# Naive Set Theory: Functions

## Restrictions

It is sometimes handy to "shrink" down a function to only pertain to a subset of its domain. Such a shrunken function is called a restriction.

Formally, the restriction of a function $f : A \rightarrow B$ to a subset $A' \subseteq A$ is a new function $f|A' : A' \rightarrow B$ whose rule is $\{(a,f(a)) : a \in A_0\}$.

## Problems

1. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function whose rule is $f(x) = \cos(x)$. Graph the following:

1. $f|\mathbb{R}_{[-\pi,\pi]}$
2. $f|\mathbb{R}_{[-\frac{3\pi}{4},-\frac{\pi}{4}]}$
3. $f|\mathbb{R}_{[-\frac{pi}{2},\frac{3\pi}{4}]}$
2. Let $f : A \rightarrow B$ be a function, and let $A'' \subseteq A' \subseteq A$. Show that $f|A'' = (f|A')|A''$.

The restriction $(f|A')|A''$ is a function whose codomain is $B$ and whose rule is $\{(a, f|A'(a)) : a \in A''\}$. In turn, the restriction $f|A'$ is a function whose codomain is $B$ and whose rule is $\{(a, f(a)) : a \in A'\}$. It follows that for all $a \in A''$, $f|A'(a) = f(a)$. Therefore we can rewrite the rule for $(f|A')|A''$ as $\{(a, f(a)) : a \in A''\}$. But this is exactly the rule for $f|A''$. Because $(f|A')|A''$ and $f|A''$ have the same codomain and rule, the two restrictions are equal.