0
$\begingroup$

I am reading Jech's book on set theory. In chapter 2, ex 2.4 says:

(Without axiom of infinity). Let $\omega =$ least limit $ \alpha \neq 0$ if it exists, $\omega = Ord$ otherwise. Prove that the following statements are equivalent:

  1. There is an inductive set

  2. There exists an infinite set.

  3. $\omega$ is a set.

For $(2) \Rightarrow (3)$, apply replacement to the set of all finite subsets of $X$.

My questions are:

  1. Isn't (2) exactly axiom of infinity. If so, when i prove $(2) \Rightarrow (3)$ can I assume I have axiom of infinity?

  2. I guess I can use the following facts?: All other axioms, $\mathbb N$ is the intersection of all inductive sets.

  3. Can I use anything else?

  4. Can I assume the "names" of elemtnts in a set are not important? For example, Is $\{0,1,2,3,4,...\}$ identical to $\{ \emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \},... \}$ which is identical to $\{ a, \{ a \}, \{ a, \{ a \} \}... \}$?

Another question: How do I make new lines here in the stackexchange editor window? is there a special command for that?

Thank you

$\endgroup$
  • $\begingroup$ Concerning 1.: No. The precise formulation of the axiom of infinity is given on page 13. In particular, the axiom of infinity is (1), i.e. that there exists an inductive set. $\endgroup$ – William Sep 15 '14 at 7:13
  • $\begingroup$ Note that in (2), I believe being infinite here means not in bijection with a finite cardinal. $\endgroup$ – William Sep 15 '14 at 7:15
2
$\begingroup$

The axiom of infinity states "There exists an inductive set". An infinite set is a set which is not finite. Inductive sets are not finite, so the axiom of infinity tells us that there is an infinite set. But specifically, it tells us that there is an inductive set.

Of the three implications, $1\implies 2$ is trivial, by noting that no finite set is inductive; and $3\implies 1$ is trivial by noting that $\omega$ is an inductive set. So the only part which left is $2\implies 3$.

You have to show now, why given that there exists an infinite set, we can deduce the existence of $\omega$. If you already know what a transitive closure is, and what is the rank of a set, then this should be sufficient for a quick proof. If you don't you might have to do this slightly more "by hand".

Here's a small hint: Let $X$ be an infinite set. Use the power set and separation axioms to conclude that the set $\{A\subseteq X\mid A\text{ is finite}\}$ exists. Now use replacement to prove that $\omega$ exists.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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