Must there be INFINITE negative elements on a succession which is NOT bounded below?

Question

We have that $\{a_n\}_{n\in N}$ is a succession of real numbers that is NOT bounded below.

(1) Must all the elements of $a_n$ be negative?

(2) If there are positive elements, those must be finite?

(3) Must there be INFINITE negative elements?

To answer 1 and 2 I find a succession that is not bounded below, that has infinite positive elements:

$$\{a_n\}_{n\in N}=\{-1,2,-3,4,-5,6,...\}$$ $$a_n=n(-1)^n$$

Which is not bounded (below), and has infinite positive and negative elements.

Now, the third question I'm not sure how to prove that there MUST be infinite negative elements on a succession which is not bounded below.

What I thought of until now:

$\{a_n\}_{n\in N}$ Is not bounded below, then, by definition:

$\nexists m\in R , \forall n\in N : a_n \geq m$

This means that there will always be an item on the succession that is smaller than a previous item:

$\forall n\in N , \exists k\in N : a_{n+k}\leq a_n$

And as $N$ is not bounded, the items smaller than the previous one are infinite?

Answer 1

$(a_n)_{n \in \mathbb{N}}$, real, is not bounded below.

Assume that there are only finitely many negative elements of $(a_n)_{ n\in \mathbb{N}}$:

$a_{n_1}, a_{n_2},.......a_{n_k}$, i.e. $k$ elements.

The smallest of these elements is a lower bound for $(a_n)_{ n \in \mathbb{N}}$.

A contradiction.

Answer 2

You proved correctly the first two assertions.

Asserting that the sequence is not bounded below means that$$(\forall M\in\mathbb R)(\exists n\in\mathbb{N}):a_n<M.$$So, take $M=0$. There is some $n_1\in\mathbb N$ such that $a_{n_1}<0$. But $a_{n_1}\in\mathbb R$. So, there is some $n_2\in\mathbb N$ such that $a_{n_2}<a_{n_1}<0$. And so on. So, each $a_{n_k}$ is negative.