How to understand real analysis proofs. Doubt at Step 2, Appendix to Chapter 1 in Baby Rudin

Question

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 $q<p$.

Answer 1

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)