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

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:

1. Is part 1 correct?
2. 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.

• The statement is incorrect. $x_n = (-1)^n\cdot n$ is an unbounded sequence, yet neither of the two properties holds.
– 5xum
Dec 12, 2018 at 12:45
• @5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence" Dec 12, 2018 at 12:54
• No, it's not translation issues. No matter which country you are in, the definition $$\lim_{n\to\infty}x_n = \infty \stackrel{\text{def}}{\iff} \forall \epsilon > 0 \exists N \in \Bbb N :\forall n > N \implies |x_n| > \epsilon$$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,\dots$ converges to $\infty$. The absolute values should not be in the definition.
– 5xum
Dec 12, 2018 at 12:56
• @5xum you are right, i've messed things up Dec 12, 2018 at 13:05
• HINT. Consider that $\lim_{n\to \infty}|x_n|=\infty$ means that for every $r>0,$ the set $\{n: |x_n|\leq r\}$ is finite . So what happens if $\neg (\lim_{n\to \infty}|x_n|=\infty)?$ Dec 12, 2018 at 14:04

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:

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 which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.

• Thank you for pointing that out, i've reworked the OP Dec 12, 2018 at 13:33
• @roman But the statement you are "proving" is still false!
– 5xum
Dec 12, 2018 at 13:33
• I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem Dec 12, 2018 at 13:45
• @roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $\lim_{n\to\infty} |x_n| =\infty$, so you might be misunderstanding some part of the problem.
– 5xum
Dec 12, 2018 at 13:53