# Spivak's Calculus Theorem 7-1 not understanding case when $f(\alpha) > 0$

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?

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.
• 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.
• @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. Commented Feb 10 at 2:04