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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.

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    $\begingroup$ The statement is incorrect. $x_n = (-1)^n\cdot n$ is an unbounded sequence, yet neither of the two properties holds. $\endgroup$
    – 5xum
    Dec 12, 2018 at 12:45
  • $\begingroup$ @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" $\endgroup$
    – roman
    Dec 12, 2018 at 12:54
  • $\begingroup$ 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. $\endgroup$
    – 5xum
    Dec 12, 2018 at 12:56
  • $\begingroup$ @5xum you are right, i've messed things up $\endgroup$
    – roman
    Dec 12, 2018 at 13:05
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    $\begingroup$ 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 $\neg (\lim_{n\to \infty}|x_n|=\infty)?$ $\endgroup$ Dec 12, 2018 at 14:04

1 Answer 1

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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:

$$\exists C\forall N\exists n>N: n>N\land x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.

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  • $\begingroup$ Thank you for pointing that out, i've reworked the OP $\endgroup$
    – roman
    Dec 12, 2018 at 13:33
  • $\begingroup$ @roman But the statement you are "proving" is still false! $\endgroup$
    – 5xum
    Dec 12, 2018 at 13:33
  • $\begingroup$ 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 $\endgroup$
    – roman
    Dec 12, 2018 at 13:45
  • $\begingroup$ @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. $\endgroup$
    – 5xum
    Dec 12, 2018 at 13:53

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