# Confusion about proof in Spivak?

I am currently reviewing some of the proofs in Spivak's Calculus, and I have come across this statement as a part of the proof for Extreme value theorem.

if $f(x)$ is continuous on $[a, b]$ then for every $\epsilon > 0$ there is $x$ in $[a, b]$ with $L - f(x) < \epsilon$ where $L$ is the least upper bound.

Now this basically mean that its a one sided limit where $f(x)$ approaches $L$ but I wonder why there were no justifications for this step in the book, usually nothing is taken for granted in Spivak's proofs but this seems to be something non-trivial by first year analysis standard that is stated as true. So is this some how "obvious" or is Spivak asking the reader to justify this step instead?

Edit: There were no mentioning of any real number properties what-so-ever. This proof is on page 136 of the third edition.

-

For any set of real numbers $S$, if $L=\sup S$ is the least upper bound of $S$, then for every $\epsilon>0$ there is an $s\in S$ such that $L-s<\epsilon$. (This is one of the first theorems in baby Rudin if I remember correctly.) In this case, $S=\{f(x)\mid x\in[a,b]\}$.
The proof is simple: if for some $\epsilon>0$ we had $L-s\geq\epsilon$ for all $s\in S$, then for every $s\in S$ we have that $s\leq L-\epsilon$, contradicting the assumption that $L$ is the least upper bound.
The assumption that $f$ is continuous on $[a,b]$ is to ensure that the least upper bound of $S=\{f(x)\mid x\in[a,b]\}$ exists; this is because the continuous image (which here is $S=f([a,b])$) of a compact set (which here is $[a,b]$) is compact, and because $S$ is compact it must be bounded, and therefore have an upper bound, and hence a least upper bound.
@Mark, this is really the definition of $\sup$: anything less that the $\sup$ is not an upper bound. – lhf May 10 '11 at 2:29