If it's possible, I'm wondering if someone can clarify the following for me as part of the proof of the Intermediate Value Theorem by Spivak in the 4th edition of his Calculus (proof and auxiliary theorem given at the bottom of the post). The line I am focusing on is "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 $A$)." In particular, I am not entirely convinced that I am internalizing the part in brackets.
My question is, why does the fact that $x_0<\alpha$ (where $\alpha$ is of course the supremum of the set, which exists as established earlier in the proof) necessarily require that $x_0$ is in $A$?
Just for example, $a-1 < \alpha$, but $a-1$ is not in $A$. Now this is a cheeky example, since it can (or so it seems to me) easily be established that we can find a $\delta$ so that this $x_0$ is between $a$ and $\alpha$. But even then, I feel like I am missing something. Why must this $x_0$ be in $A$ based on the definition of suprema? The proof, and the part I am confused about, intuitively make sense to me based on my hazy characterization of the supremum of a set as having the property that any number less than it (and greater than some other element in the set – here $a$ for example is known to be in the set and $x_0$ is greater than $a$) should be in the set, but that seems to me to roughly rest on the notion that the set contains all numbers (again, and unfortunately, roughly speaking) between that number known to be in the set and the supremum.
This question is admittedly closely related to this one, but I thought it was sufficiently different because it seems I am more confused about properties of suprema than about the proof as a whole.
Theorem 7-1 (Intermediate Value Theorem):
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:
$$A=\{x : a \le x \le b, \text{ and } f \text{ is negative on the interval } [a,x]\}.$$
Clearly $A \neq \varnothing$, 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.
In doing the proof, Spivak uses Theorem 6-3, given below:
Theorem 6-3:
Suppose $f$ is continuous at $a$, and $f(a)>0$. Then $f(x)>0$ for all $x$ in some interval containing $a$; more precisely, there is a number $\delta > 0$ such that $f(x)>0$ for all $x$ satisfying $|x−a|<\delta$. Similarly, if $f(a)<0$, then there is a number $\delta>0$ such that $f(x)<0$ for all $x$ satisfying $|x−a|<\delta$.
Best Answer
It seems that your confusion sort of stems from the fact that you didn't completely see how $x_0$ was chosen.
Assuming I understand what your mistake is, let me first make the following point.
Spivak is not saying that: if $x_0 \in \Bbb R$ satisfies $\alpha - \delta < x_0 < \alpha$, then $x_0 \in A$.
In fact, that statement would really be false.
What Spivak really is saying is that: Given any $\delta > 0$, there exists some $x_0 \in A$ such that $\alpha - \delta < x_0 < \alpha$.
This can be seen as follows:
Since $\delta > 0$, we have that $\alpha - \delta < \alpha$.
Since $\alpha$ is the least upper bound of $A$, we must have that $\alpha - \delta$ is not an upper bound of $A$.
What that means is that there exists $x_0 \in A$ such that $\alpha - \delta < x_0$.
Finally, since $x_0 \in A$, we must have that $x_0 \le \alpha$.
However, he has given a strict inequality, so you should be able to argue that you can actually choose $x_0 \neq \alpha$. (For this, you would need more properties about $A$. It will not just follow from the definition of least upper bound.)
Showing that $x_0 \neq\alpha$:
We do this by showing that $\alpha\notin A$. (Since we already know that $x_0 \in A$, this would give us that $x_0 \neq \alpha$.)
To see this, suppose that $\alpha \in A$. In this case, we have that $f$ is negative on $[a, \alpha]$. Moreover, by Theorem 6-3 (as quoted in the question), we see that there exists $\delta > 0$ such that $$f(x) < 0\quad\text{for all } \alpha - \delta < x < \alpha + \delta.$$ Moreover, since $\alpha < b$, we can choose $\delta$ small enough such that $\alpha + \delta < b$.
In particular, we have that $f(x) < 0$ for all $x \in \left[\alpha, \alpha + \frac{1}{2}\delta\right]$. Since $f$ is already negative on $[a, \alpha]$ (by assumption), we now see that $$f(x) < 0 \text{ for all } x \in \left[a, \alpha+\frac{1}{2}\delta\right].$$ Since $\alpha+\frac{1}{2}\delta < b$, we see that $\alpha+\frac{1}{2}\delta \in A$.
This contradicts that $\alpha$ is an upper bound since $\alpha < \alpha+\frac{1}{2}\delta$.