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Many arguments about the foundations or philosophy of mathematics centre on the question of whether or not there exist objects or entities (such as certain sets) which are not "finite".

(For instance, Doron Zeilberger, although he is fond of April Fool jokes, does not seem to be joking here, or here.)

For such arguments to take place at all, the participants must share some understanding of what the word "finite" means.

I have never taken part in any such argument, but if I were to have to do so, I would nervously have to admit that I do not know what this shared understanding is, and I would ask if the others would mind spelling out (if only for my benefit) what it is that they think they are arguing about.

Just to anticipate two possible lines of ensuing discussion (assuming that I wasn't just greeted by stunned or embarrassed silence):

If (assuming for the moment that there was no dispute as to what a "set" is) someone were to say that a set is finite if and only if it can be put into one-to-one correspondence with a set of the form $\{1, 2, \ldots, n\}$ for some natural number $n$, this would obviously be open to the objection that it takes for granted a common understanding of the existence of a unique set of natural numbers, known to all the participants in the discussion. It might perhaps also be objected to because it apparently implies performing a possibly non-terminating search for such a number $n.$ But even if that is not a problem, it surely cannot be the case that a person cannot even be said to know what the word "finite" means unless they already accept the existence of the (uniquely defined) infinite set of all natural numbers.

(Zeilberger, for one, would presumably be inclined to make some objection(s) along these lines.)

If, on the other hand, someone were to put forward Dedekind's definition that a set is finite if and only if it cannot be put into one-to-one correspondence with a proper subset of itself, this could be objected to because it seems to imply that you can only directly show a set to be finite by proving something about the collection of all mappings of that set into itself, whereas this can hardly be what anybody has in mind when they insist that all mathematical objects must be finite, therefore it cannot be the agreed-upon common understanding of what the word "finite" means.

(Tarski's definition that a set is finite if and only if every non-empty collection of its subsets contains a maximal element is open to a similar objection. So too is Staekel's definition that a finite set is one that can be doubly well-ordered.)

So, some other definition of "finite" would have to be agreed upon; but what might it be?

Of course, the participants, whatever their other disagreements, might silently agree upon silence as the only appropriate response to such a ridiculously naive and ignorant question; but what if it had to be discussed, say for the benefit of a child, whose naivety and ignorance could be excused?

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    You ask what people mean by "finite", but you seem to know the answer? – GFauxPas Jun 01 '15 at 16:22
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  • I do not understand the Tarski definition of Finitude you discuss. Consider the set of all integers S = $(\infty, 0]$. Each subset has a maximal element yet S is clearly infinite. – letsmakemuffinstogether Jun 01 '15 at 16:23
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    I feel that the existence of so many unnatural definitions of being finite just means that this notion is far from being as elementary as we want it to be. – lisyarus Jun 01 '15 at 16:24
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    @letsmakemuffinstogether not "every subset has a maximal element", but "every collection of subsets has a maximal element" (the latter maximal is meant in the sence of set inclusion). – lisyarus Jun 01 '15 at 16:25
  • @GFauxPas I'm oddly schizophrenic (perhaps almost literally so) about mathematics. I can usually at least go through the motions of doing it, and sometimes I even seem (even to myself) to know what I'm doing; but if I stop to think about it (which is sadly often), I am utterly bewildered, and the answers I come up with to my own questions seem always to raise further questions, and worse. You are right if you suspect that this question is addressed as much to myself as to anyone else. – Calum Gilhooley Jun 01 '15 at 16:28
  • The most basic version of finiteness in math is this: "Property $A(x)$ is true for finitely many values of $x\in\Bbb N$". This is equivalent to "There exists an $N\in\Bbb N$ such that $\lnot A(x)$ when $x>N$", so that can be taken as our basic definition of finiteness. The definitions of finiteness you've discussed are basically trying to extend this concept to sets. (Kinda like how the basic definition of exponentiation is "repeated multiplication", and then things like $e^{\pi i}$ use an extended definition.) – Akiva Weinberger Jun 01 '15 at 21:18
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    Why can't anyone have the Dedekind definition in mind when discussing finitude? – jwodder Jun 01 '15 at 21:30
  • @columbus8myhw Can your comment be fairly paraphrased as: (1) all concepts of finiteness, for sets, reduce to the concept of a finite set of natural numbers; and (2) a finite set of natural numbers is any set of natural numbers that is a subset of the set ${1, 2, \ldots N},$ for some natural number $N$? Such a rephrasing might allow even a disbeliever in $\mathbb{N}$ as an infinite set to adopt your definition. This in turn suggests that I may have been mistaken in asserting that the first definition I listed implies belief in $\mathbb{N}$ as an infinite set. Perhaps I'm learning something. – Calum Gilhooley Jun 01 '15 at 21:43
  • @jwodder Your question deserves a better answer than I can give it at the moment. For now, I'll just say that although I don't have any particular problem with Dedekind's definition in the context of pure mathematics (I've nothing to add to standard remarks about, for example, the need for the AoC in proving equivalence of various definitions), it doesn't seem to represent what we mean when we talk of finiteness outside that context: for instance, in simply counting a set, or asserting that there are finitely many stars in the universe, or that some formula is made up of finitely many symbols. – Calum Gilhooley Jun 01 '15 at 22:02
  • @CalumGilhooley Yeah, I suppose you could say that. – Akiva Weinberger Jun 02 '15 at 00:06

