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This might be a long-winded way of proving something so obvious, but I want to have it checked if it holds up.

Claim: There does not exist a sequence $\{{a}_{n \in \mathbb{N}}\}$ of natural numbers such that $a_{n+1} < a_{n}$ for every $n$.

Proof: This amounts to showing that the set $M = \{ m \in \mathbb{N} : m \leq a_{0}\}$, where $a_{0}$ is the first member from any sequence in infinite descent, is finite. Suppose for the sake of contradiction that there exists an $a_{0}$ such that the set $M$ is infinite. We suppose further that $a_{0}$ is minimal. Clearly $a_{0} \neq 0$, for if it were, the set $M = \{ m \in \mathbb{N} : m \leq 0\}$ = $\{0\}$ is finite.

Consider the set $N = M - \{n \in \mathbb{N}: a_{1} < n \leq a_{0}\}$, which is just the set $M$ with the finitely many naturals deleted. From elementrary set theory, $N$ must be infinite.

But $N = \{ n \in \mathbb{N} : n \leq a_{1}\}$, and we can suppose that $a_{1}$ is the first term in a natural number sequence in infinite descent. Thus we have found another infinite set with the desired property, but with $a_{1} < a_{0}$. This contradicts the minimality of $a_{0}$.

Maxis Jaisi
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  • I'm still insecure about my proof, so I'm going to wait until another answer arrives which might point out potential weaknesses in my reasoning, subtle invalid assumptions, improvements, etc. The other two answers provide alternative proofs which might be handy to those who don't like my verbose proof. – Maxis Jaisi Mar 04 '17 at 10:13
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    I don't understand your proof at all, notably what you already assume known about the natural numbers. Do you know that they are well-ordered, i.e., that every nonempty subset has a minimal element? If not, then you cannot "suppose (further) that $a_0$ is minimal". If you do, there is almost nothing to prove, since for an infinite decreasing sequence ${, a_i\mid i\in\Bbb N,}$ obviously has no minimal element, so contradicts what you know. – Marc van Leeuwen Mar 04 '17 at 13:33
  • @MarcvanLeeuwen , you're right. I hereby "retract my claim". :) I assumed that the naturals are well ordered. The proof presented in Tao's Analysis I textbook uses mathematical induction (he uses induction to arrive at the contradiction $a_{n} \geq k$ for every $n,k$). But it's unclear why one would arrive at the idea. I then thought of using well ordering, but assumed what I wanted to prove (finiteness of the set M implicitly assumes no infinitely descending sequence of naturals). – Maxis Jaisi Mar 04 '17 at 13:41
  • @MarcvanLeeuwen (cont) However, I do wonder, for an arbitrary infinite descending sequence, is it safe to form the set consisting of all terms in the sequence? – Maxis Jaisi Mar 04 '17 at 13:41
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    @MaxisJaisi: In the current standard mathematical foundations (ZFC set theory) you can do so for any sequence. In particular, a sequence from $S$ indexed by $\mathbb{N}$ is simply a function $f$ from $\mathbb{N}$ to $S$, and you can construct ${ x : x \in S \land \exists n \in \mathbb{N}\ ( f(n) = x ) }$ by the axiom of specification. – user21820 Mar 04 '17 at 14:37

6 Answers6

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I would have done it by induction on sequences whose first term $a_0\le n$ instead and the property that they are stationnary from a certain point.

$n=0$ only one sequence, the null sequence, it is stationnary.

For $a_0\le n+1$ then if $a_0\le n$ then we apply induction hypothesis, else since $a_1<a_0$ then $a_1\le n$ and we apply induction hypothesis on the truncated sequence.

Historical note :

Yet we can wonder if the induction principle itself doesn't use the infinite descent to be proved ? In fact Frege demonstrated that it is a consequence of second order logic given the definition of the zero, of a number and the successor operation as defined by Peano.

I found this in this article $§6.4$ (though in french, sorry) : le raisonnement par recurrence, quel fondement ?.

So in the above proof the whole thing holds effectively on these things :

  • the definiton of $0$ is used when we say that if $a_0\le0$ then the whole sequence is the null-sequence. (i.e. there is no natural <0).

  • the well ordering of $\mathbb N$ is used when we say that $(a_0\le n+1)\land (a_1<a_0)\Rightarrow a_1\le n+1-1=n$, making full use of the successor definition.

In your proof if we want to be scrupulous, when you say finitely many natural deleted, then it is a bit annoying, because you are biting your tail : you want to proove there are no infinite descending sequence of naturals but you invoque an argument of finitness...

Addendum : (from comments)

In all these proofs which are very close to axiomatic material, we should in theory check everything we write to avoid the biting one's tail issue (i.e. using a result proved by mean of the theorem we want to prove).

