1
$\begingroup$

Firstly, I would also like to ask the following:

  1. I solved pre-calculus and calculus from the Differential and integral calculus by N. Piskunov. This is the first time, I am studying Real Analysis. Are there any video lectures out there that closely follow Rudin?

  2. How do I study a proof, and how do I know I've understood it? Does that mean, I should be able to come up with the proof on my own?

I am having considerable difficulty following the proof of the construction of the real numbers set $\mathbb{R}$ from $\mathbb{Q}$ in Baby Rudin theorem 1.19. I am stuck at step 2.

Theorem. There exists an ordered field $\mathbb{R}$ which has the least upper bound property. Moreover, $\mathbb{R}$ contains $\mathbb{Q}$ as a sub-field.

Step 1. The members of $\mathbb{R}$ will be certain subsets of $\mathbb{Q}$ called cuts. A cut by definition is any set $\alpha\subset\mathbb{Q}$ with the following properties:

(I) $\mathbb{\alpha}$ is not empty and $\alpha\notin\mathbb{Q}$.

(II) If $p\in\alpha$, $q\in\mathbb{Q}$ and $q<p$ then $q\in\alpha$

(III) If $p\in\alpha$, then $p<r$ for some $r\in\alpha$.

The letters $p,q,r$ denote rational numbers and $\alpha,\beta,\gamma$ denote cuts.

(III) says that $\alpha$ has no largest member.

(II) implies the following two facts that will be used freely.

If $p\in\alpha, q\in\alpha\implies{p<q}$.

By contraposition, if $p\notin\alpha, {p<q}\implies{q\notin\alpha}$.

Step 2. In step 2, we define $\alpha<\beta$ to mean that $\alpha$ is a proper subset of $\beta$, for any two cuts $\alpha,\beta$.

I construe that we are trying to prove here : $\mathbb{R}$ is an ordered set and that's why we are checking the transitivity and trichotomy requirements

If $\alpha<\beta$ and $\beta<\gamma$, it is clear that $\alpha<\gamma$. (A proper subset of a proper subset is a proper subset).

It is also clear that at most one of the three relations

$\alpha<\beta$, $\alpha = \beta$, $\beta<\alpha$

can hold true for any pair.

To show that, atleast one holds, assume that the first two fail. Then $\alpha$ is not a subset of $\beta$. Hence, there is a $p\in\alpha$ with $p\notin\beta$. If $q\in\beta$, it follows that $q<p$(since $p\notin\beta$), hence $q\in\alpha$ by (II).

I did not follow why $q<p$.

$\endgroup$
  • 1
    $\begingroup$ I suggest you ask question 1 and 2 in a new question. $\endgroup$ – user99914 Sep 2 '16 at 4:38
  • 1
    $\begingroup$ That's on of the properties/definitions of cuts. q in Beta if p <= q, then p would also be in Beta. $\endgroup$ – fleablood Sep 2 '16 at 4:39
  • 1
    $\begingroup$ It's worth noting that the construction of the reals is not a strict requirement to understanding the heart of analysis. Press on with the subject and you may find other topics less challenging $\endgroup$ – eepperly16 Sep 2 '16 at 4:50
  • 2
    $\begingroup$ Rudin himself put the construction in an appendix and considered it tedious and abstract. The most important thing is to know that the Reals have distinct and well stated properties and we can no longer make naive high school like assumptions about numbers. That the reals are an ordered field containing the rationals with the least upper bound and what that means is more important than the construction of it. $\endgroup$ – fleablood Sep 2 '16 at 4:59
1
$\begingroup$

That follows from the definition of a cut (and trichotomy of $\mathbb Q$). If instead $q >p$, since $q\in \beta$, then $p \in \beta$ (See (II) in the definition of a cut)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.