Construction of Real Number by Dedekind Cuts

Question

I was studying Axiomatic Set Theory, and I have 2 questions about the construction of real numbers using Dedekind cut:

1. We define a real number using the Dedekind cut: $x_{\mathbb{R}} = \left \{ p \in \mathbb{Q} | p < x \right \}$, how we can show that this set is unique, in other words, if we have $x_{\mathbb{R}} = y_{\mathbb{R}}$ how we can show that $x = y$.
2. We define the addition of two real numbers as: $x_{\mathbb{R}} + y_{\mathbb{R}} = \left \{ q + r | q \in x_{\mathbb{R}} \wedge r \in y_{\mathbb{R}} \right \}$, how we can show that: $x_{\mathbb{R}} + y_{\mathbb{R}} = \left \{ p \in \mathbb{Q} | p < x + y\right \}$

Answer 1

"I was studying Axiomatic Set Theory, and I have 2 questions about the construction of real numbers using Dedekind cut:

We define a real number using the Dedekind cut: xR={p∈Q|p<x},"

NO! We don't!! A Dedekind cut is defined as any set of rational numbers that

1. is bounded above and
2. if x is in the set and y< x then y is also in the set.

It can be shown that if x is a rational number, then xR={p∈Q|p<x} is a Dedekind cut but the whole point us that there exist Dedekind cuts that are NOT of that form.
(If you allow x to be any real number then any Dedekind cut is of that form but obviously that cannot be used to define the real numbers!)

"how we can show that this set is unique, in other words, if weve xR=yR how we ca0n show that x=y." Suppose x and y were not equal. Then one is larger than the other and we can, without loss of generality, assume it is y: x< y. Then there exist a rational number. p, such that x< p< y. That contradicts the assumption that xR= yR.

"We define the addition of two real numbers as: xR+yR={q+r|q∈xR∧r∈yR}, how we can show that: xR+yR={p∈Q|p<x+y}"

Again, that is not the definition of a Dedekind cut! We prove, instead, that {q+r|q∈xR∧r∈yR} is a Dedekind cut by showing

1. Since xR is a Dedekind cut it has an upper bound, p, and since yR is a Dedekind cut it has an upper bound, q. Then it follows that p+ q is an upper bound for {q+r|q∈xR∧r∈yR}.

2. If x is in {q+r|q∈xR∧r∈yR} then x= a+ b where a is in xR and b is in yR. Then there exist u in xR, u< a and there exist v in yR, v< b so there exist y in {q+r|q∈xR∧r∈yR} such that y< x;