# Understanding this statement for the properties of continuous functions

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 $$n \in \Bbb N$$.
• 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?
• 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. Oct 26, 2018 at 23:56