I studied line-by-line of the proof of the Theorem 1.19 in the book of Principles of Mathematical Analysis (3rd ed) by Rudin. It shows that there is a proper superset of $\mathbb{Q}$ such that it is an ordered field and has the least-upper-bound property. It doesn't prove that this proper superset is $\mathbb{R}$.
Am I missing something? or, the constructed $R$ can be a proper subset of $\mathbb{R}$?
On
Since the ordered field with the least upper bound property is unique up to order preserving isomorphisms, Rudin defines $\mathbb R$ as $R$ assuming this.
A sketch proof of this result goes as follows: we assume $\mathbb R$ has been constructed (using theorem 1.19) and let $R$ be another ordered field with the least upper bound property.
Since $1 \in R,$ we get $n = 1 + 1 + \dots + 1 \in R$ also. This gives a map $\varphi : \mathbb N \rightarrow R.$ Since $\varphi(n) < \varphi(m)$ whenever $n < m,$ this map is an injection.
Since $-1 \in R$ defined by $1 + (-1) = 0,$ and $(-1) \times n = -n \in R$ for any $n \in \mathbb N,$ this map $\varphi$ extends to a map $\mathbb Z \rightarrow \mathbb R.$ Again using the ordered field property, this remains an injection.
If $p/q \in \mathbb Q$ with $p,q \in \mathbb Z$ and $q \neq 0,$ we get $\varphi$ extends to a map $\mathbb Q \rightarrow R$ by sending $\varphi(p/q) = \varphi(p)/\varphi(q).$ Note that this map is an injection by construction. Note $\varphi(p/q) = 0$ if and only if $\varphi(p)=0,$ so the map remains an injection. Also $\varphi$ can be checked to be a order-preserving homomorphism of fields, so $\varphi(a)<\varphi(b)$ if $a<b,$ $\varphi(ab) = \varphi(a)\varphi(b),$ $\varphi(a+b) = \varphi(a)+\varphi(b)$ and $\varphi(c^{-1}) = \varphi(c)^{-1}$ whenever $c \neq 0.$
Up until now what we did was completely algebraic, and showed any such field contains a copy of $\mathbb Q$ by the above construction. We now want to extend this to all of $\mathbb R$ and show the extension is bijective.
If $x \in \mathbb R,$ recall that using Rudin's definition via Dedekind cuts, $x$ satisfies, $$ x = \{a \in \mathbb Q : a < x \}. $$ Then we define $\overline{\varphi} : \mathbb R \rightarrow R$ by sending, $$ \overline{\varphi}(x) = \sup\{ \varphi(a) : a \in \mathbb Q, a < x \}. $$ Note this makes sense because $R$ has the least upper bound property, so we can take suprema (and by the archimedian property of $\mathbb R,$ this is bounded above by $\varphi(n)$ with $n > x$ an integer).
To show $\overline{\varphi}$ is injective, it is useful to show that for any $A \subset \mathbb Q$ bounded above we have, $$ \sup \varphi(A) = \overline{\varphi}(\sup A). $$ Using this, it shouldn't be too hard to show that $\overline{\varphi}$ is also an order preserving field homomorphism $\mathbb R \rightarrow R$ which agrees with $\varphi$ when restricted to $\mathbb Q.$
Finally to show surjectivity, let $y \in R.$ By the Archimedian property, the set, $$ A = \{a \in \varphi(\mathbb Q) : a < y \} $$ is bounded from above, so $\sup A = z \in R$ exists. Then $z \leq y,$ and we wish to show they are equal. If not, by the archimedian property (applied to $1/(y-z)$) there is $n \in \mathbb N$ such that $z + \varphi(n)^{-1} < y.$ But this implies that, $$ A = \{ a + \varphi(n)^{-1} : a \in \varphi(\mathbb Q), a < y \}. $$ But then $a + \varphi(n)^{-1} < y$ whenever $a \in \varphi(\mathbb Q)$ such that $a< y,$ so iteratively applying this gives $a + \varphi(m)\varphi(n)^{-1} < y$ for all $a \in A$ and $m \in \mathbb N.$ But this contradicts the archimedian property of $R$, which asserts that there is $N \in \mathbb N$ such that $y < N.$ Hence $y = z,$ so if we let, $$ B = \{b \in \mathbb Q : \varphi(b) < y\}, $$ we get $\overline{\varphi}(\sup B) = y.$ So $\overline\varphi$ is an isomorphism.
I don’t have Rudin book at hand.
However an important theorem that he probably uses is that an ordered Archimedean field having the upper bound property is unique up to isomorphism.