# Set theory

1. Sep 18, 2009

### Fredrik

Staff Emeritus
I'm thinking about buying https://www.amazon.com/exec/obidos/tg/detail/-/0412606100/. Both books are getting excellent reviews at Amazon, especially Goldrei.

I would like to learn about the ZFC axioms, cardinals and ordinals, etc. I assume both books will cover those topics. I'm also very curious about something I heard for the first time today:
Am I reading this right? Every set in ZFC theory can be constructed from the empty set by iterating the power set operation?! That sounds weird. I know you can define the integers this way, and infinite ordinals I guess, but everything else?

0={}
1={0}
2={0,1}
...

If this one axiom leads to all the ZFC axioms, then I should have heard about this before, so I feel that I must have misunderstood something.

Do both of these books explain these things?

Last edited by a moderator: Apr 24, 2017
2. Sep 18, 2009

### Hurkyl

Staff Emeritus
Let V denote the class of all sets that are contained in some transfinitely-iterated power of the empty set. (As in the Wiki article)

Suppose the set S0 is not in V. Then S0 must contain an element, S1, not in V. Iterating this argument (and using the axiom of choice), we can construct an infinite chain $\cdots \in S_2 \in S_1 \in S_0$. This contradicts the axiom of foundation, therefore S0 cannot exist.

Now, this fact does not imply all the axioms of ZFC.

This fact cannot tell you "how many" ordinal numbers are available for use -- the class of ordinals is, roughly speaking, one of the "inputs" to this construction.

This fact cannot tell you much about what kinds of sets exist and how complex they can be. e.g. it cannot tell you whether or not there exist sets whose cardinality lies between those of Vw+1 and Vw.

This fact cannot even tell you that two sets are equal iff they have the same members!

3. Sep 18, 2009

### Fredrik

Staff Emeritus
If "this fact" refers to the idea that all sets are constructed by iterating the power set operation, then you have convinced me that it can't replace the ZFC axioms (probably not even a single one of them). You saved the best argument for last:

I should have realized that myself. I guess I didn't really think because I started the thread to ask what book I should get, and ended up throwing in a question that seems silly now.

I don't understand this. I'm not sure I understand the power set construction either, but it seems to me that the first steps are

V0={}
V1={{}}={0}
V2={{},{0}}={0,1}
V3={{},{0},{1},{0,1}}={0,1,{1},2}
V4={0,1,{1},{2},2,{0,{1}},{0,2},{1,{1}},{1,2},{{1},2},3}

Suppose your V is V4. You said that if S0 isn't in V, it must contain an element that isn't in V. But {0,{1}} isn't in V (it's a subset, but not a member) and it doesn't contain anything that isn't in V. So your proof seems to fail at the first step.

But I have no doubt that you'll explain what I have misunderstood and make me feel silly again.

4. Sep 19, 2009

### Hurkyl

Staff Emeritus
The (von Neumann) universe is constructed by iterating the power set operation. The theorem is that all sets are elements of the universe.

As for your example, the set you chose is a member of V4, so I don't understand what you were getting at.

Incidentally, the contrapositive of my claim is (hopefully) more obvious:
If every element of S is in V, then S is in V​

This requires all of V -- if you stop iterating at some point, then it breaks down. For example, a set whose elements all lie in V4 is not necessarily a member of V4. (But it is necessarily a member of V5)

5. Sep 19, 2009

### Fredrik

Staff Emeritus
Sorry about the {0,{1}}. How about {0,{2}}? That's not in V and doesn't contain anything that isn't in V.

Also, my notation is inconsistent. I should have written 0,1,2,... on the left-hand sides, or V0,V1,V2,... on the right-hand sides. It doesn't make much sense to mix two notations the way I did.

6. Sep 19, 2009

### Hurkyl

Staff Emeritus
Equivocation fallacy. When I made that statement, I was using V to refer to the entirety of the Von Neumann universe. But you're using V to refer to the V4 subclass.

7. Sep 19, 2009

### Landau

Last edited by a moderator: Apr 24, 2017
8. Sep 19, 2009

### Fredrik

Staff Emeritus
OK, that wasn't at all clear, I think. But now that you've cleared that up, your proof is easy enough to understand. I needed some help from Wikipedia to understand why that sequence contradicts the axiom of foundation, but I think I got it now.

9. Sep 19, 2009

### Fredrik

Staff Emeritus
That doesn't seem like a bad thing. But the reviews at Amazon suggest that Potter is using some "new" approach to the subject. It doesn't seem to be about ZFC at all, but rather about some other set theory. If that's right, it's certainly not the best place to learn about ZFC, but I might still be interested in the things he is talking about. Have you read this book? Can you tell me what it's about?

10. Sep 20, 2009

### Fredrik

Staff Emeritus
I'm still about this. If we define V0={}, V1={V0}, V2={V0,{V0}}={V0,V1}, and so on, Vn is a set with 2n-1 members for n=1,2,3,... If these are the only sets we allow, there are no sets with 3 members. Is {1,2,3} not a set?!

11. Sep 20, 2009

### Hurkyl

Staff Emeritus
{1,2,3} is an element of the universe because it's an element of V5.

V is not the class whose elements are the Va: V is the class which is the union of all the Va.