the infinity axiom depends on axiom of regularity We know that axiom of infinity guarantees the existence of at infinite set.
But it is true if we can prove that $a$ $\not=\ a^+$.
we know that $a\subset a^+$,so we must show $a^+\not \subset a $ and for this we need axiom of regularity!!!
it means that the infinity axiom depends on axiom of regularity!
am i right?
 A: Usually stated, the Axiom of Infinity guarantees the existence of an inductive set, that is, a set $A$ with the following properties:


*

*$\varnothing \in A$; and

*$x \cup \{ x \} \in A$ for each $x \in A$.


Even though without Regularity there might be sets satisfying $x = \{ x \}$, and so $x \cup \{ x \} = x$, note that we can prove (without Regularity) that $\varnothing \cup \{ \varnothing \} = \{ \varnothing \} \neq \varnothing$.  And also that  $\{ \varnothing \} \cup \{ \{ \varnothing \} \} = \{ \varnothing , \{ \varnothing \} \} \neq \{ \varnothing \}$. And so on.  (Basically, given natural numbers $m$ and $n$, if the $m^\text{th}$ and $n^\text{th}$ sets "inductively constructed from $\varnothing$"1 are equal, then $m = n$.)
1Most texts would refer to these as the $m^\text{th}$ and $n^{\text{th}}$ successors of $\varnothing$.
A: No. The axiom of infinity does not depend on the axiom of regularity.
We can define the notion of an inductive set without referring to the axiom of regularity. Then we can prove that even if an inductive set includes some elements which satisfy $x\in x$, or something which contradicts regularity, the chain $\{\varnothing,\{\varnothing\},\ldots\}$ does satisfy well-foundedness.
The proof, as expected, is by induction. 
A: I don't think so. By the way that $a^+$ is defined = $a \bigcup ${$a$} you know that  $a^+\not \subset a$: the fact this complies with regularity is just as well, but not a consequence of regularity.
You need the axiom of infinity to garauntee that the sequence $a^+$, $a^{++}$ ... never ends: how else would you be sure of this, you can't write an infinite specification for the sequence ?
