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 $f(y) = 0$.


1 Answer 1


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?


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