Theorem 7-1: If $f$ is continuous on $[a,b]$ and $f(a) < 0 < f(b)$, then there is some number $x \in [a.b]$ such that $f(x) = 0$.
I have rewritten Spivak's proof using my own understanding of the proof, so I just want confirmation I understand it correctly if this proof and my justifications for each implication are correct. Thanks.
Proof: We define $A = \{ x : a \le x \le b, \text{ and } f \text{ is negative on the interval } [a,x]\}$. We first observe that $A$ is not empty because $f(a) < 0$ on $[a,a]$. Then notice that $b$ is an upper bound for $A$, as $A \subseteq \{ x : a \le x \le b \}$. By a previous theorem, we know that there is some $\delta$ such that whenever $a \le x < a + \delta$, we have that $f(x) < 0$. Similarly, there is some $\delta$ such that if $b - \delta < x \le b$, then $f(x) > 0$.
We let $\alpha = \sup A$. Note that $a < \alpha < b$, as $f(b) > 0$ and thus $b \notin A$.
We will show that $f(\alpha) = 0$ by showing that $f(\alpha) < 0$ and $f(\alpha) > 0$ are impossible. Suppose that $f(\alpha) < 0$. Since $f$ is continuous, $f$ must be continuous at $\alpha$, so by the previous theorem, there is some interval such that whenever $-\delta + \alpha < x < \delta + \alpha$, we have that $f(x) < 0$. Recall that $\alpha$ is the least upper bound for $A$, so there is some $x_0$ which satisfies $\alpha - \delta < x_0 < \alpha$. If there wasn't an $x_0$ which satisfies this, then $\alpha$ cannot be the least upper bound, since we should be able to find a smaller $x$ in a nonempty continuous set. We see that $x_0 \in A$, since it is smaller than $\alpha$ which is the least upper bound of $A$. Since $x_0 \in A$, we must have that $f([a,x_0])$ is a set of negative numbers. Now choose $x_1$ which satisfies $\alpha < x < \alpha + \delta$. By the previous theorem, we must have that $f(x_1) < 0$, (being in the interval around $\alpha$ for which $f$ is negative) implies $f(x) < 0$ on $[x_0,x_1]$.
However, we have that $\alpha$ was the least upper bound for $A$, yet since $f(x) < 0$ on $[x_0,x_1]$, we must have that $f(x) < 0$ on $[a,x_1]$ which implies that $x_1 > \alpha$. Therefore, a contradiction has been reached as $\alpha$ was shown to not be the least upper bound of $A$.
Now assume that $f(\alpha) > 0$. Again there is an interval such that whenever $\alpha - \delta < x < \alpha + \delta$, we have that $f(\alpha) > 0$. Like before, $\alpha$ is the least upper bound of $A$, which means we have to be able to find $x_0 \in A$ such that $\alpha - \delta < x < \alpha$. By the definition of $A$, we have that $f$ is negative on $[a,x_0]$, but we know that $f(x_0) > 0$, which is a contradiction. Therefore, $f(\alpha) > 0$ cannot be the case.
We have shown that $f(\alpha) \not< 0$ and $f(\alpha) \not> 0$, therefore we are left with only the possibility that $f(\alpha) = 0$.
I was quite verbose, but I wanted to make sure my comments are clear, because if they're not then I don't really understand each step. Also looking for general feedback.
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