If $(X,\leq)$ is a set with a total order, how can I show that there is $Y\supset X$ s.t. $Y$ has the supremum property?

Question

In my course, they defined $\mathbb R$ as the smallest set that contain $\mathbb Q$ and that has the supremum property, i.e. that all upper-bounded set has a supremum.

1) My problem, it's that I don't know how I can be sure that such a set indeed exist. I'm not so sure how to construct it.

2) Also, if I can build such a set, how can I define $x\leq y$ if $x,y\notin \mathbb Q$.

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I know that this question is not as general than my title. But at the end, I wonder how to do in the very general case.

Answer 1

You are right, one has to construct this set. Call a pair $(L, R)$ of proper subsets of $\mathbb{Q}$ a Dedekind cut if :

1. $L \cup R = \mathbb{Q}$
2. $\forall x \in L, \ \forall y \in R, \ x < y$

One can then define $\mathbb{R}$ as the union of $\mathbb{Q}$ with the set $D$ of all Dedekind cuts. One can then define the order $<'$ on $\mathbb{R}$ as follows :

• for $q, l \in \mathbb{Q}$, $q <' l :\Longleftrightarrow q < l$
• for $c = (L, R) \in D$, $q \in \mathbb{Q}$,
  $c <' q :\Longleftrightarrow \forall x \in L, \ x < q$ and
  $q <' c :\Longleftrightarrow \forall x \in R, \ q < x$
• for $c = (L,R) \in D$ and $d = (L', R') \in D$,
  $c < d :\Longleftrightarrow \exists x \in L', \ \forall y \in L, \ y < x$