I am still working through Lindstrom's real analysis book Spaces, and am trying to understand and fill in the details of his proof of the Intermediate Value Theorem. He states the theorem in a special case:
Intermediate Value Theorem: Assume that $f : [a, b] \to \mathbb{R}$ is continuous and that $f(a)$ and $f(b)$ have opposite sign. Then there is a point $c \in (a, b)$ such that $f(c) = 0$.
He proceeds as follows:
Proof: We shall consider the case where $f(a) < 0 < f(b)$; the other case can be treated similarly. Let $A = \{x \in [a, b] : f(x) < 0\}$ and put $c = \sup A$. We shall show that $f(c) = 0$. Observe first that since $f$ is continuous and $f(b)$ strictly positive, our point $c$ has to be strictly less than $b$...
I am trying to understand why $c$ must be strictly less than $b$. Since $f$ is continuous, there exists a $\delta > 0$ such that $|x - b| < \delta \implies |f(x) - f(b)| < f(b)$ for any $x \in [a, b]$. Thus, we have $f(x) > 0$ for any $x$ in a small enough interval up to $b$. Clearly, $c \leq b$, so I imagine the fact that $c = \sup A$ somehow contradicts the possibility of $c = b$. I am having a hard time finding the right way to do this though. Perhaps the $\varepsilon$ definition of supremum plays a role here? Any help would be greatly appreciated.