Equivalence of Definitions for "Not-Increasing" Sequences

Question

I'm studying the concept of "not-increasing" sequences and encountered two definitions:

1. Definition 1: $$ \exists n, m \in \mathbb{N}, \text{ such that } n < m \text{ and } a_n \geq a_m $$

2. Definition 2: $$ \exists n \in \mathbb{N}, \text{ such that } a_n \geq a_{n+1} $$

My instructor stated that these definitions are equivalent, meaning one implies the other.

However, I find this claim unclear.

Definition 1 seems more general, as it allows for $$ a_n \geq a_m $$ for some $$ m > n $$, without requiring consecutive terms to satisfy this condition.

In contrast, Definition 2 focuses on a specific pair of consecutive terms. It might be the case that no element in the sequence is larger than or equal to the element right next to it but will eventually be greater than or equal to some other element down the sequence.

I attempted to prove that Definition 2 implies Definition 1 but was unsuccessful.

Could someone provide a clear explanation or proof demonstrating the equivalence of these definitions?

Answer 1

It may be easier to think about the definition of a (strictly) increasing sequence: a sequence is increasing if either:

i) for all $n\in \mathbb N$ we have $a_n< a_{n+1}$,

or

ii) for all $n,m \in \mathbb N$ we have $a_n< a_m$ whenever $n<m$.

The two definitions of "not increasing" are then the negations of $i)$ and $ii)$ respectively.

Now $i)$ follows from $ii)$ by taking $m=n+1$. On the other hand, $ii)$ follows from $i)$ using induction: if $m \geq n$ then we may write $m=n+k$, for some integer $k \geq 0$, hence we must show $a_n \leq a_{n+k}$ for all $k \geq 0$. The case $k=0$ is trivial while the case $k=1$ is given by $i)$. But if we know $a_n < a_{n+k}$ by induction, then since $i)$ applied to $n+k$ shows that $a_{n+k}< a_{n+k+1}$ it follows that $a_n< a_{n+k+1}$ as required.

Answer 2

From comments:

A sequence satisfying Definition 2 implies that that sequence satisfies Definition 1 by letting $m=n+1$. I.e. if Definition 2 is met by some $n$ and if $m=n+1$ then both $n < n+1 \implies n < m$ and $a_n \ge a_{n+1} \implies a_n \ge a_m$.

In the reverse direction, if Definition 1 is satisfied then there are integers $n, m$ with $n < m$ and $a_n \ge a_m$, so consider the sequence $a_n, a_{n+1}, a_{n+2} ,\ldots , a_{m-1}, a_m$. Either $a_n < a_{n+1}< a_{n+2} <\ldots < a_{m-1}< a_m$ implying $a_n \lt a_m$ due to the transitivity of "$<$" but contradicting $a_n \ge a_m$ in the example taken from Definition 1, or there is some integer $k$ where $n \le k < m$ with $a_k \ge a_{k+1}$ so Definition 2 is satisfied and is therefore implied by Definition 1.

Answer 3

WTS for (1) implies (2) that: $$[\exists n,m: (n<m)\wedge (a_n\geq a_m)]\implies[\exists k: a_k \geq a_{k+1}].$$

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Suppose not, then $[\forall k: a_k < a_{k+1}]$.

If $(n<m)$, then the above statement implies: $(a_n < a_{n+1} < ... < a_{m})$.
Applying Transitivity of the relation '$<$' (as Henry hinted), we have $(a_n<a_m)$.

Reiterating, $\forall n,m:[(n<m)\implies (a_n < a_m)]$.

Applying the tautology $p\rightarrow q \Leftrightarrow \neg p\vee q$:

$\Leftrightarrow\forall n,m:\neg (n< m)\vee (a_n<a_m)$

$\Leftrightarrow \forall n,m:\neg (n<m) \vee \neg (a_n \geq a_m)$

$\Leftrightarrow \neg [\exists n,m:(n<m)\wedge (a_n\geq a_m)]$.

This contradicts the hypothesis. $\square$

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