# Dedekind's Theorem

## Theorem

Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.

Then there exists a unique real number which is a producer of $\tuple {L, R}$.

Thus it is proved that the totally ordered set $\R$ is Dedekind complete, and that is why it is referred to as the continuum.

### Corollary

Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.

Then either $L$ contains a largest number or $R$ contains a smallest number.

## Proof 1

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.

$\blacksquare$

## Proof 2

### Proof of Uniqueness

Aiming for a contradiction, suppose both $\alpha$ and $\beta$ produce $\tuple {L, R}$.

By the Trichotomy Law for Real Numbers either $\beta < \alpha$ or $\alpha < \beta$.

Suppose that $\beta < \alpha$.

From Real Numbers are Densely Ordered, there exists at least one real number $c$ such that $\beta < c$ and $c < \alpha$.

Because $c < \alpha$, it must be the case that $c \in L$.

Because $\beta < c$, it must be the case that $c \in R$.

That is:

$c \in L \cap R$

But by the definition of Dedekind Cut, $\tuple {L, R}$ is a partition of $\R$.

That is, $L$ and $R$ are disjoint.

That is:

$L \cap R = \O$

Similarly, $\alpha < \beta$ also leads to a contradiction.

It follows that $\alpha$ is unique.

$\Box$

### Proof of Existence

Let $\alpha = \sup L$.

Let $u_1 \in \R$ such that $u_1 < \alpha$.

Aiming for a contradiction, suppose $u_1 \in R$.

Suppose there is no real number $s \in L$ such that $u_1 < s \le \alpha$.

Then $u_1$ would be an upper bound of $L$ which is less than $\sup {L}$.

So there exists at least one real number $s \in L$ such that $u_1 < s \le \alpha$.

But this is also a contradiction because all the elements of $R$ are greater than all the elements of $L$.

It follows that $u_1 \in L$.

Let $u_2 \in \R$ such that $\alpha < u_2$.

Aiming for a contradiction, suppose $u_2 \in L$.

There exists $s' \in \R$ such that $\alpha < s' < u_2$.

But $s' \notin L$ and thus $s' \in R$.

This is a contradiction because all the elements of $R$ are greater than all the elements of $L$.

It follows that $u_2 \in R$.

Hence, $\alpha$ produces the cut $\tuple {L, R}$.

$\blacksquare$

## Proof 3

### Proof of Uniqueness

Aiming for a contradiction, suppose both $\alpha$ and $\beta$ produce $\tuple {L, R}$.

By the Trichotomy Law for Real Numbers either $\beta < \alpha$ or $\alpha < \beta$.

Suppose that $\beta < \alpha$.

From Real Numbers are Densely Ordered, there exists at least one real number $c$ such that $\beta < c$ and $c < \alpha$.

Because $c < \alpha$, it must be the case that $c \in L$.

Because $\beta < c$, it must be the case that $c \in R$.

That is:

$c \in L \cap R$

But by the definition of Dedekind Cut, $\tuple {L, R}$ is a partition of $\R$.

That is, $L$ and $R$ are disjoint.

That is:

$L \cap R = \O$

Similarly, $\alpha < \beta$ also leads to a contradiction.

It follows that $\alpha$ is unique.

$\Box$

### Proof of Existence

Let $\gamma$ be the set of all rational numbers $p$ such that $p \in \alpha$ for some $\alpha \in L$.

It is to be verified that $\gamma$ is a cut.

Because $L$ is not empty, neither is $\gamma$.

Suppose $\beta \in \R$ and $q \notin \beta$.

Then because $\alpha < \beta$, we have that $q \notin \alpha$ for any $\alpha \in L$.

Thus $q \notin \gamma$.

Thus $\gamma$ satisfies criterion $(1)$ for being a cut.

Suppose $p \in \gamma$ and $q < p$.

Then $p \in \alpha$ for some $\alpha \in L$.

Hence $q \in \alpha$.

Hence $q \in \gamma$.

Thus $\gamma$ satisfies criterion $(2)$ for being a cut.

Suppose $p \in \gamma$.

Then $p \in \alpha$ for some $\alpha \in L$.

Hence there exists $q > p$ such that $q \in \alpha$.

Hence $q \in \gamma$.

Thus $\gamma$ satisfies criterion $(3)$ for being a cut.

Thus, by definition of the real numbers by identifying them with cuts, $\gamma$ is a real number.

We have that:

$\alpha \le \gamma$

for all $\alpha \in L$.

Aiming for a contradiction, suppose there exists $\beta \in R$ such that $\beta < \gamma$.

Then there exists some rational number $p \in\ Q$ such that $p \in \gamma$ and $p \notin beta$.

But if $p \in \gamma$ then $p \in \alpha$ for some $\alpha\ in L$.

This implies that $\beta < \alpha$.

But this contradicts the definition of Dedekind cut:

$(3): \quad \forall x \in L: \forall y \in R: x < y$

Thus $\gamma \le \beta$ for all $\beta \in R$.

That is, $\gamma$ is a producer of $\tuple {L, R}$.

$\blacksquare$

## Also known as

Dedekind's Theorem is also known as the completeness theorem for the real numbers.

## Source of Name

This entry was named for Julius Wilhelm Richard Dedekind.