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If two people are arguing over whether or not “everything” is “finite”, then I'd say the difference between “cannot be put into one-to-one correspondence with a proper subset of itself” and “has a bijection to some set $\{1,\ldots,n\}$”, or any other definition of “finite”, is basically irrelevant. It's extremely unlikely they would suddenly agree if they meticulously chose a common definition, so why bother? (Consider an analogous situation with people arguing over the “existence” of “God”.)

Remember, even if a formal mathematical definition was assigned a particular natural-language name (like “finite” or “smooth”), it's only because it is felt to capture some aspect of that concept (at least according to the namer). I'd say that people arguing over whether “everything” is “finite” have a disagreement regarding the natural-language concept of “finite” itself; they don't need to have agreed about which mathematical approximation to this concept they like the most.

Zev Chonoles
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    I don’t think that Zeilberger’s understanding of finite is significantly different from mine, despite his (in my view absurd) ultrafinitism. I’ve read only a little of his writing on the subject – I find it silly, frankly – but it seems to me that we disagree not on the meaning of finite but rather on the meaning of exists (at least in mathematics). – Brian M. Scott Jun 01 '15 at 18:05
  • That's a really nice little insight @BrianM.Scott. Its probably fair to say that "exists" is one of the most contentious word in all of mathematics. This contention seems to be at the root of most foundational disagreements, including the finite/infinite debate and the classical/constructive debate. – goblin GONE Feb 29 '16 at 13:57
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Well. This is a tough cookie to answer properly.

The reason is simple, though. Finite is one of those words which has a mathematical definition, but also a natural language definition and those are so close that we might confuse the two.

This is similar to what does a set mean. Is a set some predefined notion, is it an element of a model of $\sf ZF$, or $\sf Z$, or $\sf NF$ or $\sf KP$, or maybe an object in the category $\bf Set$. Do every set has a power set? Do every definable subset of a set is a set?

These are notions which are fuzzy, specifically because they are taken as somewhat of primitive notions in mathematics.

But suppose that you have happened to agree upon some notion of "set", and let's agree to stipulate that it satisfied some naive set theory which is close in flavor to $\sf ZF(C)$.

Now you have several options:

  1. Claim that the natural numbers are not sets. They are urelements, or some atomic entities which satisfy the second-order axioms of $\sf PA$. Therefore the question what are the natural numbers is moot. And a set is finite if it can be mapped bijectively with a bounded set of natural numbers.

