Suppose that $(k_n\mid n\in\mathbb N)$ is a decreasing sequence in $\mathbb N$. Then $(k_n\mid n\in\mathbb N)$ is eventually constant.

Question

Please check if I miss anything!

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Suppose that $(k_n\mid n\in\mathbb N)$ is a decreasing sequence in $\mathbb N$ (i.e. if $n\leq m$, then $k_n\geq k_m$). Then $(k_n\mid n\in\mathbb N)$ is eventually constant (i.e. there exists $N\in\mathbb N$ such that if $m\geq N$ then $k_m=k_N$).

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Proof:

Assume the contrary: $(k_n\mid n\in\mathbb N)$ is not eventually constant. Thus, for all $n\in\mathbb N$, there exists $m\in \mathbb N$ such that $n<m$ and $k_n>k_m$. Let $C(n)=\min\{m\in\mathbb N\mid n<m$ and $k_n>k_m\}$. Define $C^{n}$ as $\underbrace{C \circ \dots \circ C}_{n\:\text{times}}$.

We generate sequence $(h_n\mid n\in\mathbb N)$ as follows: $h_0=k_0$ and $h_n=k_{C^{n}(0)}$. It is easy to prove that $(h_n\mid n\in\mathbb N)$ is strictly decreasing. Thus there is an infinite number of elements from sequence $(h_n\mid n\in\mathbb N)$ that are less than $h_0$. This contradicts the fact that the set $\{m\in\mathbb N\mid m<h_0\}$ is finite.

Thus $(k_n\mid n\in\mathbb N)$ is eventually constant. $\tag*{$\blacksquare$}$

Answer 1

Here is a simpler proof.

Let $A = \{ k_n : n\in\mathbb N\} \subseteq \mathbb N$. Then $A$ is nonempty and so has a minimum element, $k_N$. Then, for $m \ge N$, we have $k_N \ge k_m$ because the sequence is decreasing and $k_m \ge k_N$ because $k_N$ is the minimum value. Therefore, $k_m = k_N$ for all $m \ge N$.

Answer 2

Assume that, for any $n \in \mathbb{N} $, there is a $m(n) \in \mathbb{N}$ such that $m(n) > n$ and $k_{m(n)} < k_n$.

For a given $n$, define $n_0 = n$ and $n_{j+1} = m(n(j))$. Then $n(j+1) > n(j)$ and $k_{n(j+1)} < k_{n(j)}$. This last implies that $k_{n(j+1)} \le k_{n(j)}-1$.

By induction, this implies that $k_{n(j+i)} \le k_{n(j)}-i$. Now set $j=0$ and $i=k_n+1$.

Then $k_{n(k_n+1)} \le k_{n}-k_n-1 \lt 0$, which can not hold for members of $\mathbb{N}$.

Therefore the initial assumption is false, so, eventually, $k_n$ is constant.