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I am working through some basic set theory and I am wondering if the proof that a set is unique in ZF are all similar. For example, if I wanted to show that the empty set exists and is unique I would do the following: $$\text{let } \psi(x)=\lnot(x=x), \text{then using the comprehension axiom I would form the set } \phi=\{x:\psi(x)\}$$ I think this is enough to show that such a set exists. To show that it is unique I would then do the following: $$\text{ assume } A =\phi \text{ and } B=\phi \text{ with } A ≠ B, \text{ then given some } x\text{ }, x\in A \iff \lnot(x=x) \iff x\in B$$

By the extension axiom this would imply that A$=$B.

Is this the right technique for showing set uniqueness? They all seem to involve a similar type of proof where you show the set exists, assume two distinct sets have the same property as the set shown to exist, and then chain the two sets together using the formula that was used to define the set. I have done this with a number of sets, like union and intersection, and they seem to follow the same scheme. I appreciate any help.

Newman
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    You know that ZFC doesn't have full comprehension, I hope. That's why there is an axiom just stating that there is some set. Apply comprehension to this set to construct the empty set. (I know, that's not the question, but it'd be good to be aware of that.) Also, when you write $A=\phi$ and $B=\phi$, be aware of that. Unless you have shown that $\forall x(\psi(x)\rightarrow x\in C)$ for some set $C$, there is no guarantee that $A$ and $B$ are sets. (If I interpret the meaning of $A=\phi$ correctly, and your notation correctly.) – martin.koeberl Mar 17 '17 at 21:38
  • This is why I prefer to refer to the axiom of comprehension as "the axiom of subset selection": it makes it completely clear that we are only guaranteed to be able to select subsets of a set, rather than select globally. – Patrick Stevens Mar 18 '17 at 07:59

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First, you need the axiom of Existence : $\exists x\;(x=x).$ All the other axioms except the axiom of Infinity begin with "$\forall$" so without Existence and without Infinity we cannot prove anything exists.

Second, Comprehension does NOT state the existence of a set $\{x:P(x)\}$ for every $P.$ This leads to paradoxes, for example Russell's Paradox. What it states is $$\forall y\;\exists z\; \forall x\;(x\in z\iff (x\in y\land P(x)\;).$$ That is, any $y$ that does exist has a subset $z$, whose members are all and only those $x\in y$ that satisfy $P(x).$

So by Existence and Comprehension we have $$\exists x \;[\;x=x\land \exists y\; \forall z(z\in y\iff (z\in x\land z\ne z\;))\;].$$ Therefore $\exists y\;\forall z\in y\;(z\ne z).$ Now Extensionality implies $\forall z\;(z=z).$

Therefore $\exists y\;\forall z\;(z\not \in y).$

DanielWainfleet
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