Consider a continuous real-valued function $f$ over the interval $[a, b]$. I am trying to prove that there exists at least one absolute max and one absolute min.
More formally:
$$\exists c,d \in [a,b] : \forall x \in [a,b], ~f(c) \leq f(x) \leq f(d)$$
I don't even know where to begin because it just seems so obviously true. How could you not have a max or min?
On
It is a very important theorem.
It has to do with both the compactness of [a,b] and the continuity of your function.
There is much more going on than it seems at the first glance.
The conclusion will not necessarily be true if the interval is not closed or the function is not continuous.
For example $f(x)=1/x$ on $(-1,1)$ does not have an absolute maximum or minimum.
First show that $f([a,b])$ is bounded: Assume that if it were not, choose $(x_{n})\subseteq[a,b]$ such that $f(x_{n})>n$. Now pick a convergent subsequence $(x_{n_{k}})\subseteq(x_{n})$ such that $f(x_{n_{k}})>n_{k}$, this will violate the continuity of $f$.
Now take $M=\sup f([a,b])$. Assume that $M\notin f([a,b])$, consider that $g=1/(M-f)$, then $g$ is continuous on $[a,b]$ and hence bounded by the previous argument. But $f(\eta_{n})\rightarrow M$ for some $(\eta_{n})\subseteq[a,b]$, then $g(\eta_{n})\rightarrow \infty$, a contradiction.