self learning - Proof of the infinite descent principle

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}

Thanks :)

Answer

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.