Dedekind's Theorem/Proof 2

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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 producer of the $\tuple{L,R}$.

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


Proof

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 $\tuple {L, R}$.


Aiming for a contradiction, suppose $\beta \ne \alpha$ also produces $\tuple {L, R}$.

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 $c < \alpha$.

Then $c \in L \cap R$.

But by the definition of Dedekind Cut, $L$ 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.

$\blacksquare$