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Theorem: Let $f$ be a continuous real-valued function on the closed interval $[a,b]$. Then $f$ is a bounded function. Moreover, $f$ assumes its maximum and minimum values of $[a,b]$.

The proof is given, however, I don't really understand what it is saying.

Proof:

By way of contradiction, assume $f$ is not bounded on $[a,b]$. Then, to each $n\in N$, there corresponds an $x_n \in [a,b]$ such that $|f(x_n)|\gt n$. (the proof goes on).

I don't understand this first line of the proof. Does it mean there must exist some $x_n$ that must be in $[a,b]$ where the product value, $|f(x_n)|$, is greater than the amount of $x_n$ that exist within the interval???

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First we fix $n$. Then, since $f$ is not bounded, we can certainly find some $x \in [a,b]$ such that $|f(x)| > n$. This $x$ is what we call $x_n$ from now on. We keep doing this and get an $x_n$ for every single $n \in \Bbb N$.

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  • $\begingroup$ I'm not too good at analysis. So, are you saying that for any value n (arbitrary), we can find an x within the interval $[a,b]$ where $|f(x)|$ is always going to be bigger than that n value? For example, let $f(x)=x$ be continuous between [1,10]. How would this apply? $\endgroup$ Commented Oct 26, 2018 at 23:52
  • $\begingroup$ That's just what I was trying to say, yeah. For $f(x)=x$ this does not apply though. That's because this is a proof by contradiction: we start out by assuming that $f$ is not bounded, i.e. we try to find a function that is unbounded and show how that implies that $f$ is not continuous. $\endgroup$ Commented Oct 26, 2018 at 23:56

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