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Let $n\in\omega$ and $F\subset n$.

How do i prove that $F$ is finite?

I know that this can be proven by constructing a function in a following way: take $f(0)$ as the least element of $F$ and then take $f(1)$ as the least element of $F\setminus \{f(0)\}$ and continue this process. Since $F\subset n$, $F$ has a maximal element. Thus this process must be done in a finite step.

However, i don't know how to prove this precisely. How do i write down "this process must be done in a finite step" in first-order logic and so on. Please help :)

John. p
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    What definition of "finite" in first-order logic are we going to use? – hardmath Mar 30 '14 at 11:43
  • In general, the property of "being finite" is not expressible in f-o logic. Thus, you must assume (some portion of) set theory with a definition of "finite" (e.g. not Dedekind infinite). See also this post – Mauro ALLEGRANZA Mar 30 '14 at 11:55
  • @Mauro You are right. But isn't it obvious that i meant dedekind finite since i take $n\in\omega$? Since this set is also Tarski-finite, i think that is not a matter. – John. p Mar 30 '14 at 11:57
  • John, what exactly do you mean by Tarski finite? (There are two definitions for the term) – Asaf Karagila Mar 30 '14 at 11:59
  • You can use Patrick Suppes, Axiomatic Set Theory (1960 - Dover reprint); page 100, he defines : "$A$ is finite iff every non-empty family of subsets of $A$ has a minimal element", tahis Tarski's definition. Then (page 107), he proves : "If a set is finite, then it is Dedekind finite". – Mauro ALLEGRANZA Mar 30 '14 at 12:05
  • @Asaf I meant the one Mauro wrote above. #Mauro As far as i know, whatever a definition of "finite" is, $n(\in\omega)$ is finite. Isn't it? I'm sorry if a word "first-order logic" bothers you. I just meant set theory. – John. p Mar 30 '14 at 12:09
  • @Mauro That is, i was asking how to prove this using only formulas, not some informal words such ad process etc. – John. p Mar 30 '14 at 12:13
  • You want to write a proof completely in the language of set theory? It's not impossible. It's just so very very long. The "informal words" allow us to utilize theorems like the induction theorem, which are in fact meta-theorems in the sense that they provide us with a schema and a recipe for cooking a very long formula from a given formula. Reiterate all that, unfold all the definitions. I guess it will take you 300 pages or so? (In particular if you include all said proofs). – Asaf Karagila Mar 30 '14 at 12:15
  • The point is, John, that proofs, while informal are still precise if you're clear about everything. So first you need to provide a definition of "finite", then you should show that if $A\subseteq B$ and $B$ is "finite" then $A$ is "finite" as well. This can be done by induction, or it can be done otherwise -- depending on your toolbox of theorems, and the chosen definition. – Asaf Karagila Mar 30 '14 at 12:17
  • @Asaf I didn't mean that much precise. Your proof for the pigeon hole principle (in the link) is completely fine to me. What i meant is this: it's OK to write a proof somewhat informally, but one should know how that informal sentence could be written in formulas. That is it. Since i didn't know how to write this process must be done in a finite step in formal language, i asked this one. – John. p Mar 30 '14 at 12:20
  • Then you should clarify that in your question. This is a particular instance, but the question itself is a much more general than just this particular instance, is it now? – Asaf Karagila Mar 30 '14 at 12:20
  • @Asaf I exactly meant the first sentence in your last comment. I get what you are saying. Thank you! :) – John. p Mar 30 '14 at 12:21
  • @Asaf oh yes.. it is my mistake. – John. p Mar 30 '14 at 12:22
  • In Suppes' book, Th.26, page 100, we have : "If $A$ is finite and $B \subseteq A$, then $B$ is finite." This may be enough, assuming that we can prove that $n$ is finite. For this, we may use Th.24 (page 100) : "$\emptyset$ is finite", Th.25 : "${ x }$ is finite", and Th.29 : "If $A$ is finite, then $A \cup { x }$ is finite". – Mauro ALLEGRANZA Mar 30 '14 at 12:29
  • @Mauro: Last semester this is exactly the proof that we gave the students in class. – Asaf Karagila Mar 30 '14 at 12:33

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When we write something in the flavor of "the process must be done in a finite number of steps", what we often mean to say is that we are hiding some inductive construction, and that induction cannot proceed through all the natural numbers.

That is, we define some construction $\varphi(n)$ by recursion. We have $\varphi(0)$ defined somehow, and then given $\varphi(k)$ we have some pool of available resources, which "shrinks" from each steps, and if it hasn't dried out we do something nontrivial to define $\varphi(k+1)$.

So when we say that the process can only go through a finite number of steps, we mean to say that the pool is necessarily shrinking from time to time, and that the pool is finite (or necessarily becomes finite at some step), so that shrinkage is bound to stabilize and the recursive definition becomes trivialized (i.e. $\varphi(k+1)=\varphi(k)$).

In the case of proving that a subset of a finite set is finite, we define some bijection between a subset and a finite ordinal; we begin by ordering the finite set, then our construction is a function from the subset to some other finite ordinal, and the recursion does exactly that. It picks "the next element", but since there are finitely many elements in the larger set, the recursion must trivialize at some finite point.

As a whole, this sort of sentence is tricky, and you have to gain a lot of experience before you can use it without [too many] mistakes. Until such time, my suggestion is to try and understand what is it that you define, and how the induction and recursion methods are applied in this sort of claim. Then you can write them explicitly.

After some time, your brain will do that automatically and you can resort to "this is a finite process, so it must terminate." again.

Asaf Karagila
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