  2. Define the finite ordinals, claim that the class of finite ordinals is "definable" (either as a set, or as a proper class if you want to reject the axiom of infinity). Then prove that the finite ordinals satisfy $\sf PA$, so they are worthy of being called "The Natural Numbers", and we are reduced to the previous case.

  3. Use one of the many notions of finiteness which do not appeal to the natural numbers. These include, but not limited to, the following:

    • Every self injection is a surjection.
    • Every self surjection is a bijection.
    • Every non-empty chain of subsets has a maximal element.
    • Every non-empty collection of subsets has a maximal element.

    Be forewarned, though, that apart of the last one, the axiom of choice is generally needed to prove that this is equivalent to the first suggested definition.

You may claim that the fact that there are definitions which are non-equivalent in the absence of the axiom of choice means that finiteness is not well-defined. And this is true. You can argue that you reject both the axiom of choice (and in fact, the axiom of countable choice), and the usual definition of finiteness. But you can also reject the axioms of induction in $\sf PA$ and claim that they are inconsistent, and you can reject the soundness of propositional calculus.

You can do all these things, but mathematics is a joint effort. If you are unwilling to agree on primitive notions like set, like finiteness, like natural number, then the problem lies in a deeper level than just this.

Asaf Karagila
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    I'm afraid I cannot always decipher downvotes-based communication. If someone can translate them to English, or even Hebrew, I'd be glad to hear what's wrong with this answer! – Asaf Karagila Jun 01 '15 at 20:33
  • How do you say "downvote" in Hebrew, out of curiosity? – Akiva Weinberger Jun 01 '15 at 20:38
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    Well, "downvote" isn't really a word in any language. So you either translate it as "vote it down" in one way or another, or you let me think a little bit and try to come up with a suitable portmanteau (or contact the Academy of Hebrew Language and submit a query which may take several months or years to be processed, debated and approved). – Asaf Karagila Jun 01 '15 at 20:42
  • I'm thinking "הצבעה למטה" but I don't know if that sounds weird (or is hopelessly ungrammatical). …Sorry for bringing this up, I realize it's completely off-topic. – Akiva Weinberger Jun 01 '15 at 20:44
  • That sounds more like "voting downstairs" :-D – Asaf Karagila Jun 01 '15 at 20:48
  • More on-topic — ZFC is a first-order theory, which means that it has nonstandard models. One of them (used in IST, a conservative extension of ZFC that can be used for nonstandard analysis) basically includes what I'd love to call "infinite" natural numbers, but are technically finite by definition. (The author uses the term "unlimited".) So this could be an example where the formal definition doesn't necessarily match the intuitive definition. (Alternatively, you could say that I'm giving a semantic interpretation to something that could be considered to be purely syntactic.) – Akiva Weinberger Jun 01 '15 at 20:54
  • Internal definitions means that you are working within the model, and it means that you're unable to "recognize" that the model is non-standard. If you want to work externally then you're not working in set theory, you're working in some meta-theory. – Asaf Karagila Jun 01 '15 at 20:56
  • IST actually introduces a new predicate "standard", which allows us to discriminate between unlimited, nonstandard naturals, and limited, standard ones. All nonstandard $N\in\Bbb N$ are greater than every standard $n\in\Bbb N$. I'm just saying that, intuitively, one would want to say that a number is finite iff it's standard, but this is not the case. (IST is weird. $0$ is standard, and $n$ being standard implies $n+1$ is standard… but there exist nonstandard numbers. It turns out that the induction theorem doesn't actually apply to this new predicate "standard".) – Akiva Weinberger Jun 01 '15 at 21:05
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    If I'm not mistaken, IST is more of a non-standard analysis sort of thing; less of a set theory thingamabob, and adding a "standard" predicate will necessarily violate the axiom of regularity, as well the fact that "the finite ordinals" make the smallest inductive set. So I'm prettttty sure that it's not really a set theoretic whatchamccallit. And if you want to model your notion of finiteness using analysis, that's also fine, but analysis has proved itself to be uninteresting when it comes to modeling anything other than analysis (since RCF is complete and recursively enumerable). – Asaf Karagila Jun 01 '15 at 21:09
