38
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In order for $x$ to be an element of $B$ it needs to satisfy two conditions:

  1. $x\notin x$, and
  2. $x\in A$.

If $B\notin A$, then it is not possible that $B\in B$ anyway. If $B\in B$, then we derive the usual contradiction, by concluding that $B\notin B$. And if $B\in A$ but $B\notin B$, then we also derive a contradiction. So we can only conclude that $B\notin A$, and so we're fine (and in particular, $B\notin B$).

So Russell's paradox is now transformed to the following theorem:

If $A$ is a set, then $\{x\in A\mid x\notin x\}\notin A$. In particular, $\mathcal P(A)\nsubseteq A$.

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