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I made it through the first section fine, but I am stuck on section 2 The Axiom of Specification. Particularly the last paragraph that asks

Can it be that $B\in A$ ?

The axiom defines $A$ as:

Axiom of specification: To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds

$B$ is defined as:

$B = \{ x \in A : S(x) \}$

If we consider $S(x) = x \notin x$

then $B = \{ x \in A : x \notin x \}$ He claims it follows that: note this is equation (*) but that wasn't rendering in the quote so I changed it to (1)

(1) $y \in B$ if and only if (y $\in$ A and y $\notin$ y)

The final conclusion is:

If $B \in B$ then by (1) the assumption $B \in A$ yields $B \notin B$ - a contradiction. If $B \notin B$ then by (1) again the assumption $B \in A$ yields $B \in B$ - contradtion again. This completes the proof that $B \in A$ is impossible, so that we must have $B \notin A$.

I don't understand why if we are considering $B \in A$ why the $S(x)$ part matters in the proof. Or put another way why if $B \in A$ it must be the case that $B \notin B$?

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  • $\begingroup$ Does (1) really say "$(y \in A : y \notin y)$"? That's not standard notation. $\endgroup$ – Eric Wofsey Oct 27 '18 at 23:47
  • $\begingroup$ In any case, it's unclear to me what part of the proof you don't understand. Is it just the first sentence of the "final conclusion" paragraph that you don't understand? $\endgroup$ – Eric Wofsey Oct 27 '18 at 23:58
  • $\begingroup$ @EricWofsey sorry that was a typo in (1) I updated it. It should have been and not :. The part I don't understand is why if we are considering, B is an element of A, how B is an element of B fits into that. $\endgroup$ – achyrd Oct 28 '18 at 0:07
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(i). Assume $B\in A.$ Then $B$ is either a member of $A$ that belongs to itself or a member of $A$ that doesn't belong to itself.

Now (I). Suppose $B\not \in B.$ Then $B$ is a member of $A$ that doesn't belong to itself, so it satisfies the sufficient conditions for membership in $B.$ So it belongs to $B$. So $B\in B.$

And (II). Suppose $B\in B.$ Then $B$ is a member of A that belongs to itself, so it does not satisfy the necessary conditions for membership in $B.$ So it does not belong to $B$. So $B \not\in B.$

(ii). So we find that $B\in A\implies [B\not \in B\iff B\in B].$ We conclude that either $B$ doesn't exist or $B \not \in A.$ And we know that $B$ exists.

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  • $\begingroup$ So Specification (a.k.a. Comprehension ) implies that every $A$ has a sub$set$ $B$ that is not a member or $A$. Note this is different from Foundation, which asserts that every non-empty $A$ has a $member$ $a$ which is disjoint from $A. $ $\endgroup$ – DanielWainfleet Oct 28 '18 at 5:55
  • $\begingroup$ Thanks I tried to upvote, but I don't have the rep. This breakdown was helpful $\endgroup$ – achyrd Oct 28 '18 at 18:05
  • $\begingroup$ On re-reading the last part of my answer, I noticed that if $B$ doesn't exist then we still have $B\not \in A$...... In the absence of Comprehension we could say that IF $B$ exists then $B\not \in A$. $\endgroup$ – DanielWainfleet Oct 31 '18 at 16:26
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That proof is hard to follow. Note:
$y\notin B$ $\iff$ $y\notin A$ or $y\in y$.

Since $B \notin B$ has been established,
from the above: $B \notin A$ or $B \in B$.

By the axiom of foundations, $A = B$.

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