  • Axiom of regularity is still there, as well as all axioms of ZFC. And the finite ordinals are still the smallest inductive set. :) It is a set theory, and it sounds like you're not as familiar with it as you think you are. (If you're confused as to how we don't get a contradiction from my second sentence, why not continue in chat?) – Akiva Weinberger Jun 01 '15 at 21:12
  • (+1) to compensate for the downvote, because this is a fair answer. If and when I wish to comment in response (to this or other answers), I'll expand the OP. – Calum Gilhooley Jun 01 '15 at 21:16
  • @columbus8myhw: I know very little about IST, that is true. I find the entire thing a bit baffling, and I'm not sure what is gained by the new predicate and the new axioms that is not doable otherwise. Except, perhaps, a formalization of [ultra]finitism as everything which happens on the standard integers, and therefore on a bounded set of integers. – Asaf Karagila Jun 01 '15 at 21:41
  • A brief, non-mathematical, pedagogical comment on the last paragraph of your answer: I've tried putting philosophical worries aside, and returning, as a mature student, to the formal, assessed study of mathematics, on the principle, "Carry on, and faith will come", but although I did carry on, and got good results in coursework and exams, my doubts remained untouched, as if in a kind of doublethink. Similarly, I participated energetically here in MathSE, a few weeks ago, and could quite happily ignore my doubts while asking or answering questions, but they were always there to ambush me again. – Calum Gilhooley Jun 01 '15 at 22:22
  • @Calum: Next we will be discussing why are the natural numbers are natural, and why are the real numbers real. Those are terms meant to encompass a mathematical definition. If you have troubles connecting it with the philosophy, don't. It's doublethink to think that it should have anything to do with the natural meaning of the word, if anything, and then it stops being a mathematical discussion and becomes a linguistic-philosophical-semantical-nonsense discussion. You either agree that in mathematics "finite" means a particular definition, or you don't. There is no try. – Asaf Karagila Jun 01 '15 at 22:25
  • @AsafKaragila I promise not to discuss why the natural numbers are natural, and the real numbers are real. – Calum Gilhooley Jun 01 '15 at 22:28
  • @Calum: I was being facetious, of course. But this is the same principle. You're confusing something mathematical which is called "finite" with your natural understanding of the term "finite". And this dissonance causes friction, but it always has and always will be a cause for friction, because it is in fact in 99.99% of the cases dissonance in the two meanings of the word. But mathematics is not ambiguous. There are definitions, what satisfies them gets to be called "X", and that which does not is called "not X". – Asaf Karagila Jun 01 '15 at 22:33
  • The standard natural numbers do not form a bounded set… because they don't form a set. We cannot form the set ${x\in\mathbb N:x\text{ standard}}$. The reason is that "$x\text{ standard}$" is not a formula of ZFC; the ZFC axiom that lets us form sets like that doesn't mention the word "standard". (This explains how we have a bounded collection of natural numbers with no least upper bound — ZFC only guarantees this for sets. Similarly, they don't form an inductive set, so the induction theorem doesn't apply to them.) – Akiva Weinberger Jun 02 '15 at 00:26
  • Every theorem of IST (that doesn't use the word "standard" is a theorem of ZFC. IST's main benefit is offering shorter/easier-to-find proofs of ZFC theorems — especially in analysis, but also topology, and at least one theorem in graph theory. (This really should be discussed elsewhere… sorry.) – Akiva Weinberger Jun 02 '15 at 00:28
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    'I don't know what you mean by "glory",' Alice said. Humpty Dumpty smiled contemptuously. 'Of course you don't — till I tell you. I meant "there's a nice knock-down argument for you!"' 'But "glory" doesn't mean "a nice knock-down argument",' Alice objected. 'When I use a word,' Humpty Dumpty said, in rather a scornful tone, 'it means just what I choose it to mean — neither more nor less.' 'The question is,' said Alice, 'whether you can make words mean so many different things.' 'The question is,' said Humpty Dumpty, 'which is to be master — that's all.' – Steven Alexis Gregory Jun 02 '15 at 01:12
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I don't see any interesting philosophical inquiry here. Your objections are purely argumentative.

