Problem: Let $f$ be a continuous function whose domain includes the closed interval [$a$,$b$]. If $f(a)$ $\lt$ $0$ and $f(b)$ $\gt$ $0$, then there is a number $x$ between $a$ and $b$ such that $f(x)$ = $0$
Proof:
Let $f$ be a continuous function whose domain includes the closed interval [$a$,$b$]
Let $f(a)$ $\lt$ $0$ and $f(b)$ $\gt$ $0$
Then if $S$ is any open interval containing the number $f(x)$, then there is an open interval $T$ containing the number $x$ such that if $t$ $\in$ $T$, and $t$ is in the domain of $f$, then $f(t)$ $\in$ $S$ (our class' defintion of continuous)
Then it has a left-most point $a$ and a right-most point $b$
I'm not sure where to go from here. I need to show that there is a number $x$ between $a$ and $b$ such that $f(x)$ = $0$
Best Answer
According to your "class" definition of continuous, you could come up with the following solution :
Since the set $C = \{x\in [a,b] : f(x) < 0\}$ is bounded, it must have a least upper bound(roughly, a number which is greater than every element of $C$, but is the smallest number with that property). If you like I can elaborate here.
Let $c$ be that least upper bound(also called the supremum and denoted $c = \sup C$). We claim $f(c) = 0$.
What if $f(c) < 0$? Then let $S$ be the interval $\left(\frac{3f(c)}{2},\frac{f(c)}{2}\right)$. Note that $S$ contains strictly negative numbers. Now, we know that there is an interval $T$ around $c$ which is such that if $t \in T$ then $f(t) \in S$. Now, an interval around $c$ contains at least one point larger than $c$, say $c_0$, but then $f(c_0) < 0$ so $c_0 \in C$. This contradicts the fact that $c$ was the supremum of $C$.
Similarly, I leave you to see that $f(c) > 0$ will also lead to a contradiction. Hence, you can conclude that $f(c) = 0$. Clearly, $c \in (a,b)$ since $c < b$.