1. Intervals
Let $a,b \in {\mathbb{R}}$, $a < b$.
| Type of interval | Notation |
|---|---|
| bounded (length $\ell = b - a$) | $\lbrack a,b\rbrack = \left\{ x \in {\mathbb{R}}\mkern6mu \middle| \mkern6mu a \leq x \leq b \right\}$, $(a,b) = \left\{ x \in {\mathbb{R}}\mkern6mu \middle| \mkern6mu a < x < b \right\}$, $\lbrack a,b)$, $(a,b\rbrack$ |
| unbounded | $\lbrack a, + \infty) = \left\{ x \in {\mathbb{R}}\mkern6mu \middle| \mkern6mu x \geq a \right\}$, $(a, + \infty)$, $( - \infty,a\rbrack$, $( - \infty,a)$, $( - \infty, + \infty) = {\mathbb{R}}$ |
Remember
For $r > 0$:
$$|x - a| < r \Leftrightarrow x \in (a - r,a + r)$$
$$|x - a| \leq r \Leftrightarrow x \in \lbrack a - r,a + r\rbrack$$
$$|x - a| > r \Leftrightarrow x \in ( - \infty,a - r) \cup (a + r, + \infty)$$
2. Bounded sets. Minimum and maximum
Let $A \subset {\mathbb{R}}$, $A \neq \varnothing$.
Definitions
- $M \in {\mathbb{R}}$ is an upper bound of $A$ if $x \leq M,\mkern9mu\forall x \in A$; $A$ is bounded above if it has at least one upper bound.
- $m \in {\mathbb{R}}$ is a lower bound of $A$ if $m \leq x,\mkern9mu\forall x \in A$; $A$ is bounded below if it has at least one lower bound.
- $A$ is bounded if it is bounded above and bounded below $\Leftrightarrow \exists M > 0$ such that $|x| \leq M,\mkern9mu\forall x \in A$.
- $\max A$ is an element of $A$ that is an upper bound; $\min A$ is an element of $A$ that is a lower bound.
- $A$ is not bounded above $\Leftrightarrow \forall M \in {\mathbb{R}},\mkern9mu\exists x \in A$ such that $x > M$.
- Example: $A = (0,1\rbrack$ is bounded, $\max A = 1$, but $A$ has no minimum.
3. The bounds of a set
Definitions
- $\sup A$ (the supremum) = the least upper bound of $A$.
- $\inf A$ (the infimum) = the greatest lower bound of $A$.
- Cantor's axiom: every nonempty set of real numbers that is bounded above has a supremum in $\mathbb{R}$ (likewise, every nonempty set that is bounded below has an infimum).
The $\varepsilon$ characterization
$$s = \sup A \Leftrightarrow \text{ (1) }x \leq s,\mkern9mu\forall x \in A\mkern9mu\text{ and (2) }\forall\varepsilon > 0,\mkern9mu\exists x_{\varepsilon} \in A:\mkern9mu x_{\varepsilon} > s - \varepsilon$$
$$i = \inf A \Leftrightarrow \text{ (1) }x \geq i,\mkern9mu\forall x \in A\mkern9mu\text{ and (2) }\forall\varepsilon > 0,\mkern9mu\exists x_{\varepsilon} \in A:\mkern9mu x_{\varepsilon} < i + \varepsilon$$
- $A$ has a maximum $\Leftrightarrow \sup A \in A$, and then $\max A = \sup A$; $A$ has a minimum $\Leftrightarrow \inf A \in A$, and then $\min A = \inf A$.
Maximum or supremum?
| $\max A$ | $\sup A$ | |
|---|---|---|
| What it is | the largest element of $A$ | the least upper bound of $A$ |
| Does it belong to $A$? | yes, always | not necessarily |
| Does it exist? | not always (e.g. $(0,1)$) | yes, if $A$ is bounded above |
Worked example
$A = \left\{ \frac{n}{n + 1}\mkern6mu \middle| \mkern6mu n \in {\mathbb{N}} \right\}$.
- Since $\frac{n}{n + 1} = 1 - \frac{1}{n + 1}$, we have $0 \leq x < 1,\mkern9mu\forall x \in A$.
- $0 \in A$ (for $n = 0$) and $0$ is a lower bound $\Rightarrow \min A = \inf A = 0$.
- $1$ is an upper bound; for $\varepsilon \in (0,1)$: $\frac{n}{n + 1} > 1 - \varepsilon \Leftrightarrow n > \frac{1}{\varepsilon} - 1$, which has solutions in $\mathbb{N}$ $\Rightarrow \sup A = 1$.
- $1 \notin A$ $\Rightarrow$ $A$ has no maximum.
4. The extended real line
Remember
- $\overline{\mathbb{R}} = {\mathbb{R}} \cup \left\{ - \infty, + \infty \right\}$, with $- \infty < x < + \infty,\mkern9mu\forall x \in {\mathbb{R}}$.
- If $A$ is not bounded above: $\sup A = + \infty$. If $A$ is not bounded below: $\inf A = - \infty$.
- Examples: $\sup{\mathbb{N}} = + \infty$, $\inf{\mathbb{N}} = \min{\mathbb{N}} = 0$; $\inf{\mathbb{Z}} = - \infty$; $\inf(1, + \infty) = 1$, $\sup(1, + \infty) = + \infty$.