If you are a finitist and believe that nothing infinite exists, then you have one very simple definition for finite: everything.

If you are not a finitist and accept the axiom of infinity, then you can use any suitable definition. With choice, (I think) they are all equivalent, and without choice you just need to be more specific as to what kind of finite you are talking about.

Jonny
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  • Are you implying that [ultra]finitism is neither true nor false, but entirely without content, because it merely asserts that everything is something? Or, if you believe that ultrafinitism has content (but perhaps believe that it is false - as I do, personally, although I have no clear idea how to argue against it), then what is that content? – Calum Gilhooley Jun 01 '15 at 19:24
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    No, I simply demonstrate that 'finite' is definable regardless of whether you accept the axiom of infinity, or assert its negation. – Jonny Jun 01 '15 at 22:38
  • I don't follow, but I also don't wish to be "purely argumentative"! (It's a real sin, I know - in mathematics, as elsewhere. Oddly enough, at the same time as you were posting your answer, I was chiding my daughter for being too argumentative - and she probably gets some of it from me.) So I'll not pursue this. – Calum Gilhooley Jun 01 '15 at 23:00
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    Haha, I thats not exactly what I mean --see here: http://en.wikipedia.org/wiki/Argumentative – Jonny Jun 02 '15 at 01:30
  • At least you didn't say "Irrelevant, incompetent and immaterial!" Then I'd break down, sobbing, and confess everything! Oddly, a "witness" was what I had in mind: in the sense that a witness to a set's finiteness according to the Dedekind definition would have to be a proof (of non-existence of an injective map to a proper subset), whereas in everyday non-mathematical practice, a witness to a set's finiteness is likely to be some kind of trace of a counting procedure (but not necessarily one which matches elements of the set to numbers or numerals). Past my bedtime, hope I'm making some sense! – Calum Gilhooley Jun 02 '15 at 02:03
  • Well, regardless of the definition, a proof is generally required to demonstrate any object has any property. – Jonny Jun 02 '15 at 08:27
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Joel David Hamkins has argued that we do not have an absolute notion of the finite and has discussed this at length both in blogs and in publications (see e.g., the collection "a question for the oracle"). Namely the symbol $\mathbb N$ that we commonly use to denote an apparently clearly defined notion turns out to depend on the intended interpretation hypothesis. The stormy reception that question has received (14 upvotes, 12 downvotes) shows that the issue touches on some deep-seated beliefs among mathematicians.

Mikhail Katz
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Your question strikes me as odd, because it asks "What do people mean by finite?" while simultaneously discussing two perfectly good definitions of the term.

A set $S$ is finite if there exists a bijection from $S$ to a section of the natural numbers. Equivalently, $S$ is finite if every injection from $S$ to itself is a bijection. These definitions are logically equivalent, so it makes no difference which one you use as a definition and which one you prove as a theorem.

You raise the objection that the natural numbers definition requires a common understanding of what the natural numbers are. That's a reasonable point, which is why mathematicians typically either define the natural numbers axiomatically or construct them as part of a larger axiomatic system. Historically, the natural numbers were defined axiomatically using the Peano axioms, but in the modern foundations of mathematics they can be constructed explicitly using the ZFC axioms for set theory. What this means is that the only common notions that are required to discuss mathematics precisely are the rules for symbolic logic.

