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???
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 $n \in \Bbb N$.