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I have a question regarding the following proof in Spivak's Calculus.

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.
      Suppose, on the other hand, that $f(\alpha) > 0$. Then there is a number $\delta > 0$ such that $f(x) > 0$ for $\alpha - \delta < x < \alpha + \delta$. Once again we know that there is an $x_0$ in $A$ satisfying $\alpha - \delta < x_0 < \alpha$; but this means that $f$ is negative on $[a, x_0]$.

$$\cdots$$

It's about the second case, when we assume $f(\alpha) > 0$. Specifically, the conclusion that:

"but this means that $f$ is negative on $[a, x_0]$"

I can't see how this is the case. By Theorem 6-3, there is a $\delta > 0$ such that $f(x) > 0$ for $\alpha - \delta < x < \alpha + \delta$. Since $x_0$ is in that range, we have that $f(x_0) > 0$, so the conclusion that "but this means that $f$ is negative on $[a, x_0]$" is incorrect. What am I missing?

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1 Answer 1

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You're missing that $\alpha$ is defined as the least upper bound of $A$, so that any number $x \lt \alpha$ must satisfy $x \in A$, meaning $f(x) \lt 0$. This contradicts the statement previously derived that $f(x) \gt 0$. This contradiction means that the assumption $f(\alpha) \gt 0$ must have been false.

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  • $\begingroup$ Thank you for your answer, I would like to clarify some things. If the knowledge that $\alpha$ is the least upper bound of $A$ gives us the ability to state that $\forall x < \alpha, f(x) < 0$. Then, I don't see why Spivak needs to use Theorem 6-3 to disprove the case $f(\alpha) > 0$. He could just proceed as your answer. $\endgroup$
    – nz_
    Commented Feb 10 at 1:43
  • $\begingroup$ @Naz He needs Theorem $6.3$ to rule out the possibility that $f$ "jumps" from a positive value to negative values. You need $f$ continuous for the claim to be true, so look at how the claim can fail if $f$ isn't continuous. That will help you see why you need to use the Theorem, which in turn requires continuity. $\endgroup$ Commented Feb 10 at 2:04

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