Decreasing sequence of non-negative integers Let $ a_n , \ n=0, 1, 2, \dots\ $ be a strictly monotonically decreasing sequence consisting of non-negative integers. The sequence also has a lower bound C such that $ a_n \ge C &gt;-\infty.$
Intuitively, we know, that $a_n$ has to have finite amount of entries, but can this be proven exactly?
Well, this seems kind of obvious, but let's do it anyway. We know three things:
1: The sequence is strictly decreasing, i.e, $a_{n+1}&lt;a_n.$
2: $a_n \in \mathbb{Z} ~~\forall n.$
3: $a_n\geq 0 ~\forall n$.
Condition (1) implies that all of the terms of $a_n$ are distinct, i.e, $\nexists m,n\in\mathbb{N}:a_m=a_n$. Now using condition (2), let $a_0=N$, some positive integer. We know, using condition (1), that all subsequent terms are $&lt;N$, and using condition (3), are all $\geq0.$  Now suppose $a_n$ had infinitely many terms. That would imply that there are infinitely many integers in the range $[0,N]$, which is clearly false. Thus $a_n$ can only have finitely many terms.
Can we ever have this kind of sequence ? As it is strictly monotonically decreasing so we have $a_1 \geq a_n \geq 0$ for all $  n\geq 2 $  but there are only finitely many non negative integers satisfying this .so the sequence is eventually constant which contradicts the fact it is  strictly monotonically decreasing.