## Domains and Rules

When are two functions equal? A function is a kind of relation, which means it’s a kind of set. So two functions being equal means, fundamentally, that each is a subset of the other. That’s inconvenient, and it doesn’t really jibe with our understanding of functions as encoding *rules*.

**Theorem About When Functions Are Equal** (TAWFAE)**.** Let $f$ and $g$ be functions. Then $f=g$ if and only if:

- $\operatorname{Domain}(f)=\operatorname{Domain}(g)$, and
- $\forall x\in \operatorname{Domain}(f), f(x)=g(x)$.

**Proof. **This theorem sort of looks like it will be complicated to prove, but let’s keep our wits about us and rely on logic to help us out.

Globally, this is a biconditional proof so it has two directions:

($\Rightarrow$) Assume $f=g$. We need to prove two things:

(*clause 1*) This is the claim that two sets are equal, so we have to prove each is a subset of the other:

($\subseteq$) Let $t\in \operatorname{Domain}(f)$. Then $\exists y: (t,y)\in f$. Since $f=g$, this means $(t,y)\in g$. So $t\in\operatorname{Domain}(g)$.

($\supseteq$) Let $t\in \operatorname{Domain}(g)$. Then $\exists y: (t,y)\in g$. Since $f=g$, this means $(t,y)\in f$. So $t\in\operatorname{Domain}(f)$.

(*clause 2*) This is a universal claim, so we start the same way as always: Let $x\in \operatorname{Domain}(f)$. Then $\exists y: (x,y)\in f$. We can rewrite this as $y=f(x)$. On the other hand, we know from clause 1 that $\operatorname{Domain}(g)=\operatorname{Domain}(f)$, so $\exists z: (x,z)\in g$. We can rewrite this as $z=g(x)$. But since $g=f$, we actually have $(x,z)\in f$.

Since $(x,y)\in f$ and $(x,z)\in f$, and $f$ is a function, $y=z$.

So $f(x)=y=z=g(x)$.

($\Leftarrow$) Assume both clauses 1 and 2 hold. In this direction, we need to prove two sets are equal.

($\subseteq$) Let $Q\in f$. Then $Q=(x,y)$ for some $x$ and some $y$. $x\in\operatorname{Domain}(f)$, so by clause 1, $x\in\operatorname{Domain}(g)$. Then there’s some $z$ with $(x,z)\in g$. We can rewrite in function notation as $y=f(x)$ and $z=g(x)$. By clause 2, we know $f(x)=g(x)$, so $y=z$. That is, $Q=(x,y)\in g$.

($\supseteq$) This is so similar to $\subseteq$, I’ll let you work it out.

Since we proved both directions, we’re done. $\Box$.

This theorem gives us a new way to prove two functions are equal, should we ever need it: we establish clause 1 (“they have the same domain”) and clause 2 (“they have the same rule”).

### Restrictions and Extensions

In middle and high school algebra courses, we often ask questions like “What’s the domain of $f(x)=\frac{sin(x)}{x^2-25}$?”. According to the TAWFAE, this is actually a dumb question: TAWFAE says in order to know what a function is, we need to know what the domain is. So a question like doesn’t actually contain enough information to be answerable.

What the question is really asking, though is: *what’s the most natural*, or maybe *what’s the largest possible* domain on which this rule makes sense?

As a middle-school student, I was very confused by questions like this one:

Find the domain of $f(x)=\frac{x^2-9}{x+3}$.

My answer would usually go like this: that function is really the same as $x-3$, the domain of which is all of $\mathbb{R}$. And I would always get that question wrong. Why?

The functions $\psi:t\mapsto \frac{t^2-9}{t+3}$ and $\varphi:x\mapsto x-3$ do indeed have the same rule, in the sense that, whenever $\psi(t)$ makes sense, we have

$$ \psi(t)=\frac{t^2-9}{t+3}=\frac{(t-3)(t+3)}{t+3}=t-3=\varphi(t)$$

**But **the natural domain of $\psi$ is $\mathbb{R}\setminus\{-3\}$ and the natural domain for $\varphi$ is $\mathbb{R}$; so that means the domains are different — hence by the TAWFAE, we know $\psi\neq\varphi$. Nevertheless, the two functions *are* related.

**Definition.** If $f:X\rightarrow Y$ and $A\subseteq X$, we define the *restriction* of $f$ to $A$ as

\begin{align*}f|_{A}:A&\rightarrow Y\\ a&\mapsto f(a)\end{align*}

That is, $f|_A$ has the same rule as $f$, just we only care about inputs that come from $A$.

Equipped with this notion, we can describe the relationship between $\psi$ and $\varphi$:

$$\psi=\varphi|_{\mathbb{R}\setminus\{-3\}}$$

**Definition.** If $f:X\rightarrow Y$, and $X\subseteq Z$, and there is $g:Z\rightarrow Y$ with $g|_X=f$, we call $g$ an *extension *of $f$.

Here’s an illustration: if we let $f$ denote the function displayed in green, $g$ denote the function displayed in blue, and $h$ the function displayed in red, then we have that:

- $f|_{[-2,\infty)}=g$
- $f|_{[3,\infty)}=h$
- $g|_{[3,\infty)}=h$
- $f$ is an extension of $g$, from $[-2,\infty)$ to all of $\mathbb{R}$
- $g$ is an extension of $h$, from $[3,\infty)$ to $[2,\infty)$