I need to clarify my understanding of the extreme value theorem.

Question

I have had some difficulty in trying to understand the extreme value theorem but I think I might understand it correctly now but would like to clarify that I am thinking about this correctly. If we take for example the function 1/x, this is a continuous function over the reals without $0$. Clearly, the closer x approaches $0$, the larger 1/x becomes so 1/x is not bounded as x approaches $0$. However, $0$ is not part of the domain of 1/x so we can not apply the extreme value theorem since we require a closed interval for the theorem to hold is this correct?

Answer 1

The extreme value theorem is a special case of the fact that a continuous function on a compact set attains a maximum and minimum.

A compact set in $\mathbb{R}$ is a set $S$ that is closed and bounded. The closed part you already know. In the case that $S$ is an interval, it means that the interval must contain the endpoints. More generally, it means that if $S$ contains points arbitrarily close to some point $x$, it must also contain $x$. In the case of $T = \mathbb{R} \setminus \lbrace 0 \rbrace$ that you considered, $T$ contains points arbitrarily close to $0$ (like $.01$, $.001$, and so on) but does not contain $0$, so it is not closed.

I should also point out that this $T$ is not an interval. An interval is a set such $S'$ that if $x \in S'$, $y \in S'$, and $x \leq a \leq y$, then $a \in S'$. But the fact about continuous functions attaining a maximum and minimum is true for all compact sets, not just compact intervals.

The bounded part means that all the points in the interval have absolute value less than or equal to some fixed finite number. That is, $S \subseteq [-M,M]$ for some finite positive number $M$. This excludes the $(-\infty, \infty)$ case that Michael Hoppe brought up.