Issue with Spivak's Solution

Here was the problem:

Here is the solution from his solutions book:

This is barely a proof. How can he just say let $f(c) = 0$? How do you prove that $f(c) =0$ and how do you prove that $f(d) = 0$?

How can I use the IVT to prove that criterion? Thanks!

$$c:=\max\{x\in[a,x_0]|\ f(x)=0\}$$

By definition $c\in\{x\in[a,x_0]|\ f(x)=0\}$. Therefore $f(c)=0$. Similarly for $f(d)=0$.

More interesting is to show that such maximum exists. We can define instead $$c':=\sup\{x\in[a,x_0]|\ f(x)=0\}$$ which always exists. By definition of supremum there is a sequence $x_n\in\{x\in[a,x_0]|\ f(x)=0\}$ such that $x_n\to c'$. But, because $f$ is continuous, then $f(c')=\lim f(x_n)=\lim 0=0$. Therefore $c'\in\{x\in[a,x_0]|\ f(x)=0\}$, i.e. $c'$ is actually a maximum.

The other interesting part is to show that $f(x)>0$ for $x\in[c,d]$. Assuming $f(x)=0$ for some $x\in[c,d]$ is a contradiction with the definition of $c$ and $d$. Assuming that $f(x)<0$ for some $x\in[c,d]$ gives a point in $[x,x_0]$ where $f$ vanishes. This again contradicts the definition of $c$ and $d$.

Apart from the fact that the labels $c$ and $d$ in the graph have been interchanged, there’s nothing wrong with the argument. The set of $x\in[a,x_0]$ such that $f(x)=0$ is non-empty because it contains $a$, and it’s closed because $f$ is continuous, so it has a largest element, which we’ll call $c$. $f(x_0)\ne 0$, so $c<x_0$. The existence of $d$ follows by similar reasoning applied to the interval $[x_0,b]$.

If you don’t know about closed sets at this point, the problem is harder: you have to show that $A=\{x\in[a,x_0]:f(x)=0\}$ actually has a maximum. It certainly has a supremum (least upper bound) $u$, since it’s bounded above by $x_0$. And if $u\notin A$, there must be a sequence $\langle y_n:n\in\Bbb N\rangle$ in $A$ converging to $u$. Continuity of $f$ then implies that $\langle f(y_n):n\in\Bbb N\rangle$ converges to $f(u)$ and hence that $f(u)=0$, i.e., that $u\in A$ after all.