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The Claim:

From a conversation on Twitter, from someone whom I shall keep anonymous (pronouns he/him though), it was claimed:

[T]he existence of natural numbers and the fact that given a natural number $n$, there is always a successor $(n+1)$, do not imply the existence of an infinite set. You need an extra axiom for that.

It was clarified that he meant the Axiom of Infinity.

The Question:

Is the claim true? Why or why not?

Context:

I like how, if true, it goes against the idea that, if you just keep adding one to something, you'll get something infinite.

This is beyond me. Searching for an answer online lead to some interesting finds, like this.

To add context, then, I'm studying for a PhD in Group Theory. I have no experience with this sort of foundational question. I'm looking for an explanation/refutation.

To get some idea of my experience with playing around with axioms, see:

What is the minimum number of axioms necessary to define a (not necessarily commutative) ring (that doesn't have to have a one)?

I have included as Peano Arithmetic seems pertinent.

Shaun
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    The existence of $n+1=n\cup{n}$ follows from the axioms of union and pairing (and the existence of $n$), but this is not enough to show that a set containing every $n$ exists, and the axiom of infinity is needed. To see that note that $V_\omega$, the class of hereditarily finite sets, is a model for ZFC minus the axiom of infinity – Alessandro Codenotti Mar 22 '24 at 17:32
  • I would appreciate some feedback in the Helpful Commentary chatroom. – Shaun Mar 22 '24 at 17:47
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    Actually , we know that there must be infinite many natural numbers , if every $n$ has a successor since if there were a largest natural number , it would have no successor. – Peter Mar 22 '24 at 18:39
  • Apparently not, @Peter, according to the other responses so far. What have I missed? – Shaun Mar 22 '24 at 18:42
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    @Shaun For me , this is a valid formal enough proof just using this one axiom. What we cannot rule out that there are objects not being natural numbers , but the existence of infinite many natural numbers can clearyl be proven this way. – Peter Mar 22 '24 at 18:44
  • I don't understand, @Peter; I'm sorry. What do you make of the responses here so far, then? – Shaun Mar 22 '24 at 18:46
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    @Shaun We can prove without the axiom of infinity that no finite set of natural numbers contains all of the natural numbers, but we can't prove that there is any set that contains all of the natural numbers. – spaceisdarkgreen Mar 22 '24 at 18:57
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    As an analogy, we can prove that the set of all sets doesn't exist in ZFC. We can show there is infinitely many natural numbers, while not showing that there exists an infinite set of natural numbers. This was @Peter's point here. – Jakobian Mar 22 '24 at 19:10
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    As a group theorist, you might profit from the popular analogy between ZFC and the theory of groups. Both are first-order theories. Group theory is about the class of structures that are equipped with an operation satisfying the group axioms. ZFC is about the class of directed graphs ("universes of sets") that satisfy certain properties ("the axioms of ZFC"), and no more than that. "There is an infinite set" is a shorthand for a precise statement about these structures. An analogous situation in group theory might be "associativity cannot be deduced from the other group axioms". – Izaak van Dongen Mar 22 '24 at 20:38
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    Jakobian and Peter already mention the point, but I would like to extend with an example. You may prove that two sets $A$ and $B$ exists, but you need the Axiom of Union to guarantee the existence of the set ${A, B}$. The same is happening here. You may prove the existence of infinitely many natural numbers but you need an axiom of Infinity to guarantee the existence of a set containing exactly those numbers. – jjagmath Mar 22 '24 at 22:33
  • I really do not see why this deserves, at present, six downvotes. – Shaun Mar 22 '24 at 22:37
  • I welcome feedback in this comment. – Shaun Mar 22 '24 at 22:38
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    You need to define what you mean by a natural number - so as to exclude (for example) the numbers on the face of a clock, where the successor of 12 is 1. – Martin Kochanski Mar 23 '24 at 09:39
  • Duplicate of Why is the Axiom of Infinity necessary? and many others. – Bill Dubuque Mar 23 '24 at 20:15
  • That first link doesn't answer my question at all and the second one is just a link to the highest scoring questions in set theory here with search term "axiom infinity hereditarily finite", @BillDubuque; they're not even remotely satisfactory for my understanding. – Shaun Mar 23 '24 at 20:20
  • @Shaun You should read more carefully. Your answer (and more) are in the linked post(s). Please search for answers before posting questions. – Bill Dubuque Mar 23 '24 at 21:56
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    @jjagmath Quibble: it's not the axiom of union that's needed for that, but rather the axiom of pairing. The axiom of union doesn't go "$A,B\leadsto A\cup B$" but rather "$\color{red}{{} A_i:i\in I\color{red}{}}\leadsto\bigcup_{i\in I}A_i$" - you have to already have formed the set of sets you're trying to union together. – Noah Schweber Mar 24 '24 at 02:58
  • @NoahSchweber. You're right, I mixed the names of the axioms. – jjagmath Mar 24 '24 at 03:27
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    Under these conditions, there is no largest set. However, Using Powersets, Unions, Succesors, etc... will only ever get you another finite set. So constructing an infinite set is impossible, Under set theoretic operations. Which is why the Axiom Of infinity is even included. And as others have noted, The Set of Hereditary Finite Sets is a Model of ZF - Infinity, meaning we cannot expect to prove the Axiom of Infinity from the Other Axioms of ZF. – Michael Carey Mar 24 '24 at 17:44
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    Perhaps an easier example to parse: Imagine the Universe of Natural Numbers- where every object is a natural number, and we satisfy the Peano Axioms. Can we prove that an infinitely large natural number exists?? – Michael Carey Mar 24 '24 at 17:48
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    Also, to be more specific: Its not just that the Set of all Hereditarily finite sets is a Model of ZF - Infinity, Its really that its a model of (ZF - Infinity) + (There is no Infinite Set) i.e. which establishes the independence of Infinity from the other Axioms. – Michael Carey Mar 24 '24 at 17:54
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    I think it is odd that seven users downvoted a question that has an answer with a $+8$ score. I would encourage users to be more open-minded about questions that are slightly out of the box. – Mikhail Katz Mar 25 '24 at 10:35

