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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 $mathbb{R}$?

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 $overlinevarphi$ 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.