Hi everyone I wonder to myself if the next proof is correct. I would appreciate any suggestion.

Proposition: There is not a sequence of natural numbers which is infinite descent.

Proof: Suppose for contradiction that there exists a sequence of natural numbers which is infinite descent. Let $(a_n)$ be such sequence, i.e., $a_n>a_{n+1}$ for all natural numbers n.

We claim that if the sequence exists, then $a_n\ge k$ for all $k, n \in N$.

We induct on $k$. Clearly the base case holds, since each $a_n$ is a natural number and then $a_n \ge 0$ for all $n$. Now suppose inductively that the claim holds for $k\ge 0$, i.e., $a_n\ge k$ for all $n \in N$; we wish to show that also holds for $k+1$ and thus close the induction. Furthermore, we get a contradiction since $a_n \ge k$ for all $k, n \in N$, implies that the natural numbers are bounded.

$a_n>a_{n+1}$ since $(a_n)$ is an infinite descent. By the inductive hypothesis we know that $a_{n+1}\ge k$, so we have $a_n>k$ and then $a_n\ge k+1$.

To conclude we have to show that the claim holds for every $n$. Suppose there is some $n_0$ such that $a_{n_0}<k+1$ so, $\,a_{n_0}\le k$. Then either $\,a_{n_0} = k$ or $\,a_{n_0} < k$ but both cases contradicts our hypothesis. Thus $a_n\ge k+1$ for all $n$, which close the induction.

Thanks :)


I would argue a different way.

By assumption, for all $n$, $a_n > a_{n+1}$, or $a_n \ge a_{n+1}+1$.

Therefore, since $a_{n+1} \ge a_{n+2}+1$, $a_n \ge a_{n+2}+2$.

Proceeding by induction, for any $k$, $a_n \ge a_{n+k}+k$.

But, set $k = a_n+1$. We get $a_n \ge a_{n+a_n+1}+a_n+1 > a_n$.

This is the desired contradiction.

This can be stated in this form: We can only go down as far as we are up.

Note: This sort of reminds me of some of the fixed point theorems in recursive function theory.

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    $\begingroup$ I think it should be $a_{n+1} \ge a_{n+2}+1$ instead of $a_{n+1} \ge a_{n+2}-1$ doesn't it? $\endgroup$ – Jose Antonio Oct 11 '13 at 20:17
  • $\begingroup$ Aha! I see what you meant. You are correct. Thanks, fixed, upvoted. $\endgroup$ – marty cohen Oct 16 '13 at 1:10
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    $\begingroup$ I think this proof is not correct. You can only say that $a_{n+a_n} \ge 0$, and therefore, $a_n \ge a_{n+a_n}+a_n \ge a_n$, which does not have a contradiction. $\endgroup$ – winston Nov 17 '18 at 14:11
  • $\begingroup$ Then set $k=a_n+1$. I'll edit the answer. Thanks. $\endgroup$ – marty cohen Aug 26 '19 at 16:41

Your argument does seem to work. Note that the last paragraph, however, is unnecessary. In the second last paragraph you have shown that $a_n \ge k + 1$ for every $n$ ($n$ was arbitrary here).

Also, your argument seems to be unneccesarily complicated. Here is how I would argue:

Fix any sequence of natural numbers $\left\{a_n\right\}$. Then by definition of natural numbers there are only finitely many $m \le a_0$, therefore we cannot choose infinitely many distinct numbers below $a_0$ and $\left\{a_n\right\}$ does not have infinite descent.

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    $\begingroup$ Doesn't "... by definition of natural numbers there are only finitely many $m \leq a_{0}$..." require proof? $\endgroup$ – Maxis Jaisi Mar 4 '17 at 7:06

THEOREM : There are only finite no. of positive integers less than any given positive integer.

The above theorem is also known as Principle of Infinite Descent.

PROOF : Let us assume that there are infinite positive integers less than any given positive integer. So there can't be any minimal or maximal element.

Consider, "a"$\in$Z such that a>0.

Let "S" be the set of positive integers less than "a"

Here,S≠∅ ( because of our assumption)

But, By WOP( Well Ordering Principle).. There must be a minimal element.


Hence our theorem is true

Hope it helps 😋


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