The Axiom of Infinity states that there is a set $S$ containing $\varnothing$ such that if $x$ is an element of $S$ then so is $x\cup\{x\}$.

Is the following variant equivalent? There exists a nonempty set $S$ such that if $x$ is an element of $S$ then so is $\{x\}$.

This one also guarantees an infinite set, and it's shorter. The fact that we use the longer, more complicated one makes me think that the shorter version isn't as strong. Is this correct?

EDIT: To be clear, I'm wondering if they're equivalent if I leave the other axioms of ZFC unchanged.

  • $\begingroup$ Are you assuming the axiom of foundation? $\endgroup$ – Andrés E. Caicedo Aug 7 '15 at 3:44
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    $\begingroup$ I think it might be beacuse a common construction of the natural numbers follows easily first axiom. $\endgroup$ – ignoramus Aug 7 '15 at 3:46
  • $\begingroup$ @ignoramus I know. I'm curious if it's possible to construct the natural numbers from the second, as well. $\endgroup$ – Akiva Weinberger Aug 7 '15 at 3:48

I'm not sure it is shorter. To say a non-empty set you need $\exists x (x \in S)$ which adds to your statement. To prove equivalence, you define a function $f(x)=\emptyset, f(\{x\})=\{\emptyset \}\dots$ and use replacement on your $S$ to show the existence of the canonical $\omega$.

  • $\begingroup$ How do you define the function, though? Don't you need $\mathbb N$ first in order to show that inductively-defined functions exist? $\endgroup$ – Akiva Weinberger Aug 7 '15 at 4:10
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    $\begingroup$ @columbus8myhw: Nah, you can express the recursive definition directly in the language of set theory. For instance, you can say $f(y)=z$ iff there exists a transitive set $A$ containing $x$ and $y$ and a function $g$ with domain $A$ such that $g(y)=z$, $g(x)=\emptyset$ and whenever $a\in A$ and $\{a\}\in A$, $g(\{a\})=g(a)\cup\{g(a)\}$. It is tedious but not too hard to prove that this $f$ defines a function on the smallest set containing $x$ and closed under $y\mapsto\{y\}$. $\endgroup$ – Eric Wofsey Aug 7 '15 at 4:15
  • $\begingroup$ @EricWofsey Ah, that seems to work. Thank you! $\endgroup$ – Akiva Weinberger Aug 7 '15 at 4:18

In the absence of the Axiom of Foundation they are not equivalent. Without Foundation it is consistent that x={x} for some x. The Axiom of Foundation is that any non-empty x has a member y which is disjoint from x.

  • $\begingroup$ If there's an $x=\{x\}$, then doesn't $x\cup\{x\}=x$ as well? In any case, I'm mostly curious about whether they're equivalent if I leave the other axioms unchanged. Sorry if I forgot to mention that in the question. $\endgroup$ – Akiva Weinberger Aug 7 '15 at 3:55
  • $\begingroup$ If x={x} then of course xU{x}=x, but it's hardly an infinite set.Its only member is x. $\endgroup$ – DanielWainfleet Aug 8 '15 at 22:56
  • $\begingroup$ That doesn't mean they're not equivalent. It just means that neither of them deserves the name "Axiom of Infinity." $\endgroup$ – Akiva Weinberger Aug 9 '15 at 5:05

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