Axion of completeness

Axiom of completeness

Every nonempty set of real numbers that is bounded above has a least upper bound.

Upper bound

Definition

$A$ set $A \subseteq \mathbb{R}$ is bounded above if there exists a number $b \in \mathbb{R}$ such that $a \leq b$ for all $a \in A$. The number $b$ is called an upper bound for $A$.

Lower bound

Definition

The set $A$ is bounded below if there exists a lower bound $l \in \mathbb{R}$ satisfying $l \leq a$ for every $a \in A$.

Supremum, Least upper bound

Definition

A real number $s$ is the least upper bound for a set $A \subseteq \mathbb{R}$ if it meets the following two criteria:

• $s$ is an upper bound for $A$

• if $b$ is any upper bound for $A$, then $s\leq b$

Notation

The Supremum $s$ of a subset $A$ is written as:

$$ s = \sup \left( A \right) $$

Remark

A less common notation is: $s = \text{lub} \left( A \right)$
Meaning least upper bound.

Corollary

A set can have a lot of upper bounds but it can only have at least one least upper bound.

Proof

If $s_1$ and $s_2$ are both least upper bounds for a set $A$, then we can assert $s_1 \leq s_2$ and $s_2 \leq s_1$. The conclusion is that $s_1 = s_2$ and least upper bounds are unique.

Minimum and Maximum

Definition

A real number $a_0$ is a maximum of the set A if $a_0$ is an element of $A$ and $a_0 \geq a \quad \forall a \in A$.
Similarly, $a$ number $a_1$ is $a$ minimum of $A$ if $a_1 \in A$ and $a_1 \leq a \quad \forall a \in A$.

Lemma

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = \sup (A)$ if and only if, for every choice of $\varepsilon > 0$, there exists an element $a \in A$ satisfying $s - \varepsilon < a$.

Proof

Firstly, we need to prove that if $ s = \sup(A) $, then for every choice of $ \varepsilon > 0 $, there exists an element $ a \in A $ such that $ s - \varepsilon < a $.
Secondly, we need to prove that if for every choice of $ \varepsilon > 0 $, there exists an element $ a \in A $ satisfying $ s - \varepsilon < a $, then $ s = \sup(A) $.

1. Assume $ s = \sup(A) $. We need to show that for every $ \varepsilon > 0 $, there exists an element $ a \in A $ such that $ s - \varepsilon < a $.

   Since $ s = \sup(A) $, by definition, $ s $ is the least upper bound of $ A $. This means:

   • $ s $ is an upper bound for $ A $, i.e., $ a \leq s $ for all $ a \in A $.

   • For any $ t < s $, there exists some element $ a_t \in A $ such that $ a_t > t $.

   Let $ \varepsilon > 0 $. Consider $ t = s - \varepsilon $. Since $ t < s $, by the property of the supremum, there must exist an element $ a_{\varepsilon} \in A $ such that $ a_{\varepsilon} > t $. Therefore, we have:

   $$ a_{\varepsilon} > s - \varepsilon. $$

   Thus, for every $ \varepsilon > 0 $, there exists an element $ a_{\varepsilon} \in A $ such that $ s - \varepsilon < a_{\varepsilon} $.

2. Assume that for every $ \varepsilon > 0 $, there exists an element $ a \in A $ satisfying $ s - \varepsilon < a $. We need to show that $ s = \sup(A) $.

   First, we show that $ s $ is an upper bound of $ A $. Assume for contradiction that $ s $ is not an upper bound. Then there exists some $ b \in A $ such that $ b > s $, which contradicts the assumption that $ s $ is an upper bound. Hence, $ s $ must be an upper bound.

   Next, we show that $ s $ is the least upper bound. Assume for contradiction that there exists some $ u < s $ which is also an upper bound of $ A $. Choose $ \varepsilon = s - u > 0 $. By assumption, there exists an element $ a_{\varepsilon} \in A $ such that:

   $$ s - \varepsilon < a_{\varepsilon}. $$

   Substituting $ \varepsilon = s - u $, we get:

   $$ s - (s - u) < a_{\varepsilon}, \quad u < a_{\varepsilon}. $$

   This contradicts the assumption that $ u $ is an upper bound of $ A $. Therefore, no such $ u < s $ can be an upper bound, and $ s $ must be the least upper bound.

   Thus,

   $$ s = \sup(A) $$

Combining both parts, we conclude that $ s = \sup(A) $ if and only if for every $ \varepsilon > 0 $, there exists an element $ a \in A $ such that $ s - \varepsilon < a $.

$\square$