# Least Upper Bound Property

## 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 1

Suppose that $S \subseteq \R_{\ge 0}$ has the positive real number $U$ as an upper bound.

Then $\R_{\ge 0}$ can be represented as a straight line $L$ whose sole endpoint is the point $O$.

Let $l_0 \in \R_{\ge 0}$ be the standard unit of length.

There exists a unique point $X \in L$ such that $U \cdot l_0 = OX$.

Furthermore, if $x \in S$, then:

- $\map f x = x \cdot l_0$

where $\cdot$ denotes (real) multiplication.

### Segments of Finite Lines are Finite

No line segment of $OX$ is infinite.

For suppose that the segment $s$ of $OX$ is infinite.

Then $s$ is greater than every line segment, including any line four times greater than $OX$.

Therefore the less contains the greater: which is impossible.

$\Box$

### Existence of Second Endpoint of a Segment of $OX$ beginning at $O$

More precisely, every line segment $s$ of $OX$ having $O$ as one of its endpoints must have another endpoint within $OX$.

The second endpoint of $s$ of $OX$ must exist.

For if the second endpoint does not exist, then $s$ can be continued to any length however great and still remain a segment of $OX$.

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But then $s$ can be made over four times as great as $OX$, and still remain a segment of $OX$: which is impossible.

Therefore the second endpoint exists.

$\Box$

### Both Endpoints of a Segment of a Line lie Within the Line

Every point of a segment of a straight line $ab$ lies within $ab$.

This second endpoint must be within $OX$.

$\Box$

### Formation of the Set $S^*$ Corresponding to the Set $S$

Therefore for every $x \in S$, there is a unique point $\map w x$ such that:

- $\map f x = O \cdot \map w x = x \cdot l_0$

Thus let $S^*$ be the corresponding set of all respective line segments:

- $\map f x = O \cdot \map w x$

for all $x \in S$.

We have that $OX$ is greater than or equal to every line segment of $S^*$.

Therefore $OX$ contains every line segment of $S^*$.

And for any two line segments $P, Q$ of $L$ with an endpoint at $O$, either:

- $P, Q$ are identical
- $P \subset Q$ but $Q \not \subset P$

or:

- $Q \subset P$ but $P \not \subset Q$.

Also:

- $P \subset Q, Q \not \subset P \iff P > Q$

The same can be proven for any other upper bound $OY$ in $S^*$.

### Definition of $\Lambda$

Let $\Lambda$ be the union of all line segments of $S^*$.

### Existence of $\Lambda$

Let $x \in S$.

Then:

- $\map f x = O \cdot \map w x$

The point $O$ is an element of $O \cdot \map w x$.

Therefore there is a point $p$ contained in at least one line segment of $S^*$.

But then there must exist an exhaustive and complete figure $F$ containing only all of those points $p$ contained in at least one line segment of $S^*$.

Set theory shows that this figure $F$ is precisely $\Lambda$.

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$\Box$

### Continuity of $\Lambda$

$\Lambda$ is everywhere continuous.

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For, given $p, q \in \Lambda$, such that $p$ and $q$ do not coincide, either $Op > Oq$ or $Op < Oq$.

Without loss of generality, let $p$ be less distant from $O$ than $q$.

Then:

- $\exists x, h \in \R_{\ge 0}: x \in S \land x + h \in S \land p \in O \cdot \map w x \land q \in O \cdot \map w {x + h}$.

But:

- $O \cdot \map w x \subset O \cdot \map w {x + h}$

Therefore:

- $p, q \in O \cdot \map w {x + h}$

Therefore $O \cdot \map w {x + h}$ contains every point in between $p, q \in \Lambda$.

Thus suppose $r$ is between $p, q$.

Therefore:

- $r \in O \cdot \map w {x + h}$

But:

- $O \cdot \map w {x + h} \in S^*$

Also, $\Lambda$ contains all $p$ in at least one $O \cdot \map w x \in S^*$

Therefore:

- $r \in O \cdot \map w {x + h}$

Therefore:

- $\forall p, q \in \Lambda: \forall r: p < r < q: r \in \Lambda$

Therefore $\Lambda$ is everywhere continuous.

