Dedekind's Theorem
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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 field $\R$ is complete, and that is why it is 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 $\displaystyle \frac {l + r} 2$ 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.
Source of Name
This entry was named for Richard Dedekind.
