Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently working "the elements of advanced mathematics" by steven g. krantz, currently on Chapter 5.

I came to "Axiom of Infinity" which roughly states:

$$\exists A \; s.t. \; \phi \in A \; and \; \forall a\in A, a\cup \left \{ a \right \} \in A$$

Now, doesn't this mean: $$\exists B=\left \{ \phi,\left \{ \phi \right \},\left \{ \phi,\left \{ \phi \right \} \right \},\left \{ \phi,\left \{ \phi \right \},\left \{ \phi,\left \{ \phi \right \} \right \} \right \},... \right \}$$ which results in: $$B \in B$$ ??? (the 'last' element of B will be B itself...)

did I get this wrong, or are there some "premises" that I'm missing?

Thanks :D

p.s: Any good source for learning ZFC??? the book seems to fly off in a hurry and doesn't explain much, and my google-fu isn't giving me any "ZFC-for-dummies"


OK, so is this a good summary of the answers?

Notice that

B[0] = $\phi$

B[1] = $\left \{ \phi \right \}$

B[2] = $\left \{ \phi,\left \{ \phi \right \} \right \}$

and so on (thus, for B[n], as n++, B[n] -> B)

BUT, similar to "limits doesn't mean the value actually approaches lim", B[n] never reaches B

share|improve this question
1  
I like Hrbacek & Jech, Introduction to Set Theory. –  Brian M. Scott Jun 8 '13 at 1:16
    
Yes. So the axiom of infinity states that the class $\{B[n]\mid n\in\Bbb N\}$ is actually a set. –  Asaf Karagila Jun 8 '13 at 1:41
    
No, that's not a good summary of the answer. There is no limit here (it is not even meaningful), B is not a limit of a sequence of B[n]. B is a set that contains all of those B[n], similar to the comment above (but technically, it could contains extra stuff). –  anonymous Jun 8 '13 at 3:26
    
@anonymous: It is in fact the limit. It's the union of all the $B[n]$'s. –  Asaf Karagila Jun 8 '13 at 7:24

2 Answers 2

up vote 3 down vote accepted

But there is no last element to $B$. Indeed the next "step" would be $B\cup\{B\}$, but that set is not equal to $B$ anymore.

share|improve this answer

Notice that $a$ have a finite number of element, then $a\bigcup\{a\}$ still have a finite number of element. And since you started with $\emptyset$ which is certainly finite, all the element in $B$ that the axiom asserted to be in are all set with finite number of element.

On the other hand, $B$ is infinite. So this axiom do NOT assert that $B\in B$.

The axiom do not assert that $B\notin B$ either, but this is done by axiom of regularity.

This is just a set-theoretic way of expressing certain axiom from Peano's arithmetic. If $a$ is a number, then $a\bigcup\{a\}$ is the successor. So at the minimum $\mathbf{N}$ is included in $B$, though the axiom did not deny the possibility of extra stuff.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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