Really, really basic question in set theory

[IamNameless]

Very simple question

Are the Pairing Axiom and the Union axiom in the Zermelo–Fraenkel set theory the same?

I have a book that states them as the following:
Pairing Axiom: For any sets u and v, there is a set having as members just u and v.
Union axiom: For any sets a and b there exists a set whose members are those belonging to either a or b.

Also in the book, they give these definitions in the form of a logic definition (I'd post but I can't find some of the symbols in any LaTex reference), the definitions are completely identical.


So are they different and if so what is the difference (and what would I be able to prove with one but not the other).


Thanks!

 

[HallsofIvy]

You didn't post the "logic definitions" so I can't speak for those being identical (sometimes it's easy miss two symbols being interchanged, etc.) but the two definitions you give are clearly not completely identical!

 

[IamNameless]

Ignore what I just wrote, they were not correct. I'll edit in one minute.

 

[IamNameless]

Hmm. Just ignore this post. I can't put the logic symbols on here without a great struggle, I'm going to ignore one of these axioms and continue with the book.

 

[Dragonfall]

Suppose you have two sets A={a, b, c} and W={x, y, z}.

Pairing says: {{a, b, c}, {x, y, z}, ...} exists.

Union says: {a, b, c, x, y, z, ...} exists.

See the difference? Note that the axioms do not say that the set with specifically those elements exist, but rather a set with those elements exist (that set may contain other stuff). You will need to use comprehension after to get the specific sets {{a,b,c,},{x,y,z}} and {a,b,c,x,y,z}.

In rigorous notation,

Paring: \forall a\forall b\exists c(a\in c \wedge b\in c)

Union: \forall a\forall b\exists c(\forall x((x\in a\vee x\in b)\rightarrow x\in c)).

If you really want to ignore one of them, ignore paring, since you can prove it using replacement and infinity. I don't remember how.

 

[Hurkyl]

Union axiom: For any sets a and b there exists a set whose members are those belonging to either a or b.

That's not the union axiom; the union axiom says that given any set S, there exists another set T whose elements are precisely the elements of the elements of S.

e.g. if S = {{a, b, c}, {c, d, e}}, then the axiom of union says that {a, b, c, d, e} is a set.

If you really want to ignore one of them, ignore paring, since you can prove it using replacement and infinity. I don't remember how.

(1) Construct a set with 2 elements.
(2) Replace those 2 elements with the desired objects.

 

[Dragonfall]

Where does infinity come into this?

 

[Hurkyl]

If you don't assume the pair axiom, how else are you going to find a set with 2 elements?

Unless I'm much mistaken, if you discard the axioms of infinity and the pair set, then there is a model of the remaining axioms where the empty set is the only set. If you also discard the axiom of the null set, then there even exists a model where no sets exist!

 

[Dragonfall]

You mean if you don't assume the infinity axiom?

If you assume power set and at least the empty set, then you can certainly construct a set with 2 elements without the pair axiom. Or infinity for that matter:

0
{0}
{0, {0}}

Unless mistaken, you can get a "class" of all finite sets \mathbb{V}_{\omega}. Infinity says, of course, that \mathbb{V}_{\omega} is a set. No model of it can only include the empty set unless you change extensionality to something weaker so that 0={0}.

 

[Dragonfall]

Union: \forall a\forall b\exists c(\forall x((x\in a\vee x\in b)\rightarrow x\in c)).

Forget what I said here. It's wrong.

 

[Hurkyl]

You mean if you don't assume the infinity axiom?

If you assume power set and at least the empty set, then you can certainly construct a set with 2 elements without the pair axiom. Or infinity for that matter:

0
{0}
{0, {0}}

Unless mistaken, you can get a "class" of all finite sets \mathbb{V}_{\omega}. Infinity says, of course, that \mathbb{V}_{\omega} is a set. No model of it can only include the empty set unless you change extensionality to something weaker so that 0={0}.

You can't write {0} and {0,{0}} unless you've already proven that pair sets exist. But anyways... You're right in spirit (I think); you can show that 0 and P(0) are both elements of P(P(0)), and that they are distinct.


But what fails in a "there is only one set" model isn't quite what you would think:

I'm considering a model where only one set exists (which I will call 0), and that 0 is an element of 0.

Going down the list in the wikipedia article:

Extensionality holds.
Regularity fails.
Specification fails.
Pairing holds.
Union holds.
Replacement holds. (I think)
Infinity holds.
Power set holds.

And I don't want to try and unfold the statement of the axiom of choice.

 

[Dragonfall]

Yes I see what you mean. I was thinking about what kind of set would be a fixed point under power sets. Obviously no well-founded set can. So x={x} would be it.