Original Post
This may be a stupid question, but does there axist an axiom $\phi$ that is independent of $\mathsf{ZFC}$, and not equivalent to the axiom of $\mathsf{Infinity}$, such that $\left(\mathsf{ZFC} -\mathsf{Infinity}+\phi\right) \vdash \mathsf{Infinity}$? Can such a $\phi$ exist? Or will all $\phi$ be equivalent to $\mathsf{Infinity}$?

Edit1: Danul's answer certainly answers the question, but I am wondering if there exists a weaker axiom $\phi$.
Edit2 I realize I had no idea what I meant by weaker axiom

Edit3, Conclusions
I guess I want to know whether or not every axiom $\phi$ such that $\mathsf{ZFC}-\mathsf{Infinity} + \phi$ proves that "There exists an inductive set" is such that $\phi$ is either stronger than $\mathsf{Infinity}$ or equivalent to $\mathsf{Infinity}$ and I guess this is the case as $\left(\mathsf{ZFC} -\mathsf{Infinity} + \phi\right) \vdash \mathsf{Infinity} \Longrightarrow \left(\phi \Longrightarrow \mathsf{Infinity}\right) \lor \left(\phi \longleftrightarrow \mathsf{Infinity}\right)$

In this beginning, I really just wanted to know if we could prove that a countably infinite set existed without simply postulating that one such exists. Can we? What is the most minimal assumption that we need to make? Is it really just $\mathsf{Infinity}$?
AH, correct, Thanks Danul. I guess what I meant was: $$ \left(\mathsf{ZFC} -\mathsf{Infinity} + \phi\right) \vdash \mathsf{Infinity} \Longrightarrow \left(\phi \vdash \mathsf{Infinity}\right) $$ As the axiom of infinity cannot be derived from the rest of the axioms of $\mathsf{ZFC}$, so it must be the case that $\phi \vdash \mathsf{Infinity}$. Steven Stadniski answered my question in the comments. Thanks everyone.

I am making all sorts of reasoning errors! We have that $$(\mathsf{ZFC} - \mathsf{Infinity} + \phi) \vdash \mathsf{Infinity} \not{\rightarrow} \phi \vdash \mathsf{Infinity}$$

Because $\phi = \mathsf{Extensionality} \longrightarrow \mathsf{Infinity} $ is a counterexample.

False logic will be the end of me.

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    $\begingroup$ In a way, the axiom of infinity as usually formulated has the property. Without at least some rudimentary set theory, it doesn't really say that an inductive set exists, and it certainly doesn't say that there is an infinite set. $\endgroup$ – tomasz Apr 5 '14 at 21:04
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    $\begingroup$ How do you mean '$\phi$ doesn't already prove infinity'? If $\mathsf{ZFC}-\mathsf{Infinity}+\phi\vdash\mathsf{Infinity}$ doesn't that by definition imply that $\phi$ proves Inf? Or are you thinking in a weaker formal system than $\mathsf{ZFC}$? When you talk about relative consistency results you have to be very precise about what base systems your equivalencies are over... $\endgroup$ – Steven Stadnicki Apr 5 '14 at 21:04
  • $\begingroup$ I am really trying to make precise what I mean, but perhaps I'm doing a piss poor job of it. $\endgroup$ – Rustyn Apr 5 '14 at 21:08
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    $\begingroup$ @Rustyn: I think with a question like this, you should say exactly what you mean by "not equivalent" and what exactly you mean by "Infinity". Usually it means that there is an inductive set, but I think there are weaker formulations that imply the existence of infinite sets, for example "There exists a set with more than $1$ element which is equipotent with its own square". I might be wrong, but I think this would not imply the existence of infinite ordinals. $\endgroup$ – tomasz Apr 5 '14 at 21:10
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    $\begingroup$ @Rustyn: I think something like "if (some axiom of ZFC), then infinity" should provide a counterexample. But you need to take care so that the axiom is not one of those required for the notion of inductive set to make sense. $\endgroup$ – tomasz Apr 5 '14 at 21:20

Make $\varphi$ be Axiom of infinity plus $GCH$. Then you have that $ZFC-\text{infinity}+\varphi\vdash\text{infinity}$, but $\varphi$ is definitely much stronger than $\text{infinity}$.

Edit: I don't know what you mean by weaker axioms. But remember that $ZFC$ can produce a model (in-fact a countable transitive model) for $ZFC-\text{infinity}+\neg{\text{infinty}}$, namely the hereditarily finite sets (or $V_{\omega}$ if you prefer). So in a sense you can't really be any weaker.

Edit 2: At your current edit (You should keep track of the edits btw or else the entire conversation will become useless, so maybe not delete the original post) the question is trivial. Let $\varphi=\neg{\text{infinity}}$. Now the right hand side is trivial because of the way $\implies$ is defined.

Edit 3: Edit 2 is meant to show that your reasoning is faulty. $\neg{\text{infinity}}$ and $\text{infinity}$ are contradictory, but if you set $\varphi=\neg{\text{infinity}}$, you still get $ZFC-\text{infinity}+\varphi\vdash \text{Infinity}\implies(\varphi\implies\text{Infinity})\lor(\varphi\iff{\text{infinity}})$.

For an answer to your informal question, see the first edit for this solution.

  • $\begingroup$ That doesn't make any sense. This axiom has to combine with the other ZFC axioms to prove there exists an infinite set anyway. Otherwise your question itself doesn't make sense. $\endgroup$ – UserB1234 Apr 5 '14 at 21:03
  • $\begingroup$ You need to make your question more precise. The way you have formulated this I can give an even more trivial answer that I already did: Take $\varphi=\exists{x}x\neq{x}$. This will prove everything on its own and has nothing to do with infinity (or you don't really need to worry about models) whatsoever $\endgroup$ – UserB1234 Apr 5 '14 at 21:12
  • $\begingroup$ I have edited the question and come to a conclusion. $\endgroup$ – Rustyn Apr 5 '14 at 21:19
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    $\begingroup$ I think we have $ \mathbf{ON}^{V_\omega} = \omega $. So $ \omega $ is a proper class and $ V_\omega $ is "nothing at all", i.e. not definable. $\endgroup$ – Justus87 Apr 5 '14 at 22:48
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    $\begingroup$ It is the universe for the model and is not a regarded as a set inside the model. However it is the class $x=x$ in this case. (To see that it is not a set we can carry out the argument the class of all sets is not a set inside of $V_{\omega}$). $\endgroup$ – UserB1234 Apr 6 '14 at 14:13

If you are looking for a simpler, more intuitive alternative to the ZF axiom of infinity -- reading between the lines here -- you might consider simply assuming that there exists a set $S$ on which there is defined an injective, but not surjective function $f$, i.e. $S$ is Dedekind-infinite. Could it get more minimal? From such a set, you can extract at least one subset $N$ that satisfies Peano's axioms where $f$ serves as the successor function.


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