construction of the R set from Dedekind method

Question

I'm currently looking a way to understand how Dedeking is building R from Q set. How can i formally build R set from the Dedeking method ? I mean, from the Q set. Can we see each real number as an element of P(Q) ?

Answer 1

The idea is that a real number $x$ can be uniquely defined by the set $Q_x$ of all rational numbers less than it. In fact, if we have two real numbers $x<y$, then there is some $p\in \Bbb{Q}$ with $x < p < y$, so $Q_x \neq Q_y$. So this approach is pretty reasonable.

Starting only with $\Bbb{Q}$ (which we know to be an ordered field with the usual ordering), we can define $\Bbb{R}$ to be the set of all $x\subset \Bbb{Q}$ such that $x$ is an interval which is not bounded below, and not equal to $\Bbb{Q}$. Note: not every element of the power set of $\Bbb{Q}$ gives us a real number as you imply in the OP, only the unbounded below/bounded above intervals do.

We want $\Bbb{R}$ to be an ordered field in which every bounded above subset has a supremum. To do this, we have to define addition, multiplication, negation, and multiplicative inverses, and the order.

Given $x, y\in \Bbb{R}$, we can define $x+y$ to be the sum of $x$ and $y$ as sets; that is, $x+y = \{p+q : p\in x, q \in y\}$. You can check that $x+y\in \Bbb{R}$. Similarly, we define $$x-y = \{p-q:p\in x, q\notin y\}$$ $$x > 0 \iff \exists p\in x : p > 0$$

For multiplication, we have to be very careful so as not to include extra elements in the product. Defining the set $S^* = S\cap [0, \infty)$ for any $S\subset \Bbb{Q}$, the following definitions works for $x,y> 0$: $$xy = \{pq:p\in x^*, q\in y^*\}\cup (-\infty,0)$$ $$\frac{x}{y} = \{\frac{p}{q}:p\in x, q\notin y\}$$ Multiplication and division can then be extended to numbers of arbitrary sign in the obvious manner.

This defines our field operations and the ordering, so one just has to prove that these definitions do, in fact, provide an ordered field, and that $\Bbb{R}$ has the supremum property (note that the supremum of a bounded above $S\in \Bbb{R}$ is just the union of its elements).