# Misunderstanding in Spivak's Proof of the Intermediate Value Theorem

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.

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I agree with your assertions. Perhaps his choice of argument comes from pedagogical concerns? –  Dylan Moreland Jul 5 '11 at 23:58
I agree with @Dylan. Spivak's argument is basically the method of continuity without the topology. (The method of continuity states that if $X$ is a topological space, and if $A\subset X$ is (a) nonempty (b) connected (c) open (d) closed, then $A$ is the entirety of a connected component of $X$. The existence of $x_0$ roughly is part (d). $x_1$ is part (c). Parts (a) and (b) follow from definition. So in some sense, once you learn a bit of topology, you can just use the usual global replace of $\epsilon-\delta$ arguments with "open-set" arguments and learn a general technique based on this. –  Willie Wong Jul 6 '11 at 0:21

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 $a\leq x <\alpha$ then $f(x)<0$ ?
Yes, it is true. But how do you know it is true? You either have to demonstrate that it is true, or appeal to a previously proven Theorem. (Just "intuitively this must be the case" is not "rigorous" enough in mathematics!) In this case, Spivak showed, as part of that last paragraph, precisely that claim. The choice of $x_0$ is to justify your quote above.