# A confusion on Axiom of infinity

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 = $\phi$

B = $\left \{ \phi \right \}$

B = $\left \{ \phi,\left \{ \phi \right \} \right \}$

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

• 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

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.

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.