real number system 5 (Dedekind cut)

Given a real number line $(-\infty, \infty)$, for example a coordinate axis, we are able to cut the line in two at an arbitrary point. The point will correspond one to one with a real number. Let the left side of the point on the line be $S$, and the right side be $T$. $S$ and $T$ are nonempty sets of real numbers such that
\[ S\cup T=\mathbb{R}, \quad  S\cap T=\phi .  \]
That is, all real numbers are mutually exclusive and collectively exhaustive in $S$ or $T$. Furthermore, if $s\in S$ and $t\in T$, then $s<t$. It is said 'Dedekind cut ' expressed by $(S,T)$.

In general, a cut of one ordered set into two sets $S,T$ logically has one of following opportunities.

(1) There exist both $\max S$ and $\min T$.
(2) There exists $\max S$, and does not $\min T$.
(3) There exists $\min T$, and does not $\max S$.
(4) There do not exist both $\max S$ and $\min T$.

You should note that we have to use $\max, \min$ in substitution for $\sup, \inf$. The reason is because conditions $\max S\in S$ and $\min T\in T$ should be met.

Any Dedekind cut of the real number line can not bring the result (1) in the dense field. Because, if we put $a=\max S$, and $b=\min T$,  then $a<b$. However, as real numbers are dense, there is a real number $x$ such that $a<x<b$. it is the contrary of $S\cup T=\mathbb{R} $. Hence, the result (1) is cleared out.

Suppose that $c$ is a point of 'Dedekind cut '. If $c$ is in $S$, the result (2) occurs, or if $c$ is in $T$, (3) does. As $c$ could be any real number, an irrational number $c=\sqrt{2}$  is available.

In the past, R.Dedekind explained real numbers were a continuum, since (2) or (3) was true by assuming that irrational numbers would exist in addition to rational numbers on real number
system.Then, although we intuitively feel it true, as it is impossible to reject the result (4) from all Dedekind cut in a dense ordered field, especially real numbers, we have to decide not to accept (4).

[41''] Dedekind cut at every point of a real number line brings just only the result (2) or (3).

It's assertion is equivalent to the axiom for completeness of real numbers [41] and  means the continuity of real numbers.

How do you feel about all of this?

real number system 4

 We will return to discuss the axiom for completeness [41]. In general, given an arbitrary ordered field $A$, and a nonempty subset $M\in A$, if every $M$ which is bounded above always has a least upper bound $\sup M$, then it is said  "the field $A$ is complete."

This is the reason why the term of "completeness" is used in [41]. As we have accepted the assertion [41] as an axiom, real numbers become complete. But rational numbers are incomplete.

The completeness axiom [41] will be also written as below.

[41'] If a nonempty set $M$ of real numbers is bounded above, then there is a unique real number $\sup M $ such that
(1) $m\leq \sup M $ for all $m\in M$
(2) if $\epsilon>0$ (no matter how small), there is a $m\in M$ such that $m>\sup M - \epsilon$.

As real numbers are dense and a ordered field, $\sup M$ is unique (why?). (1) is the definition of $\sup M$. If $\sup M \in M$, (2) is also no doubt. Moreover (2) is true even though $\sup M\notin M$. We already have used (2) in the proof of Archimedean property.  (2) indicates that we should examine the cut in between  $M$ and $\sup M$.

rational numbers

It is clearly true that rational numbers are dense, because for rational numbers $a$ and $b$ with $a<b$, there is a rational number $(a+b)/2$. Furthermore this dense property of rational numbers is satisfied in real numbers.

If $a$ and $b$ are real numbers with $a<b$, then there is a rational number $p/q$ such that $a<p/q<b$.

It is said that rational numbers are dense in real numbers. That is, between any different two real numbers, there is a rational number.

[wrong answer]
For any rational number $q>0$, $qa<qb$. As real numbers are dense, there is a $p$ such that
\[ qa<p<qb \]
Therefore,
\[ a<\frac{p}{q}<b \]
But, we are not able to state assertively that $p$ and $p/q$ are rational numbers.

[almost sure]
On Archimedean property " $na>b$ ", let $a$ be $b-a$, $b$ be $1$, and $n$ be $q$. Then,
\[ q(b-a)>1 \]
In this form $a$ and $b$ are real numbers and $q$ is a integer number. Next, if we choose $p$ be a smallest integer number such that $p>qa$, then $p-1\leq qa$. Hence, as
\[ qa<p\leq qa+1<qa+q(b-a)=qb, \]
if we divide above all sides by $q$, then we get the equation which would be hoped. Obviously, $p/q$ is a rational number.

real number system 3 (Archimedean property)

For understanding the completeness axiom [41], we should find out the density of real numbers and Archimedean property.

[density of real numbers]
If $a$ and $b$ are real numbers with $a<b$, then there is a real number $x$ such that $a<x<b$.

The proof of this proposition is very simple. If we put $x=(a+b)/2$, the proposition is always satisfied. Moreover, you will immediately find that there are so many real numbers in open interval $(a,b)$. For example, let $x$ be $a+(b-a)/n, (n>1, n\in \mathbb{N})$. This property is called the density of real numbers.

 The equation $0.999\cdots=1$ which we have addressed at first post is consistent with the dense property. Because if $0.999\cdots\ne 1$, as $0.999\cdots<1$ is a natural relationship on the ordered field, then there is an $x$ such that $0.999\cdots<x<1$. However, we can never accept it.

[Archimedean property of real numbers]
If $a$ and $b$ are positive real numbers, then there is an $n\in \mathbb{N}$ such that $na>b$.

The proof is by contradiction. For any $n\in \mathbb{N}$, if $na\leq b$, the set $A=\left\{ na | n\in \mathbb{N} \right\}$ has an upper bound $b$. By axiom for completeness [41], there exists $\sup A=s $.

For a $n'\in \mathbb{N}$, if $s=n'a\in A$, $s<(n'+1)a$. It is impossible, since $s$ is an upper bound of $A$. Therefore for any $n\in \mathbb{N}$, $na<s$. That is, $s\notin A$.

However by the density of real numbers, for any $\epsilon>0$(no matter how small), as we can put $s-\epsilon<s$, $s-\epsilon$ is not an upper bound of $A$. Therefore, we can choose an $n$ such that $s -\epsilon<na$.

At this time, if $\epsilon=a$, $s - a<na$, and $s<(n+1)a$. As $s$ is the least upper bound, it is a contradiction. Thus the proposition is correct.

Real number system became a dense ordered field satisfied completeness.