I am a student who has taken the basic calculus courses but I am working through Calculus (the fourth edition) by Spivak in my spare time in order to both review the material and to gain a more rigorous understanding of the concepts. I am currently in Chapter 8 where he supplies a proof of the basis of the Intermediate Value Theorem, Theorem 7-1 (pp. 135-136 of the fourth edition). It seems mostly straightforward but I think there is some nuance of it that I don't grasp.
I am including in this post the statement Theorem 6-3 since he references it in the proof of Theorem 7-1 as well as part of Theorem 7-1 itself.
Theorem 6-3
Suppose $f$ is continuous at $a$, and $f(a) > 0$. Then $f(x) > 0$ for all $x$ in some interval containing $a$; more precisely, there is a number $\delta > 0$ such that $f(x) > 0$ for all $x$ satisfying $|x-a| < \delta$. Similarly, if $f(a) < 0$, then there is a number $\delta > 0$ such that $f(x) < 0$ for all $x$ satisfying $|x-a| < \delta$.
Problem 6-16 is also referenced. However,it is the same thing as Theorem 6-3 for one-sided limits. The following proof is the Theorem 7-1 up through the paragraph which I don't fully understand.
Theorem 7-1:
If $f$ is continuous on $[a,b]$ and $f(a) < 0 < f(b)$, then there is some number $z$ in $[a,b]$ such that $f(x) = 0$.
Proof: Define the set $A$ as follows: $$A = \{x : a \le x\le b, \mbox{ and } f \mbox{ is negative on the interval } [a,x] \}.$$ Clearly $A \ne \emptyset$, since $a$ is in $A$; in fact, there is some $\delta > 0$ such that $A$ contains all points $x$ satisfying $a \le x < a + \delta$; this follows from Problem 6-16, since $f$ is continuous on $[a,b]$ and $f(a)<0$. Similarly, $b$ is an upper bound for $A$ and, in fact, there is a $\delta > 0$ such that all points $x$ satisfying $b-\delta < x \le b$ are upper bounds for $A$; this also follows from Problem 6-16, since $f(b) > 0$.
From these remarks it follows that $A$ has a least upper bound $\alpha$ and that $a < \alpha < b$. We now wish to show that $f(\alpha) = 0$, by eliminating the possibilities $f(\alpha) < 0$ and $f(\alpha) > 0$.
Suppose first that $f(\alpha) < 0$. By Theorem 6-3, there is a $\delta > 0$ such that $f(x) < 0$ for $\alpha - \delta < x < \alpha + \delta$. Now there is some number $x_0$ in $A$ which satisfies $\alpha - \delta < x_0 < \alpha$ (because otherwise $\alpha$ would not be the least upper bound of $A$). This means that $f$ is negative on the whole interval $[a,x_0]$. But if $x_1$ is a number between $\alpha$ and $\alpha+\delta$, then $f$ is also negative on the whole interval $[x_0,x_1]$. Therefore $f$ is negative on the interval $[a,x_1]$, so $x_1$ is in $A$. But this contradicts the fact that $\alpha$ is an upper bound for $A$; our original assumption that $f(\alpha) < 0$ must be false.
The point I don't understand is in the final paragraph of the excerpt. Why does Spivak split the interval $[a,b]$ up into the subintervals $\lbrack a,x_0 \rbrack$ and $[x_0,x_1]$? Isn't it true that, because $\alpha$ is a least upper bound of $A$ and $f$ is continuous on $[a,b]$, it must be true that if $a \le x < \alpha$ then $f(x)<0$? I understand the need to pick out the $x_1$ but not the $x_0$. Especially perplexing is the statement "there is some number $x_0$ in $A$ which satisfies $\alpha - \delta < x_0 < \alpha$ (because otherwise $\alpha$ would not be the least upper bound of $A$." Wouldn't it be true that if there were some $x_0$ such that $a \le x_0 < \alpha$ and $f(x_0) \ge 0$ that $x_0$ would be an upper bound for $A$, contradicting that $\alpha$ is the least upper bound of $A$?
Thanks for any help.
