01-analysis.Rmd

# (PART) Background material {-} 

# Analysis {#analysis}

## Upper bounds and suprema

This section is drawn from @lay2005analysis.

```{theorem, label = "inequality-of-reals"}
Let $x,y\in\mathbb{R}$ such that $x\leq y+\epsilon$ for every $\epsilon>0$. Then $x\leq y$.
```

```{definition}
Let $\mathcal{S}$ be a subset of $\mathbb{R}$. If there exists a real number $m$ such that $m\geq s$ for all $s\in \mathcal{S}$, then $m$ is called an *upper bound* for $\mathcal{S}$, and we say that $\mathcal{S}$ is bounded above. If $m\leq s$ for all $s\in \mathcal{S}$, then $m$ is a *lower bound* for $\mathcal{S}$ and $\mathcal{S}$ is bounded below. The set $\mathcal{S}$ is said to be *bounded* if it is bounded above and bounded below.

If an upper bound $m$ for $\mathcal{S}$ is a member of $\mathcal{S}$, then $m$ is called the *maximum* (or largest element) of $\mathcal{S}$, and we write $m=\max \mathcal{S}$.

Similarly, if a lower bound of $\mathcal{S}$ is a member of $\mathcal{S}$, then it is called the *minimum* (or least element) of $\mathcal{S}$, denoted by $\min \mathcal{S}$.

A set may have upper or lower bounds, or it may have neither. If $m$ is an upper bound for $\mathcal{S}$, then any number greater than $m$ is also an upper bound. While a set may have many upper and lower bounds, if it has a maximum or a minimum, then those values are unique. Thus we speak of *an* upper bound and *the* maximum.
```

```{definition}
Let $\mathcal{S}$ be a nonempty subset of $\mathbb{R}$. If $\mathcal{S}$ is bounded above, then the least upper bound of $\mathcal{S}$ is called its *supremum* and is denoted by $\sup \mathcal{S}$. Thus $m=\sup \mathcal{S}$ iff

1. $m\geq s$, for all $s\in \mathcal{S}$, and
2. if $m'<m$, then there exists $s'\in \mathcal{S}$ such that $s'>m'$.

If $\mathcal{S}$ is bounded below, then the greatest lower bound of $\mathcal{S}$ is called its *infimum* and is denoted by $\inf \mathcal{S}$.
```

```{definition, name = "completeness axiom"}
Every nonempty subset $\mathcal{S}$ of $\mathbb{R}$ that is bounded above has a least upper bound. That is, $\sup \mathcal{S}$ exists and is a real number.
```

```{theorem}
Given nonempty subsets $\mathcal{A}$ and $\mathcal{B}$ of $\mathbb{R}$, let $\mathcal{C}$ denote the set
$$
\mathcal{C}=\left\{ x+y:x\in \mathcal{A}\text{ and }y\in \mathcal{B}\right\} .
$$
If $\mathcal{A}$ and $\mathcal{B}$ have suprema, then $\mathcal{C}$ has a supremum and $\sup \mathcal{C}=\sup \mathcal{A}+\sup \mathcal{B}$.
```

```{proof}
Let $\sup \mathcal{A}=a$ and $\sup \mathcal{B}=b$. If $z\in \mathcal{C}$, then $z=x+y$ for some $x\in \mathcal{A}$ and $y\in \mathcal{B}$. Thus $z=x+y\leq a+b$, so $a+b$ is an upper bound for $\mathcal{C}$. By the completeness axiom, $\mathcal{C}$ has at least an upper bound, say $\sup \mathcal{C}=c$. We must show that $c=a+b$. Since $c$ is the least upper bound for $\mathcal{C}$, we have $c\leq a+b$.

To see that $a+b\leq c$, choose any $\epsilon>0$. Since $a=\sup \mathcal{A}$, $a-\epsilon$ is not an upper bound for $\mathcal{A}$, and there must exist $x\in \mathcal{A}$ such that $a-\epsilon<x$. Similarly, since $b=\sup \mathcal{B}$, there exists $y\in \mathcal{B}$ such that $b-\epsilon<y$. Combining these inequalities, we have
$$
a+b-2\epsilon<x+y\leq c.
$$
That is, $a+b<c+2\epsilon$ for every $\epsilon>0$. Thus, by Theorem \@ref(thm:inequality-of-reals), $a+b\leq c$. Finally, since $c\leq a+b$ and $c\geq a+b$, we conclude that $c=a+b$.
```