Help with completing the proof: Any unbounded sequence is either infinitely large or has a convergent subsequential limit

Question

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.

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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: $$ \lim_{n\to\infty}x_n = \infty $$ $\Box$

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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:

1. Is part 1 correct?
2. How do I proceed with the second part?

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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: $$ \lim_{n\to\infty}x_n = \infty\ \ \text{when}\ \ \lim_{n\to\infty}x_n = + \infty \ \ \text{or} \ \ \lim_{n\to\infty}x_n = - \infty $$

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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.

Answer 1

The statement is incorrect. $x_n = (-1)^n\cdot n$ is an unbounded sequence, yet neither of the two properties holds.

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Also, your proof is wrong here:

Suppose that: $$ \lim_{n\to\infty}x_n \ne \infty $$ That means: $$\forall n\in\Bbb N\ \exists C >0 \in\Bbb R : |x_n| < C $$

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.