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I'm reading Rudin Principles of Mathematical Analysis and am having trouble understanding this point. We haven't developed the real numbers yet (on the verge of doing so as the informed reader of this question is probably aware).

if $\alpha = \sup E \in Q$ exists then $\alpha$ may or may not be an element of $E$

Take $E_1$ to be the set of all $r \in Q$ with $r< a$ and let $E_2$ be the set of all $r \in Q$ with $r\le a$

$$\sup E_1 = \sup E_2 = a$$

I need external confirmation that my understanding of this statement is sound.

Specifically, I must show that $\sup E_1 = \sup E_2 = a$ by the definition of supremum given:

Definition Suppose $S$ is an ordered set, $E \subset$ and $E$ is bounded above. Suppose there exists an $\alpha \in S$ with the following properties

(I) $\alpha$ is an upper bound of $E$

(II) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E$

Then $\alpha$ is called the least upper bound of $E$ and we write

$$\alpha = \sup E$$

(Rudin, Walter. Principles of mathematical analysis. Vol. 3. New York: McGraw-Hill, 1964. 4. Print)

First, I'll verify that $a$ is an upper bound of $E_1$ and $E_2$. If $a \in Q$ is an upper bound then for all $b \in E_1, E_2$, $b \le a$. Inspecting how $E_1$ and $E_2$ are constructed verifies this.

What I do not understand is why the second part of the definition of supremum holds for $E_1$ and $E_2$. Please tell me if my reasoning is correct.

Here's my attempt:

For $E_1$:

If $\gamma < a$, we can choose $\gamma^\prime = \gamma + \frac{a-\gamma}{2} <a \implies \gamma^\prime \in E_1$ and $ a > \gamma^\prime > \gamma$. Therefore $\gamma < a$ is not an upper bound of $E_1$. This completes the lemma that $\sup E_1 = a$

Note 1 The reason the $E_1$ case is more complicated is because it is not clear that $\gamma < a \implies \exists x\in E_1 < \gamma$.

For $E_2$:

The verification is simple to see from taking $\gamma < a$.

Secondly, I'm hoping there is a better way of writing the verification than by my awkward construction of $\gamma^\prime$. In general, are moves like this bad or necessary. I just cannot see any other way to do it, re note 1.

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  • $\begingroup$ I checked Rudin's text and updated my response. Hopefully you find it helpful. $\endgroup$ – Matt A Pelto Mar 22 '17 at 23:15
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This is the (roughly) correct approach.

The average of two unequal numbers always lies strictly between them, and the average of two rational numbers is always rational.

These two assertions together tell you that there is some rational number ($\gamma'$) lying between $\gamma$ and $a$ which falls in the interval $(-\infty, a)$, if $a$ is a rational number. If you are not given that $a$ is rational, you need to find a rational number between the two, which can be easily justified using the density of $Q$ in $R$ or some moderately complicated algebra with floor functions.

This is not an awkward construction at all. Quite the opposite, taking the average of two numbers is the standard way to produce a number between them.

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  • $\begingroup$ Am I formally allowed to use this average of numbers if the real numbers, fields, etc. have not yet been defined. If so, I'd need to construct the real numbers before revisiting the proof of this little lemma. $\endgroup$ – theideasmith Mar 22 '17 at 22:01
  • $\begingroup$ @theideasmith tbh that wildly depends on the course and the professor. $\endgroup$ – Stella Biderman Mar 22 '17 at 22:06
  • $\begingroup$ I'm just reading the book on my own. What I really want to know is if the book itself was allowed to do this before formally defining the properties of real numbers. For example, the whole discussion of $\sqrt{2}$ to motivate the chapter was done without the axioms of arithmetic. However, we haven't proven that the average actually exists in $Q$ $\endgroup$ – theideasmith Mar 22 '17 at 22:08
  • $\begingroup$ @theideasmith if this problem is about rational $a$, then my answer is sufficient. Otherwise, it implicitly assumes that the reals exist when it says that the sup exists. If you haven't constructed the reals yet, the $a$ is probably intended to be rational because those are the only numbers known to exist. I would have to read the book again to be sure of the context though. $\endgroup$ – Stella Biderman Mar 22 '17 at 22:11
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$*$Updated$*$

I checked Rudin's text and answered my question. I see there is an example involving zero.

So we consider the case $a=0$. Assuming you still want to proceed by contradiction, we shall assume $\hat{\gamma}$ is an upper bound for $E_1$ but $\hat{\gamma} < 0$. By the Archimedean Principle, we may choose $N \in \mathbb{N}$ so that \begin{equation}\displaystyle N>-\frac{1}{\hat{\gamma}} \;.\end{equation} Since $-\frac{1}{N}$ is an element of $E_1$ we have reached a contradiction. Thus $0$ satisfies $(ii)$ from the definition of $\sup E_1$.

$*$for future reference$*$

Notice $(i)$ and $(ii)$ are equivalent to saying, for each $\varepsilon>0$ there exists $x \in E$ such that $\,\; \alpha - \varepsilon < x \leq \alpha$.

$\forall \varepsilon>0 \exists x \in E : \big\{ \alpha - \varepsilon < x \leq \alpha \big\}$

Knowing $\mathbb{Q}$ is dense in $\mathbb{R}$ is probably helpful in any problem dealing with a set such as $E_1$. Because it is helpful for getting your hands on this positive number $\varepsilon$ from this more concise definition.

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  • $\begingroup$ As I'm generally beginning my formal maths career, is the technique you used above to construct $\hat{\gamma}$ commonly used in proofs? Furthermore, is there a place to find a list of common techniques used in proofs like this one as well as the averaging approach used in Stella Biderman's answer $\endgroup$ – theideasmith Mar 23 '17 at 0:00
  • $\begingroup$ I would say it is. Though not every writer will mention how we know that we can find this positive integer $N$. Well you have to assume that this different upper bound is rational. Otherwise it essentially just boils down to the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, as Stella mentions. $\endgroup$ – Matt A Pelto Mar 23 '17 at 0:34
  • $\begingroup$ Am I expected to know what dense means at this point in the book? $\endgroup$ – theideasmith Mar 23 '17 at 2:16
  • $\begingroup$ I don't think so, but you will soon. $\endgroup$ – Matt A Pelto Mar 23 '17 at 2:20
  • $\begingroup$ This is the first time I've read a formal text – does Rudin assume you understand the reason for the motivating assertion above about $E_1$ and $E_2$ using either the density argument or the average argument, or is the reader expected to trust him with suspended disbelief. $\endgroup$ – theideasmith Mar 23 '17 at 2:24

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