$\Box$

### $\Lambda$ is Finite

$\Lambda \subseteq OX$.

Let $p \in \Lambda$.

Then:

- $\exists y \in S: p \in O \cdot \map w y \in S^*$

But it was proven that $OX$ contains every line segment of $S^*$.

Therefore:

- $p \in O \cdot \map w y \subseteq OX$

Therefore:

- $p \in OX$

Therefore:

- $\Lambda \subseteq OX$

Therefore $\Lambda$ has a second endpoint $Z \in OX$, such that $Z$ is between $O, X$.

Therefore $\Lambda = OZ$.

$\Box$

### $\Lambda$ is an Upper Bound on $S^*$

$OZ$ is an upper bound on $S^*$.

For from set theory it is known that the union of all the sets any given set $D$ contains every set of $D$.

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Therefore $OZ$ contains every element of $S^*$.

But then no element of $S^*$ can ever be greater than $OZ$.

Therefore $OZ$ is an upper bound on $S^*$.

$\Box$

### $\Lambda$ is the Supremum on $S^*$

$OZ$ is the supremum on $S^*$.

For if $OY$ is any upper bound on $S^*$, $\Lambda \subseteq OY$.

For if $p \in \Lambda$, there is some $y \in S$ such that $p \in O \cdot \map w y \in S^*$.

But it was remarked earlier that if $OY$ is an upper bound on $S^*$, then $OY$ contains every line segment of $S^*$.

Therefore $p \in O \cdot \map w y \subseteq OY$.

Therefore $p \in OY$.

Therefore $\Lambda \subseteq OY$.

Therefore $\Lambda \le OY$.

Therefore $OZ$ is an upper bound on $S^*$ and less than or equal to every upper bound on $S^*$.

Therefore $OZ$ is the supremum on $S^*$.

From Supremum is Unique, $OZ$ is unique.

$\Box$

### Definition of $z$

There is a unique $z \in \R_{\ge 0}$ such that:

- $\map f z = O \cdot \map w z = OZ$

### $z$ is an Upper Bound on $S$

$z$ is an upper bound on all the elements of $S$.

For if not, then suppose there had been some $x \in S$ such that $x > z$.

$L$ is a representation of the positive real number line.

But if $x \in S$, then $O \cdot \map w x \in S^*$.

Yet $O \cdot \map w x > OZ$, which is impossible because $OZ$ is the supremum on $S^*$.

Therefore $z$ is an upper bound on all the elements of $S$.

$\Box$

### $z$ is the Supremum on $S$

$z$ is the supremum on all the elements of $S$.

For if $g \in \R_{\ge 0}$ and $g < z$, then $O \cdot \map w g < OZ$.

Therefore $O \cdot \map w g$ is not an upper bound on $S^*$.

But on the contrary, there exists $\xi \in S$ such that $O \cdot \map w \xi \in S^*$ and $O \cdot \map w \xi > O \cdot \map w g$.

But then $\xi > g$ and $\xi \in S$.

Therefore no $g < z$ can be an upper bound on $S$.

Therefore $z$ is the supremum on all the elements of $S$.

$\Box$

### Conclusion

Therefore the set $S \subseteq \R_{\ge 0}$ with upper bound $U$ has the unique supremum $z$.

$\blacksquare$

## Proof 2

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.

$\blacksquare$

## Also known as

The **least upper bound property of $\R$** is also known as:

- the
**supremum principle**. - the
**continuum property**(although this is also used to encompass the Greatest Lower Bound Property, a complementary result)

## Also see

## Sources

- 1970: Arne Broman:
*Introduction to Partial Differential Equations*... (previous) ... (next): Chapter $1$: Fourier Series: $1.1$ Basic Concepts: $1.1.2$ Definitions - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Axiom $1.1.4$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.4$: The Continuum Property - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**order properties**(of real numbers): $(5)$:*Completeness property*