I noticed that if one writes $$\exists X:(\varnothing\in X\space\wedge\space(n\in X\rightarrow n\cup \{n\}\in X)) $$ As the axiom of infinity, one must define The Natural numbers as the smallest set that satisfies the above condition. But by replacing the conditional with an iff, ie: $$\exists\mathbb{N}!:(\varnothing\in\mathbb{N}\space\wedge\space(n\in\mathbb{N}\leftrightarrow n\cup\{n\}\in\mathbb{N}))$$ One could simply call the set proclaimed to exist and be unique above the natural numbers. Am I correct? Is there any reason that the Axiom of Infinity could not be replaced with this statement?

  • 1
    $\begingroup$ Even though this question has an easy negative answer, I hope it won't be downvoted. It's well formulated and allowed Henning to clear OP's confusion without any need for further clarification. (Which, unfortunately, is rare enough here on MSE.) $\endgroup$ – Stefan Mesken Aug 4 '16 at 0:20
  • $\begingroup$ Related: math.stackexchange.com/questions/1706661/… $\endgroup$ – Asaf Karagila Aug 4 '16 at 5:15

Your proposed condition does not define $\mathbb N$ uniquely.

For example, it is also satisfied by $V_\omega$, the set of all hereditarily finite sets.

We could take as an axiom that there exists an inductive set that has no inductive proper subset, and that would be unique. But that seems just to be added complexity that doesn't really buy us anything, given that we have the axiom of separation available anyway.

Alternatively we could take: $$\exists X: X=\{\varnothing\}\cup \{n\cup\{n\}\mid n\in X\}$$ which in the presence of the Axion of Regularity would pinpoint $\omega$ uniquely. (It is clearly inductive, and if $Y$ is any inductive subset of it, then $Y=X$, or else $X\setminus Y$ would violate Regularity).

  • $\begingroup$ Furthermore, given that we work in something that resembles $\operatorname{ZF}$, we have that for any infinite set $X$ there is some $X \subseteq Y$ such that there is a bijection $f \colon X \to Y$ and $Y$ is such that $y \in Y \leftrightarrow y \cup \{y \} \in Y$ - by a simple closure argument. $\endgroup$ – Stefan Mesken Aug 4 '16 at 0:16
  • $\begingroup$ @Stefan: I think you need some form of choice for that. If $X$ is amorphous, it cannot have your property. $\endgroup$ – Henning Makholm Aug 4 '16 at 0:18
  • $\begingroup$ Oh, right. Thanks! $\endgroup$ – Stefan Mesken Aug 4 '16 at 0:22
  • 2
    $\begingroup$ The OP's proposed condition is also satisfied by any non-zero limit ordinal. $\endgroup$ – DanielWainfleet Aug 4 '16 at 0:52

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.