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Here is a theorem which can be found in page 92 of Introduction to Real Analysis, Fourth Edition by Robert G. Bartle and Donald R. Sherbert.

If $(x_n)$ is an unbounded increasing sequence, then $\lim(x_n)=+\infty$.

Proof. Suppose that $(x_n)$ is an increasing sequence. We know that if $(x_n)$ is bounded, then it is convergent. If $(x_n)$ is unbounded, then for any $\alpha\in\mathbb R$ there exists $n(\alpha)\in\mathbb N$ such that $\alpha<x_{n(\alpha)}$. But since $(x_n)$ is increasing, we have $\alpha<x_n$ for all $n\geq n(\alpha)$. Since $\alpha$ is arbitrary, it follows that $\lim(x_n)=+\infty$.

Here is my problem:

By definition, $(x_n)$ is bounded if there exists a real number $M$ such that for all natural numbers $n$, we have $|x_n|\leq M$.

If we negate this we get, $(x_n)$ is unbounded if for every real number $M$, there exists a natural number $n_0$ such that $|x_{n_0}|>M$.

However, while taking into consideration the unboundedness of the sequence, the authors consider $\alpha < x_{n(\alpha)}$ not $\alpha<|x_{n(\alpha)}|$.

Why didn't the authors consider the absolute value of $|x_{n(\alpha)}|$ ?

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3 Answers 3

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If an increasing sequence $(x_n)$ has an upper bound $M$ then $x_1 \leq x_n \leq M$ for all $n$ and this implies that $(x_n)$ is bounded. Since our sequence is not bounded it follows that it has no upper bound. Hence, for any real number $\alpha$ there exits $n(\alpha)$ such that $x_{n(\alpha)} >\alpha$.

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  • $\begingroup$ Please tell me if I understand the argument correctly. We use the increasing and bounded supposition to come up with the inequality $x_1\leq x_n\leq M$. Then we leave the increasing assumption (first part of the inequality) as it is and negate only the bounded assumption. This time there is no absolute value in the inequality so even the negation has no absolute value. So, the information that $(x_n)$ is increasing lets us get away with not considering $-M\leq x_n$. $\endgroup$
    – DS2830
    Sep 16, 2019 at 7:20
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For $n$ sufficiently large, $x_n$ must be positive, since the sequence is increasing and unbounded.

Suppose not, then, you'd have $x_1\leq x_n\leq 0$ for all $n\in \mathbb{N}$, since, again, the sequence is increasing. This implies that $|x_n|\leq |x_1|$ for all $n\in \mathbb{N}$ and thus, the sequence would be bounded.

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  • $\begingroup$ You are very correct. My statement is that an unbounded increasing sequence must be eventually positive. I suppose I'll just edit that in. $\endgroup$ Sep 16, 2019 at 6:30
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If $(x_n)$ is unbounded, then for any $\alpha \in \mathbb R$ there exists $n(\alpha)\in\mathbb N$ such that $\alpha < x_{n(\alpha)}$

As you noticed, the authors claim that if the sequence is not bounded, then it is not bounded above. Your irritation is understandable: Of course there are unbounded sequences that are bounded from above (namely the sequences bounded above, but not bounded below).

But since all increasing sequences are bounded below$^1$, an increasing sequence is unbounded if and only if it is not bounded above.


$^1$ If $(x_n)$ is an increasing sequence, then $x_1$ is a lower bound.

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