Least Upper Bound Property/Proof 2

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Theorem

Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded above.

Then $S$ admits a supremum in $\R$.

This is known as the least upper bound property of the real numbers.


Proof

Let $S$ be bounded above.

Let $L$ be the set of real numbers defined as:

$\alpha \in L \iff \exists x \in S: \alpha < x$

Let $R := \relcomp \R L$, where $\complement_\R$ denotes complement in $\R$.

By construction of $L$, every element of $L$ is less than some element of $S$.

Hence no element of $L$ is an upper bound of $S$.

By construction of $R$, for every element $x$ of $R$, there exists no element of $S$ which is greater than $x$.

Hence every element of $R$ is an upper bound of $S$.

So, to prove the existence of $\sup S$, it is sufficient to demonstrate that $R$ contains a smallest number.


We verify that $L$ and $R$ fulfil the conditions for Dedekind's Theorem to hold.

We confirm that $\tuple {L, R}$ is a Dedekind cut of $\R$:

$(1): \quad \set {L, R}$ is a partition of $\R$
$(2): \quad L$ does not have a greatest element
$(3): \quad \forall x \in L: \forall y \in R: x < y$

By Union with Relative Complement:

$L \cup R = \R$

By Set with Relative Complement forms Partition, $\tuple {L, R}$ forms a partition of $\R$.

So $(1)$ holds immediately.


Let $\alpha \in L$.

Then there exists $x \in S$ such that $\alpha < x$.

Let $\alpha'$ be such that $\alpha < \alpha' < x$.

Then $\alpha' \in L$

So whatever $\alpha \in L$ is, it cannot be the greatest element of $L$.

Thus $(2)$ holds.


Let $\alpha \in L$.

Let $\beta \in R$.

Then there exists $x \in S$ such that $\alpha < x$.

By construction of $R$, $x \le \beta$.

Thus $\alpha < \beta$ for all $\alpha \in L, \beta \in R$.

Thus $(3)$ holds.


By the corollary to Dedekind's Theorem, either $L$ contains a greatest element or $R$ contains a smallest element.

We have shown that $L$ does not contain a greatest element.

Hence $R$ contains a smallest element.

Hence if $S$ is bounded above, it has a supremum.


Thus $\R$ is Dedekind complete by definition.


Now let $S$ be bounded below.

By Dedekind Completeness is Self-Dual, it follows that $S$ admits an infimum.

$\blacksquare$


Sources

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