Yet here, we can set the starting point as Peano axioms and the induction principle and prove that infinite descent as well as well-ordering principle (i.e. every set of naturals has a minimum) are sound. And in fact these can eventually become new forms of induction on their own.

This are indeed principles used in the other contributors' proofs.

zwim
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    The non existence of infinite decreasing sequences is not a first order consequence, not even of the full first order theory of the natural numbers. There are models of arithmetic where there are infinite decreasing sequences. – Jonathan Schilhan Mar 04 '17 at 09:11
  • yep, I made a mistake while reading the article, it is a second order consequence. I add a link to the article. I edit my answer consequently, thanks for the precision. – zwim Mar 04 '17 at 09:19
  • Could you expand on your proof using recurrences? It still seems cryptic to me, perhaps you could make clear what you mean by recurrences. Does induction translate to recurrence in French? I'm also not exactly sure what you mean by "truncated sequence". Anyhow, I appreciate the link you shared; I'll try to learn some French in order to read it! – Maxis Jaisi Mar 04 '17 at 14:37
  • @MaxisJaisi: Technically zwim's answer will not be applicable in the context of your question because it is basically using induction, which presumably you are not permitted to use in proving the desired claim. However, it can be viewed as the inductive version of the "minimal counter-example" principle. – user21820 Mar 04 '17 at 14:40
  • Ah ah french wording sneaking in my answer, yes, that was induction, now corrected. I added a little addendum to answer concerns. – zwim Mar 04 '17 at 17:04
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This amounts to showing that the set $M = \{ m \in \mathbb{N} : m \leq a_{0}\}$, where $a_{0}$ is the first member from any sequence in infinite descent, is finite. $ \def\nn{\mathbb{N}} $

This first line of your 'proof' is invalid. You are using merely your intuition to claim that the non-existence of an infinite descending chain of natural numbers follows from the finiteness of some set that you specified. This is circular in this case; proving the equivalence amounts to proving something of roughly the same strength as the original desired theorem.

Instead what you actually need is:

Let $S = \{ n : n \in \nn \land \text{there is a strictly decreasing sequence from $\nn$ starting with $n$ } \}$.

If $S$ is non-empty then let $m = \min(S)$ and (use your other ideas) to prove that there is a strictly decreasing sequence from $\nn$ starting with something less than $m$, which contradicts the definition of $m$.

Therefore $S$ is empty and you are done.

user21820
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  • And if your "minimal counter-example" rule involves statements rather than sets, then simply re-write the argument as "If there exists $n$ such that there is a strictly decreasing sequence from $\nn$ starting from $n$, then there exists a minimal such $n$, and then ... contradiction.". – user21820 Mar 04 '17 at 11:40
  • Your answer was very helpful, because I realised that the assertion $M = {m \in \mathbb{N} : m \leq a_{0}}$ is finite implicitly assumes that there does not exist an infinite descending sequence of naturals, which was precisely what I wanted to prove. Thank you! – Maxis Jaisi Mar 04 '17 at 14:30
  • @MaxisJaisi: You're welcome! Feel free to ask further if you need more clarification. – user21820 Mar 04 '17 at 14:38
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    Given the assumption that every non-empty subset of the naturals has a least element (which the OP says they assume as given, and which your proof needs anyway for $m = \min(S)$ to be well defined), wouldn't it be easier to just note that the set of elements of any sequence of naturals, being a subset of $\mathbb N$, must have a least element, which (being a member of the set) must be an element of the sequence, which implies that this element must either be last in the sequence (making the sequence finite) or must be followed by an equal or greater element (making it not strictly descending)? – Ilmari Karonen Mar 04 '17 at 16:22
  • @IlmariKaronen: Of course that also works, and is related to the asker's comment under the question, which I answered. My answer was just the natural way to use the ideas already present in the asker's attempt. Your way uses 'less' set-specification strength, since you don't quantify over sequences from $\nn$ in the defining formula, but it's too fine a distinction to make at this level. =) – user21820 Mar 04 '17 at 17:21
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Your proof looks good to me. As @user21820 and others point out, it actually isn't, and the other answers do a better job of explaining this. In fact, even the proof below is inaccurate as the $\dotsb$ implies induction but I haven't stated that, because it's more of an 'intuition' proof. Assuming induction and 'common-sense' properties of natural numbers is not a trivial thing for a very axiomatic question like this. (Thanks @user21820!)

Here's another I thought of: since the $a$s are natural numbers, $a_{n + 1} < a_n$ is equivalent to $a_{n + 1} \leq a_n - 1$ because there are no natural numbers between $a_n$ and $a_n - 1$. Then, let $n = a_1$ and you have $$a_{n+1} \leq a_{n} - 1 \leq a_{n-1} - 2 \leq \dotsb \leq a_1 - n = 0$$ but $a_{n+1} \geq 0$ as it is a natural number, so $a_{n+1} = 0$. Beyond that there can be no smaller numbers.

