**The supremum**

A non-empty set $S \subseteq \mathbb{R}$ is **bounded from above **if there exists $M \in \mathbb{R}$ such that

$$x \le M, \quad \forall x \in S.$$

The number $M$ is called **an upper bound** of $S$.

If a set is bounded from above, then it has infinitely many upper bounds, because every number greater then the upper bound is also an upper bound. Among all the upper bounds, we are interested in the smallest.

Let $S \subseteq \mathbb{R}$ be bounded from above. A real number $L$ is called the **supremum** of the set $S$ if the following is valid:

(i) $L$ is an upper bound of $S$:

$$x \le L, \quad \forall x \in S,$$

(ii) $L$ is the least upper bound:

$$(\forall \epsilon > 0) (\exists x \in S)(L – \epsilon < x).$$

The supremum of $S$ we denote as

$$L = \sup S$$

or

$$L = \sup_{x \in S} \{x\}.$$

If $L \in S$, then we say that $L$ is a maximum of $S$ and we write

$$L= \max S$$

or

$$ L= \max_{x \in S}\{x\}.$$

If the set $S$ it is not bounded from above, then we write $\sup S = + \infty$.

**Proposition 1.** If the number $A \in \mathbb{R}$ is an upper bound for a set $S$, then $A = \sup S$.

The question is, does every non- empty set bounded from above has a supremum? Consider the following example.

**Example 1**. Determine a supremum of the following set

$$ S = \{x \in \mathbb{Q}| x^2 < 2 \} \subseteq \mathbb{Q}.$$

**Solution**.

The set $S$ is a subset of the set of rational numbers. According to the definition of a supremum, $\sqrt{2}$ is the supremum of the given set. However, a set $S$ does not have a supremum, because $\sqrt{2}$ is not a rational number. The example shows that in the set $\mathbb{Q}$ there are sets bounded from above that do not have a supremum, which is not the case in the set $\mathbb{R}$.

In a set of real numbers the **completeness axiom **is valid:

Every non-empty set of real numbers which is bounded from above has a supremum.

It is an axiom that distinguishes a set of real numbers from a set of rational numbers.

**The infimum**

In a similar way we define terms related to sets which are bounded from below.

A non-empty set $S \subseteq \mathbb{R}$ is **bounded from below** if there exists $m \in \mathbb{R}$ such that

$$m \le x, \quad \forall x \in S.$$

The number $m$ is called **a lower bound** of $S$.

Let $S \subseteq \mathbb{R}$ be bounded from below. A real number $L$ is called the **infimum** of the set $S$ if the following is valid:

(i) $L$ is a lower bound:

$$L \le x, \quad \forall x \in S,$$

(ii) $L$ is the greatest lower bound:

$$(\forall \epsilon > 0) ( \exists x \in S) ( x < L + \epsilon).$$

The infimum of $S$ we denote as

$$L = \inf S$$

or

$$L = \inf_{x \in S} \{x\}.$$

If $ L \in S$, then we say that $L$ is the minimum and we write

$$L= \min S$$

or

$$ L= \min_{x \in S}\{x\}.$$

If the set $S$ it is not bounded from below, then we write $\inf S = – \infty$.

The existence of a infimum is given as a theorem.

**Theorem**. Every non-empty set of real numbers which is bounded from below has a infimum.

**Proposition 2**. Let $a , b \in \mathbb{R}$ such that $a<b$. Then

(i) $\sup \langle a, b \rangle = \sup \langle a, b] = \sup [a, b \rangle = \sup [a, b] = b$,

(ii) $ \sup \langle a, + \infty \rangle = \sup [a, + \infty \rangle = + \infty$,

(iii) $\inf \langle a, b \rangle = \inf \langle a, b] = \inf [a, b \rangle = \inf [a, b] = a$,

(iv) $\inf \langle – \infty, a \rangle = \inf \langle – \infty, a ] = – \infty$,

(v) $\sup \langle – \infty, a \rangle = \sup \langle – \infty, a] = \inf \langle a, + \infty \rangle = \inf [a, + \infty \rangle = a$.

**Example 2**. Determine $\sup S$, $\inf S$, $\max S$ and $\min S$ if

$$ S = \{ x \in \mathbb{R} | \frac{1}{x-1} > 2 \}.$$

**Solution**. Firstly, we have to check what are the $x$-s:

$$ \frac{1}{x-1} > 2$$

$$\frac{1}{x-1} – 2 > 0$$

$$\frac{3 – 2x}{x-1} >0$$

The inequality above will be less then zero if the numerator and denominator are both positive or both negative. We distinguish two cases:

1.) $3-2x >0$ and $x-1 > 0$, that is, $ x < \frac{3}{2}$ and $ x > 1$. It follows $ x \in \langle 1, \frac{3}{2} \rangle$.

2.) $3 – 2x < 0$ and $ x-1 < 0$, that is, $x > \frac{3}{2}$ and $ x < 1$. It follows $ x \in \emptyset$.

$\Longrightarrow S = \langle 1, \frac{3}{2} \rangle$

From the proposition 2. follows that $\sup S = \frac{3}{2}$ and $\inf S = 1$.

The minimum and maximum do not exist ( because we have no limits of the interval).

**Example 3**. Determine $\sup S$, $\inf S$, $\max S$ and $\min S$ if

$$ S = \{ \frac{x}{x+1}| x \in \mathbb{N} \}.$$

**Solution**.

Firstly, we will write first few terms of $S$:

$$S= \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \cdots \}.$$

We can assume that the smallest term is $\frac{1}{2}$ and there is no largest term, however, we can see that all terms do not exceed $1$. That is, we assume $\inf S = \min S = \frac{1}{2}$, $\sup S = 1$ and $\max S$ do dot exists. Let’s prove it!

To prove that $1$ is the supremum of $S$, we must first show that $1$ is an upper bound:

$$\frac{x}{x+1} < 1$$

$$\Longleftrightarrow x < x +1$$

$$ \Longleftrightarrow 0 < 1,$$

which is always valid. Therefore, $1$ is an upper bound. Now we must show that $1$ is the least upper bound. Let’s take some $\epsilon < 1$ and show that then exists $x_0 \in \mathbb{N}$ such that

$$\frac{x_0}{x_0 +1} > \epsilon$$

$$\Longleftrightarrow x_0 > \epsilon (x_0 + 1)$$

$$\Longleftrightarrow x_0 ( 1- \epsilon) > \epsilon$$

$$\Longleftrightarrow x_0 > \frac{\epsilon}{1-\epsilon},$$

and such $x_0$ surely exists. Therefore, $\sup S = 1$.

However, $1$ is not the maximum. Namely, if $1 \in S$, then $\exists x_1 \in \mathbb{N}$ such that

$$\frac{x_1}{x_1 + 1} = 1$$

$$\Longleftrightarrow x_1 = x_1 +1$$

$$ \Longleftrightarrow 0=1,$$

which is the contradiction. It follows that the maximum of $S$ does not exists.

Now we will prove that $\min S = \frac{1}{2}$.

Since $\frac{1}{2} \in S$, it is enough to show that $\frac{1}{2}$ is a lower bound of $S$. According to this, we have

$$ \frac{x}{x+1} \ge \frac{1}{2}$$

$$ \Longleftrightarrow 2x \ge x+1$$

$$\Longleftrightarrow x \ge 1,$$

which is valid for all $x \in \mathbb{N}$. Therefore, $\inf S = \min S = \frac{1}{2}$.