Now, one can raise the objection that the axioms of ZFC may be inconsistent, in which case our entire discussion of mathematics is, from a formal point of view, entirely fruitless. Presumably this is what radical finitists like Zeilberger believe. Though we have no way to prove that the ZFC axioms are consistent, they have been working well so far, so the onus is on the finitists to demonstrate an inconsistency.

Jim Belk
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  • It seems to me that this answer doesn't really answer what the OP is asking. This is an answer to "how do mathematicians define finite sets?", but the OP asked "what do people mean by 'finite'?". In other words, perhaps the OP is asking: "what notion of 'finite' do we all share that leads us to making that mathematical definition?" – Greg Martin Jun 01 '15 at 17:31
  • The definitions in your second paragraph are not equivalent: without the axiom of choice there can be sets that are infinite by the first definition but finite by the second. They are the so-called infinite, Dedekind-finite ses. – Brian M. Scott Jun 01 '15 at 17:32
  • @GregMartin I think my answer is precisely an answer to the OP's question. The reason that we use definitions in mathematics is to make sure that everyone agrees precisely on the meaning of the defined terms before continuing the discussion. The "notion of finiteness" that we all share is just the definition of finite. – Jim Belk Jun 01 '15 at 17:40
  • @BrianMScott That's true, but they are equivalent in ZFC, which is the context in which virtually all of modern mathematics takes place. But as you point out, without the axiom of choice there are multiple inequivalent notions of finite. – Jim Belk Jun 01 '15 at 17:42
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    I'm quite a fan of mathematical definitions, for precisely the reason you say. Yet we don't make definitions arbitrarily; rather, we make them to reflect some phenomenon that exists in our minds before the definition is created. That phenomenon is what I believe the OP is asking about. – Greg Martin Jun 01 '15 at 19:49
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Intuitively we may say that a set is finite if the process of listing its elements terminates. Formally a set is finite if it is in a bijection with 1,2,...,n for some natural number n.

  • Though this is a perfectly reasonable way to define "finite", the OP elaborated on this idea in paragraph 6. I think the idea is to bend toward a more philosophical discussion. – Ken Jun 01 '15 at 17:55
  • In spite of the downvote, this is not a bad answer to my question, in that it spells out, however informally (any answer to my question is necessarily going to be informal, so that's not meant as a criticism), a possible meaning of the word "finite", which might be the shared understanding I am asking about. Of course, it also might not be, and one cannot know for sure what people mean by a word until they say what they think they mean by it. In any case it is not unreasonable to suggest thay what people informally have at the back of their minds is some notion about terminating processes. – Calum Gilhooley Jun 01 '15 at 18:00
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Possibly, Kazimierz Kuratowski's definition is what you're looking for -- it's very straightforward.

Straightforward(!) but still a mouthful (more than other definitions), here is the Kuratowski's contribution:

Let $X$ be an arbitrary set. A family $\ L\subseteq 2^X\ $ is called a supfin family $\ \Leftarrow:\Rightarrow\ $ the following three conditions hold:

$$ (i)\quad\quad\quad \emptyset\in L $$ $$ (ii)\quad \forall_{x\in X}\quad \{x\}\in L $$ $$ (iii)\quad \forall_{A\ B\,\in\,L} \quad A\cup B\in L: $$

Let $\ \Lambda(X)\ $ be the set of all supfin families $\ L\ $ in $\ X.\ $ Then we can let

$$ \mbox{Fin}(X)\ :=\ \bigcap\Lambda $$

be -- by definition -- the set of all $X$-finite sets.

DEFINITION (K.Kuratowski)$\ \ $ Set $\ X\ $ is finite $\ \Leftarrow:\Rightarrow\ X\ $ is an $\ X$-finite set.

This definition simply says that finite sets are obtained from nothing by adding single elements just a (hm!) finite number of times -- this definition is exactly what it is meant to be.

Wlod AA
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