3 Answers3

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As noted in the comments, the structure ($V_\omega$, $\in$) satifies all the axioms of ZFC except for the axiom of existence of an inductive set, and moreover every object of it is finite under all (usual?) definitions (since choice holds), so it follows that some axiom form of infinity is in fact necessary to prove existence of an infinite set

edit to further clarification: while naïvely one may think "but of course there are infinitely many different objects available (inside such a model / provably in such a theory, etc.)", the point is that 'finite' and 'infinite' really are formal phrases; compare and contrast Skolem's paradox

ac15
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The answer of ac15 is correct and to the point. It helps to clarify the underlying philosophical terminology. Namely, there are two types of infinity, in a sense.

If one says Take $n$, form its successor $n + 1$, and its successor $(n + 1) + 1$ in turn, and proceed, the described non-ending process may be viewed as a potential infinity. This is something that never ends without being a concrete whole. Other examples include when defining a formal theory, you may think of the set of syntactic variables as being potentially infinite (there are always more variables than you need). Or you may believe the tape in Turing machines to be potentially infinite (tape never runs out but does not need to be actually infinite in size), and so on.

When one adds This does not imply the existance of an infinite set, they are talking about completed or actual infinities. The set of natural numbers taken as a whole is (has the property of being) a completed infinity, as are $\mathbb{R}$, $\aleph_{5}$, and so on.

If you lived in $\mathrm{V}_\omega$, you would see $\omega$ as a potential infinity. Instead in $\mathrm{V}$, the set of naturals $\omega$ is the smallest completed infinity. In this particular case, the two notions of infinity are also more-or-less encapsulated by saying that $\omega$ is the trivial strongly inaccessible cardinal, provided you omit the condition of uncountability.

Linear Christmas
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One way of thinking of this is in terms of Peano Arithmetic (PA). The theory PA includes, of course, the successor axiom, and therefore a way of generating arbitrarily large numbers. However, PA does not prove the existence of an infinite set. In fact, PA is equiconsistent with the theory ZFC with the axiom of infinity replaced by its negation; see wiki on this.

Mikhail Katz
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