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 $B \notin B$?

  • $\begingroup$ 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

(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.

  • $\begingroup$ 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

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, $A = B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.