I'm complementing my knowledge of set theory using other sources in addition to the book that normally I use.

Working with the Jech's book (great book, by the way), in one exercise of the first section says the next:

Jech's fragment

And I need to prove the axiom of pair, union and power set using the weaker versions.

So, my approach for the first one is something like this:

Definition 1: Axiom of Pair:

$ \forall A\ \forall B\,\, \exists C\,\, \forall x\, (\,x \in C \leftrightarrow x=A \vee x=B\,). $

Claim 1: The weaker version of the axiom of pair, implies the stronger version.

Proof: We assume the weaker version of the axiom of pair. Then, if $A$ and $B$ are sets, there exist a set $C$ such that $A\in C$ and $B\in C$.

Let $\varphi(x)$ be the the statement frame "$ x = A \vee x = B $". And we define using the axiom of separation, the set $C^{*}$ as follows:

$y \in C^{*} \leftrightarrow y \in C \wedge \varphi(y)$

So, there exist a set whose elements are exactly the set $A$ and $B$.

Definition 2: The Union axiom:

$ \forall S\, \exists U\, \forall x\, (\,x\in U \leftrightarrow \exists A\, (\,A\in S \wedge x \in A\,)\, ) $

Claim 2: The weaker version of the axiom of union, implies the stronger version.

Proof: We assume the weaker version of the axiom of Union. So, If $S$ is a set, there exists a set $U$, such that $x \in A \wedge A \in S \rightarrow x \in U$.

Let $\varphi(x)\,$ be "$\, \exists A\in S.\, x\in A\,$". We define, using the schema of separation the set $\cup S$ such that:

$y \in \cup S \leftrightarrow y\in U \wedge \varphi(y) $

So,there is a set whose elements are exactly the elements of the elements of $S$.

  • $\begingroup$ I have only the weaker version of the well-known axioms and the separation schema. $\endgroup$ – Jose Antonio Jul 31 '13 at 2:39
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    $\begingroup$ What you did is correct, and the same idea works for the other cases. If $U$ is a set as in the weak axiom of union, then $\bigcup A$ is a definable subset of $U$, meaning, separation allows us to prove that $\bigcup A$ is a set. Same with power set. $\endgroup$ – Andrés E. Caicedo Jul 31 '13 at 2:41
  • $\begingroup$ @AndresCaicedo and André Nicolas. I see, let me work in the others and put them here. But, to be honest, I have that kind of feeling of incompleteness :P $\endgroup$ – Jose Antonio Jul 31 '13 at 2:48
  • $\begingroup$ The last one is a strange argument. I suspect that I should rethink it. $\endgroup$ – Jose Antonio Jul 31 '13 at 4:41

Your argument for the axiom of pair seems to use the same symbol $C$ for two different things. At first, you use $C$ for a set of the sort given by the weak axiom of pair, so $C$ contains $A$ and $B$ and possibly other things as well. Then you correctly apply the axiom of separation to produce a set $C^*$ that contains $A$ and $B$ and nothing else. At this point, you try to prove that $C^*=C$, but this does not follow from the information you have. It is true (and, as you said, clear) that $C^*\subseteq C$, but the reverse inclusion will be false if $C$, as obtained from the weak axiom of pair, contained elements other than $A$ and $B$.

Fortunately, you don't need $C^*=C$. You have a set, namely $C^*$, whose members are exactly $A$ and $B$ (and nothing else). The axiom of pair, which you're trying to prove says exactly that such a set exists. So your proof is already complete as soon as you've produced $C^*$.

The fact that the axiom of pair uses the symbol $C$ for the desired set does not in any way obligate you to show that the desired set that you produced coincides with another set that you happened to call $C$ earlier.

To put it another way, although the weak axiom of pair implies (in the presence of the axiom of separation) the axiom of pair, it is not the case that just any $C$ as described in the weak axiom of pair also fulfils the description in the axiom of pair.

  • $\begingroup$ Ok, I think I got It. Thank you so much. I need to rethink my arguments for the proof... $\endgroup$ – Jose Antonio Jul 31 '13 at 4:56
  • $\begingroup$ please sorry if my question is too elemental. But in each case as I see it just I need to use the separation axiom and the weaker version to guarantee the existence of the" stronger" one (as I called it), does it right? $\endgroup$ – Jose Antonio Jul 31 '13 at 5:20
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    $\begingroup$ @JoseAntonio That's right. In each case, the weak axiom gives you a set containing the elements you want plus possibly some more junk, and then the separation axiom lets you throw away the junk to produce a subset containing exactly the elements you want. $\endgroup$ – Andreas Blass Jul 31 '13 at 5:56
  • $\begingroup$ Thanks again. The idea is already clear :) $\endgroup$ – Jose Antonio Jul 31 '13 at 6:03

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