Dedekind's Theorem/Proof 1

Proof
Suppose $P$ and $Q$ are two properties which are mutually exclusive.

Suppose that one of either of $P$ and $Q$ are possessed by every $x \in \R$.

Suppose that any number having $P$ is less than any which have $Q$.

Let us call the numbers with $P$ the left hand set $L$, and the ones with $Q$ the right hand set $R$.

There are two possibilities, as follows.


 * $L$ has a greatest element, or
 * $R$ has a least element.

It is not possible that both of the above can happen.

Because suppose $l$ is the greatest element of $L$ and $r$ is the least element of $R$.

Then the number $\dfrac {l + r} 2$ is greater than $l$ and less than $r$, so it could not be in either class.

However, one of the above must occur.

Because, suppose the following.

Let $L_1$ and $R_1$ be the subsets of $L$ and $R$ respectively consisting of only the rational numbers in $L$ and $R$.

Then $L_1$ and $R_1$ form a section of the set of rational numbers $\Q$.

There are two cases to think about:

Maybe $L_1$ has a greatest element $\alpha$.

In this case, $\alpha$ must also be the greatest element of $L$.

Because if not, then there's a greater one, which we can call $\beta$.

There are always rational numbers between $\alpha$ and $\beta$ from Rational Numbers are Densely Ordered.

These are less than $\beta$ and thus belong to $L$ and (because they're rational) also to $L_1$.

This is a contradiction, so if $\alpha$ is the greatest element of $L_1$, it's also the greatest element of $L$.

On the other hand, $L_1$ may not have a greatest element.

In this case, the section of the rational numbers formed by $L_1$ and $R_1$ is a real number $\alpha$.

It must belong to either $L$ or $R$.

If it belongs to $L$ we can show, like we did before, that it is the greatest element of $L$.

Similarly, if it belongs to $R$ we can show it is the least element of $R$.

So in any case, either $L$ has a greatest element or $R$ has a least element.

Thus, any section of the real numbers corresponds to a real number.