I followed the 9 steps given by Rudin but in the end I didn't fully understand. My main questions are the following (in my understanding 2. is the follow up of 1.):
since the members of R have been defined as cuts of Q (thus containing rational numbers), how does this definition of R let appear the irrational numbers in R in the end?
In step 9 (the end indeed!) Rudin writes:
It is this identification of Q with Q' which allows us to regard Q as a subfield of R
What's the link between Q->Q'->R? Is Q' actually R? I understand that Q' contains cuts and it is isomorphic to Q. R also contains cuts by definition but how is it that Q is a subfield of R?
I hope my doubts are clear and I thank you in advance!
p.s.
Q' is what Rudin writes for Q* but in the post the star symbol gave me trouble forcing italics.
Let $C=\{q\in\Bbb Q:q<0\}\cup\{q\in\Bbb Q:q^2<2\}$; then the cut $C$ is the irrational number $\sqrt2$: it appears in the right place in the linear order on $\Bbb R$, and its square, using the definition of multiplication in $\Bbb R$, is $\{q\in\Bbb Q:q<2\}$, i.e., the member of $\Bbb Q'$ corresponding to $2\in\Bbb Q$. The other irrational numbers also appear as cuts: there is one for each cut $C$ with the property that $\Bbb Q\setminus C$ does not have a smallest element.
For your second question, the assertion is not that $\Bbb Q$ is literally a subfield of $\Bbb R$; after all, it is not even a subset of $\Bbb R$, so it can hardly be a subfield. However, $\Bbb Q'$ is a subfield of $\Bbb R$, and $\Bbb Q$ is isomorphic to $\Bbb Q'$, so from a purely algebraic and order-theoretic point of view $\Bbb Q'$ is indistinguishable from $\Bbb Q$. Thus, for all practical purposes we can think of $\Bbb Q'$ as actually being $\Bbb Q$, as if $\Bbb Q$ itself were a subset (and subfield) of $\Bbb R$, even though in this construction of $\Bbb R$ that is not literally the case.