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Here is an excerpt from Ross' Elementary Analysis

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Why did he let $|f(x_n)|>n$?

Oscar Flores
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    That's the assumption to reach the contradiction: if $;f(x);$ is not bounded then... – Timbuc Nov 13 '14 at 04:43
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    Let $n$ be a positive integer. He says that since the function $f$ is unbounded, it takes values outside the interval $[-n,n]$. Thus there is an $x$ such that $|f(x)| > n$. He calls this $x$ by the name $x_n$. The reason is that you get a different $x_n$ (possibly) each time you make a new choice of $n$. – Mike Nov 13 '14 at 04:44
  • @Mike Shouldn't the interval in question be $[a,b]$? Would it be $n \in \mathbb{N}$ such that $a \le n \le b$? – Oscar Flores Nov 13 '14 at 04:56
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    @OscarFlores $[a,b]$ is the domain of $f$. The fact that $f$ is unbounded means that its range is never contained in an interval of the form $[-M,M]$. Thus this is the case when $M = n$. – Mike Nov 13 '14 at 05:06
  • @Mike So with $|f(x)|>n$, you get a function that is unbounded on the interval $[a,b]$, but is still bounded which allows the contradiction. – Oscar Flores Nov 13 '14 at 05:14
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    He wants to prove that the function is bounded. So he assumes temporarily that it is unbounded, draws some consequences from that, reaches a contradiction, and then, since he's reached a contradiction, concludes that the function must in fact be bounded. The way he uses the assumption that the function is unbounded is to assert that no matter what value of $n$ is selected, $f(x)$ takes values with absolute value greater than $n$. – Mike Nov 13 '14 at 05:18
  • @Mike All right. Let me just get it straight, since he assumed that $f$ is unbounded in the interval $[a,b]$ for any $x$ in the interval, $f(x)$ will not within a closed interval (in this case [-n,n]). Hence, we get $|f(x)|>n$. – Oscar Flores Nov 13 '14 at 05:25
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    When you say that a function $f$ is unbounded, that is a property of the function $f$ as a whole. It's not meaningful to speak of $f$ being unbounded at a point. – Mike Nov 13 '14 at 05:27

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