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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 $|x−a|<\delta$.

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    $\begingroup$ Is the solution to my problem as simple as: Suppose $x_0$ is not in $A$. Then either $x_0$ is less than the supremum but greater than all elements in $A$ or there is an element in $A$ which is greater than $x_0$ but which is (necessarily) less than the supremum. But $f(x_0)>0$, so no such element can exist since $x_0$ would be in its interval, as defined by the properties of elements in $A$ (contradiction). Thus $x_0$ is less than the supremum but greater than all elements in $A$, so $x_0$ is the supremum (contradiction). Thus $x_0$ is in $A$. Something as simple as "correct" would be perfect! $\endgroup$
    – EE18
    Commented Jul 6, 2020 at 15:17
  • $\begingroup$ It doesn't make sense to start the argument as "Suppose $x_0$ is not in $A$" since $x_0$ was chosen from $A$, to begin with! (So a better thing to ask would be - "Why does such an $x_0$ exist?".) $\endgroup$ Commented Jul 6, 2020 at 15:28

3 Answers 3

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Why must this $x_0$ be in $A$

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 $\alpha < \alpha+\frac{1}{2}\delta$.

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  • $\begingroup$ Thanks for the answer...I get confused starting at your statement that "What that means is that there exists 𝑥0∈𝐴 such that 𝛼−𝛿<𝑥0." It's clear to me that there is some $x_0$, $\alpha - \delta < x_0 < \alpha$, but it's still not clear to me why such an $x_0$ has to be in $A$ in particular from your answer. $\endgroup$
    – EE18
    Commented Jul 6, 2020 at 15:40
  • $\begingroup$ There must be such $x_0$ in $A$ since that is what you get from (negating) the definition of an upper bound. A number $u \in \Bbb R$ is said to be an upper bound of $A$ if for all $x \in A$, we have $x \le u$. Thus, if $u$ in not an upper bound of $A$, then there exists some $x_0 \in A$ such that $u < x_0$. In my answer, the role of $u$ was played by $\alpha-\delta$. $\endgroup$ Commented Jul 6, 2020 at 15:44
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    $\begingroup$ Never mind, I think I now get it...*since* $\alpha - \delta$ is not an upper bound (since it is less than the least upper bound), it follows that there is some $x_0$ in $A$ that is greater than it. And it is therefore this $x_0$ that we choose. Incidentally, what other properties of $A$ might I use to get this strict inequality? $\endgroup$
    – EE18
    Commented Jul 6, 2020 at 15:47
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    $\begingroup$ Got it, that clears things up beautifully, thanks. Just to confirm, did you mean $b$ when you wrote $\beta$? $\endgroup$
    – EE18
    Commented Jul 6, 2020 at 16:04
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    $\begingroup$ I did, indeed. Have fixed it now, thanks! $\endgroup$ Commented Jul 6, 2020 at 16:06
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If $\alpha$ is the least upper bound of $A$, then we are claiming that no matter how small $\delta > 0$ is, we can find $x_0\in A$ and $\alpha-\delta < x_0 < \alpha$. To see why this is true, let us suppose it were not true. Then, there must be some $\delta_0>0$ such that there is no $x_0\in A$ with $\alpha-\delta_0 < x_0$. In other words, every $x_0\in A$ must satisfy $x_0\le \alpha-\delta_0$. This means that the number $\alpha-\delta_0$ is an upper bound for $A$, but it is visually strictly less than $\alpha$, so we have contradicted the assumption $\alpha$ was the least upper bound of $A$.

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This question is probably no longer active but...

Another way of seeing why all $x< \alpha$ in the interval must be in $A$ is as follows:

We've defined the set A as:

$$A=\{x : a \le x \le b, \text{ and } f \text{ is negative on the interval } [a,x]\}.$$

Note especially the requirement "$f$ is negative on the interval $[a,x]$".

Suppose there exists some $x_0$ in $[a,\alpha)$ such that $f(x_0)$ is not negative.

This $x_0$ will not be in $A$. Furthermore, no $x > x_0$ can be in $A$, because the "interruption" at $x_0$ means any intervals that contain $x_0$ will not be negative over the whole interval $[a,x]$.

Therefore, $x_0$ is an upper bound of $A$. But $x_0$ is less than $\alpha$, a contradiction!

Because of the way we've defined $A$, any $x$ not in $A$ instantly becomes an upper bound on $A$.

Every point in the interval less than $\sup A$ must be in $A$.

This is not true of all bound sets. It's a consequence of the way this particular set is defined.

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  • $\begingroup$ Ah, I see now that 1729_SR pretty much already had this approach in their first comment under their question. FWIW, I think you were on the right track 1729_SR, and subsequent answers maybe failed to notice this important property of this specific set $A$. Spivak's proof really rests on this quality of $A$. $\endgroup$
    – Ben
    Commented Nov 13, 2020 at 4:31

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