I'm reading Rudin's Principles of Mathematical Analysis. It is said in the book, 1.19 Theorem (pg-8): "there exists an ordered field $\mathbb{R}$ which has the least-upper-bound property. Moreover, $\mathbb{R}$ contains $\mathbb{Q}$ as a subfield". So $\mathbb{Q}\subset\mathbb{R}$.
But in the proof of the theorem (pg-17), it is said that the members of $\mathbb{R}$ will be certain subsets of $\mathbb{Q}$.
That is to me a contradiction, I mean, if the members of $\mathbb{R}$ will be certain subsets of $\mathbb{Q}$, it can't be that $\mathbb{Q}\subset\mathbb{R}$. ¿What is the explanation for that?
Thanks.
On page. 17 it is indeed explained that the members of $\mathbb R$ are subsets of $\mathbb Q$ called cuts.
Based on that construction, you identify each element of $q \in \mathbb Q$ to a unique cut. Namely for $q \in \mathbb Q$ you take the cut $\bar{q}=\{r \in \mathbb Q \ ; \ r <q\}$. Based on that all elements of $\mathbb R$ are subsets of $\mathbb Q$. For example, the irrational number $\sqrt{2}$ is identified to the cut $$\sqrt{2}=\{r \in \mathbb Q_+ \ ; \ r^2 < 2\} \cup \mathbb Q_-\cup\{0\}$$ which is a subset of $\mathbb Q$.