Upper and Lower Bounds of Subsets of Real Numbers

If $S \subseteq \mathbb{R}$ is a subset of real numbers then elements of $x$ may all be less than a certain real number $M$ or greater than a certain real number $m$. Such sets are given special names which we define below.

Definition: Let $S \subseteq \mathbb{R}$ be a subset of real numbers. A real number $M \in \mathbb{R}$ is said to be an Upper Bound if for every $x \in S$ we have that $x \leq M$ and $S$ is said to be Bounded from Above. If no such $M \in \mathbb{R}$ exists we say that $S$ is Unbounded from Above.

Definition: Let $S \subseteq \mathbb{R}$ be a subset of real numbers. A real number $m \in \mathbb{R}$ is said to be a Lower Bound if for every $x \in S$ we have that $m \leq x$ and $S$ is said to be Bounded from Below. If no such $m \in \mathbb{R}$ exists we say that $S$ is Unbounded from Below.

Definition: Let $S \subseteq \mathbb{R}$ be a subset of real numbers. If $S$ is bounded from above and bounded from below, i.e., there exists $m, M \in \mathbb{R}$ such that for every $x \in S$ we have that $m \leq x \leq M$ then $S$ is said to be Bounded. If $S$ is not bounded then $S$ is said to be Unbounded.

Note that if $S$ is bounded above with upper bound $M$ then for all $\epsilon > 0$, $M + \epsilon$ is also an upper bound of $S$.

Similarly, if $S$ is bounded below with lower bound $m$ then for all $\epsilon > 0$, $m - \epsilon$ is also a lower bound of $S$.

Let's now look at some examples of bounded and unbounded sets of real numbers.

For example, consider the set $S = \{ 1, 2, 3, 4, 5 \}$. If $M = 6$ and $m = -2$ then for all $x \in S$ we have that $x \leq M$ and $m \leq x$. Hence $S$ is bounded above and bounded below.

For another example, consider the set of natural numbers, $\mathbb{N} = \{ 1, 2, 3, ... \}$. Then $\mathbb{N}$ is unbounded from above but is bounded below.

Definition: If $S \subseteq \mathbb{R}$ is a finite set then: a) $S$ is bounded from above. b) $S$ is bounded from below.

We only prove (a) as the proof of (b) is similar.

• Proof of a) Let $S$ be a finite subset of real numbers, say:

(1)

\begin{align} \quad S = \{ x_1, x_2, ..., x_n \} \end{align}

• Let $M = \max \{ x : x \in S \}$. Then $M = x_t$ for some $t \in \{ 1, 2, ..., n \}$ and by definition, $x_s \leq x_t$ for all $s \in \{ 1, 2, ..., n \}$. So $M$ is an upper bound for $S$ and so $S$ is bounded from above. $\blacksquare$