Dedekind's Theorem

Theorem
Let the set of real numbers $$\R$$ be partitioned into two subsets $$L$$ and $$R$$ such that $$L$$ and $$R$$ form a section.

That is:
 * Every real number belongs to one or the other (but not both) of the two sets $$L$$ and $$R$$;
 * Each of $$L$$ and $$R$$ contains at least one element;
 * Any element of $$L$$ is less than any element of $$R$$.

Then $$\exists \alpha \in \R: x < \alpha \implies x \in L, x > \alpha \implies x \in R$$.

$$\alpha$$ itself may belong to either set.

Thus it is proved that the set $$\R$$ is continuous, and can this be referred to as the continuum.

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 $$\frac {l + r} 2$$ 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 Close Packed.

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.

Comment
In this context, the section is sometimes known as a Dedekind cut.