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The axiom of infinity says that the set of natural numbers exists, while the axiom of replacement says that if an object (a member of a set) exists, then all definable mappings of that object yield objects (e.g if 0 is an object, then 0+1 is also an object).
Doesn't this mean that the axiom of infinity is redundant since one can recursively prove the existence of the set of natural numbers using the successor function of the Peano axioms?


marked as duplicate by Carl Mummert, Cameron Buie, Mauro ALLEGRANZA, dantopa, Xander Henderson Mar 18 at 21:44

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  • $\begingroup$ See the post why-is-the-axiom-of-infinity-necessary $\endgroup$ – Mauro ALLEGRANZA Mar 18 at 12:38
  • $\begingroup$ This would prove that $\{0,1,\dots,n-1\}$ exists for all $n$, but not that $\{0,1,2,\dots\}$ exists. $\endgroup$ – Mike Earnest Mar 18 at 16:55
  • $\begingroup$ @MikeEarnest In the same manner, then the peano axioms assert the existence of $0,1,\dots,n-1$ for all n but not $0,1,2,\dots$. Why is this different? $\endgroup$ – Amr Ayman Mar 18 at 17:15

No, replacement doesn't imply infinity. What you've found is that if $S$ is a set so is $\{x\cup\{x\}|x\in S\}$, hence so is $x\cup\{x\}$. Each element of $\omega$ can be proven to be a set by this method, but we can't use this on its own to prove $\omega$ is a set, even though it's just the union of its elements.

  • $\begingroup$ So the problem lies in "infinitely applying the pairwise union axiom"? I don't see how that's different from the axiom of induction of the natural numbers.. $\endgroup$ – Amr Ayman Mar 18 at 13:00
  • $\begingroup$ @AmrAyman No, the problem is you don't have a way of proving these individual items all belong to the same set $\omega$. If you actually try to write down a proof that such a set exists, you'll find none of the other axioms let you do so. $\endgroup$ – J.G. Mar 18 at 13:01
  • $\begingroup$ Ok. But why can't $\omega$ be proven to exist, and simply be defined as the union of its elements? $\endgroup$ – Amr Ayman Mar 18 at 13:29
  • $\begingroup$ @AmrAyman Because you can't glue together arbitrary sets, e.g. as $\{x|\phi(x)\}$ or the union thereof for a unary predicate $\phi$. $\endgroup$ – J.G. Mar 18 at 13:50
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    $\begingroup$ Oh, that's only possible under unrestricted comprehension. My bad. $\endgroup$ – Amr Ayman Mar 18 at 17:59

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