Given $\{x_n\}$ is an unbounded sequence, prove:
$$
\lim_{n\to\infty}x_n = \infty\ \ \text{or}\ \ \lim_{n_k \to \infty} = A
$$
In other words any unbounded sequence either infinitely large or has a convergent subsequence.
Everything before udpate section is wrong.
Part 1. Proving $|x_n| > M \implies \lim_{n\to\infty} x_n = \infty$
$\Box$ Start with the definition of an unbounded sequence:
$$
\forall M > 0 \in \Bbb R \ \exists N \in \Bbb N: \forall n > N\implies |x_n| > M
$$
Suppose that:
$$
\lim_{n\to\infty}x_n \ne \infty
$$
That means:
$$
\forall n\in\Bbb N\ \exists C >0 \in\Bbb R : |x_n| < C
$$
Now taking $M^\prime = \max\{M, C\}$ we have that:
$$
\begin{cases}
\begin{align}
\exists N \in \Bbb N : &\forall n>N \implies &|x_n| > M^\prime \\
&\forall n \in \Bbb N: &|x_n| < M^\prime
\end{align}
\end{cases}
$$
So we have arrived to a contradiction which means the assumption is wrong and:
$$
\lim_{n\to\infty}x_n = \infty
$$
$\Box$
Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.
Two questions is my mind:
- Is part 1 correct?
- How do I proceed with the second part?
I'm using the following definition.
$$
\lim_{n\to\infty}x_n = +\infty \stackrel{\text{def}}{\iff} \forall \epsilon > 0 \exists N \in \Bbb N :\forall n > N \implies x_n > \epsilon \\
\lim_{n\to\infty}x_n = -\infty \stackrel{\text{def}}{\iff} \forall \epsilon > 0 \exists N \in \Bbb N :\forall n > N \implies x_n < -\epsilon \\
$$
A sequence is considered infinitely large when:
$$
\lim_{n\to\infty}x_n = \infty\ \ \text{when}\ \ \lim_{n\to\infty}x_n = + \infty \ \ \text{or} \ \ \lim_{n\to\infty}x_n = – \infty
$$
Update.
Looks like i've messed up a lot of things. I will try once again.
As shown in comments and answers the statement holds only in case:
$$
\lim_{n\to\infty} |x_n| = \infty \iff \forall \epsilon > 0 \exists N \in \Bbb N: \forall n > N \implies |x_n| > \epsilon
$$
Consider the following:
$$
\lnot P = \lnot\left(\lim_{n\to\infty}|x_n| = \infty \right) \iff \exists \epsilon > 0\ \forall N_1 \in \Bbb N : \exists n > N_1 \land |x_n| < \epsilon
$$
But on the other hand it is given that $x_n$ is unbounded:
$$
Q = \forall M > 0\ \exists N_2 \in \Bbb N : |x_{N_2}| > M
$$
Construct a negative expression for boundedness:
$$
\lnot P = \exists M > 0\ \forall N_2 \in \Bbb N : |x_{N_2}| < M
$$
If $S = \lnot P \implies \lnot Q$ then $S = P \lor \lnot Q$
Let $\epsilon = M$, choose $N = \max\{N_1, N_2\}$ then both statements are true and:
$$
\exists \epsilon > 0\ \forall N = \max\{N_1, N_2\}: \exists n > N \land |x_n| < \epsilon
$$
Now either $P$ is true, which would mean $\lim |x_n| = \infty$, or $\lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence.
Best Answer
The statement is incorrect. $x_n = (-1)^n\cdot n$ is an unbounded sequence, yet neither of the two properties holds.
Also, your proof is wrong here:
This is false. EVERY sequence satisfies the condition $$\forall n\in\Bbb N\ \exists C >0 \in\Bbb R : |x_n| < C$$ (since you can always set $C=|x_n|+1$) however not every sequence satisfies the condition $$\lim_{n\to\infty}x_n \ne \infty$$
In fact, the condition $$\lim_{n\to\infty}x_n = \infty$$
is written as:
$$\forall C \exists N\forall n:n>N\implies x_n>C$$
which means that the negation of that is written as:
$$\exists C\forall N\exists n>N: n>N\land x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.