(I use 'natural' to mean 'positive integer' but it works for 'non-negative integer' or even 'set of integers with a lower bound'.)

shardulc
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    Actually the 'proof' is not correct. And your proof is not rigorous either; ellipsis ("$\cdots$") is never rigorous. If you attempt to make it rigorous you will have to use induction, which is (in a certain technical sense) equivalent to "no infinite descending chain" and "every non-empty set has a minimum". So for this particular question it is meaningless to use your method, precisely because in the context of the question we are not free to use induction or anything equivalent, otherwise the question itself becomes meaningless. – user21820 Mar 04 '17 at 11:35
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    -1: for the reasons stated by user21820. – Martin Argerami Mar 04 '17 at 14:24
  • @user21820 I agree, the proof implies induction and it isn't rigorous because I don't state it explicitly, using $\dotsb$ instead. However, why is that wrong? (I think you have already explained this in your other comments but I didn't understand.) – shardulc Mar 04 '17 at 17:26
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    When I said "not correct" I was referring to the asker's attempted 'proof', which you had said "looks good". This is the unambiguous error. When I said "meaningless to use your method" I meant that the question makes most sense in the context where you do not have induction, because it asks for a proof of "no infinite descending chain" from "every non-empty set has a minimum". If you use induction, then the given minimal-counterexample principle is redundant, making the question silly. Anyway you should definitely amend your first sentence because it's simply false. – user21820 Mar 04 '17 at 17:36
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    Also, there's a significant disadvantage to your method; it invokes the arithmetic properties, whereas the inductive structure of the naturals do not have anything to do with arithmetic. Furthermore, depending on the axiomatization, even your claim that $a<b \implies a \le b-1$ may need to be proven by induction. It's okay as a separate viewpoint, but one must be aware that it's not so trivial as it seems. – user21820 Mar 04 '17 at 17:40
  • Thanks for the edit; I've removed my downvote. Anyway I hope you're not the one who downvoted Martin's and my post, unless there's a mistake, in which case please tell us! – user21820 Mar 05 '17 at 04:32
  • @user21820 Thanks! No, I didn't downvote your answers. – shardulc Mar 05 '17 at 05:01
  • You're welcome! I seem to be getting a number of random downvotes lately without comment on what needs to be fixed, so it is a bit irritating haha.. – user21820 Mar 05 '17 at 05:06
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Besides the comments in the answers by zwim and user21820, I would emphasize that the proof uses in an essential way the fact (required to be proven in the most common presentations of the naturals) that every set of naturals has a least element.

Martin Argerami
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(I think this response does not reproduce an existing one, although the ideas are all quite similar.)

Suppose for the sake of contradiction that there exists such a sequence, $\{a_n\}_{n \in \mathbb{N}}$. Consider the set containing all of its elements. Since this is a nonempty set of natural numbers, it contains -- by the Well Ordering Principle -- a least element, $a_k \in \mathbb{N}$. Since the sequence is strictly decreasing, $a_{k+1} < a_k$, which contradicts the minimality of $a_k$.

Thus, our supposition was incorrect, and no such sequence exists.

Benjamin Dickman
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This is the way Tao asks the reader to prove the result in his book Analysis 1: Let us assume for the sake of contradiction that such a sequence $\{a_{n}\}_{n\in\mathbb{N}}$ exists.

$\textbf{Claim: }$ If such a sequence exists then $\forall k\in\mathbb{N}, \forall n\in\mathbb{N}$ we have $a_{n}\geq k$.

$\textbf{Proof: }$ We shall prove the claim by induction on $k$. $a_{n}\geq 0\:\forall n\in\mathbb{N}$ since $a_{n}\in\mathbb{N}$ so the claim is true for $k = 0$. Now suppose $a_{n}\geq k\:\forall n\in\mathbb{N}$ is true, we need to prove that $a_{n}\geq k+1\:\forall n\in\mathbb{N}$. Suppose this statement is false then $\exists m\in\mathbb{N}$ such that $a_{m}<k+1$. Since $a_{m}\in\mathbb{N}\implies a_{m}\leq k$ so $a_{n}\leq k\:\forall n\geq m$ by the property of the sequence which leads to a contradiction. Therefore, the claim is proved.

But we know that the claim cannot be true since $\forall n\in\mathbb{N}\:\exists a_{n}+1\in\mathbb{N}$ s.t. $a_{n}+1 > a_{n}$. So our initial assumption that such a sequence exits must be false.

Karthik Kannan
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