Dedekind's Theorem

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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 $\displaystyle \frac {l + r} 2$ 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.

$\blacksquare$


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 Close Packed, 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:

$L \cap R = \O$

This is a contradiction.

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 $\sup {L}$.

This is a contradiction.

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 Close Packed, 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:

$L \cap R = \O$

This is a contradiction.

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.


Also see


Source of Name

This entry was named for Julius Wilhelm Richard Dedekind.


Sources