# Spivak Chapter 8, 3b

I think I am missing something on Spivak's chapter 8, problem 3b. Here is the question:

The proof of Theorem 1 depended on consideration of A = $$\{x: a \leq x \leq b,\forall y\in[a,x], f(y)<0\}$$. Give another proof of Theorem 1, which depends on consideration of $$B = \{x: a \leq x \leq b, f(x)<0\}$$. Which point x in [a,b] with f(x) = 0 will this proof locate?

Here is Theorem 1:

If f is continuous on [a,b] and f(a)<0<f(b), then there is some number x in [a,b] such that f(x) = 0.

To prove Theorem 1, we showed that (1) A has a least upper bound $$\alpha$$ and (2) $$f(\alpha) = 0$$. We then noted that $$\alpha$$ is the smallest $$x$$ with $$f(x) = 0$$.

I was able to give an alternative proof of Theorem 1 using the set B defined above. I showed (1) B has a least upper bound $$\beta$$ and (2) $$f(\beta) =0$$. The solution manual says that $$\beta$$ is the largest $$x$$ with $$f(x) = 0$$. This is the part that I don't understand. I see that if $$y>\beta$$, then $$f(y)\geq 0$$, but I don't see why it couldn't be that there is a $$y>\beta$$ with $$f(y) = 0$$.

Proof : suppose there is no $$x\in [a, b]$$ such that $$f(x) =0$$ i.e $$0\notin f([a, b])$$

Then $$A=f^{-1}(-\infty ,0)$$

$$B=f^{-1}(0, \infty)$$

$$[a, b]=A \cup B$$

$$•$$ $$A\cap B=\emptyset$$

$$•$$ $$a\in A, b\in B$$

Can you split $$[a, b]$$ as a union of two non empty disjoint open sets?