Does every function have a graph?

Question

I'm having trouble understanding this. My friend said it doesn't, but I disagree, though I'm not sure.

Given a function $f:A \to B$, the graph of $f$ is defined by $G(f) = \{(x,y)| x \in A , y = f(x)\}$. Then, is it true that if a function exists, its graph exist (Even though there may be no geometric interpretation)? I think this is true, since in the "worst" case it would be the empty set, which exists...

Of course it exists, by definition, just the way you described it. It can not be empty unless $A$ itself is empty. (At least, I assume that when one uses the notation $f:A\rightarrow B$, then one is implicitely stating that the domain of $f$ is $A$) — Suzet, Jun 13 '18 at 12:28
There are cases, however, when a function cannot be sketched on a graph, like the Dirichlet function. — StubbornAtom, Jun 23 '18 at 07:59

score 2 · Answer 1 · answered Jun 13 '18 at 12:37

Actually, any relation from a set $A$ a set $B$ has a graph. Such a relation can be defined as a triple $(G,A,B)$ such that $G\subset A\times B$, and $G$ is the graph of the relation.

A function from $A$ to $B$ is just a relation $(G,A,B)$ such that, if $(x,y)$ and $(x,y')\in G$, then $\;y=y'$ (such a graph is called a functional graph).

A map from $A$ to $B$ is a function such that for each $x\in A$, there exists a (necessarily unique) $y$ such that $(x,y)\in G$.

score 1 · Accepted Answer · answered Jun 13 '18 at 12:30

1

You are right, it exists.

If you are seeking to plot a function between 2 sets, you can plot $A$ on one axis, and $B$ on the other one and pick off the points as you mentioned.

In addition, if $A$ and $B$ are finite, you can draw $A$ and $B$ as points in space, and connect them via edges. You get $V = A \cup B$ with edges $E = \{(a,b) | f(a)=b\}$, which is also a representation with a graph, albeit quite a different one from the one you are describing.

answered Jun 13 '18 at 12:30

gt6989b

54,422

How would you graph the function mapping every car in Utah to a licence plate number? Would the domain be a parking lot? – William Elliot Jun 14 '18 at 02:05
@WilliamElliot you make a representation mapping the VIN number to the index on the absciss and list the license plates as strings on the ordinate – gt6989b Jun 14 '18 at 15:27
So you use another function from cars to numbers that requires a parking lot domain so the first function would not need a parking lot. – William Elliot Jun 15 '18 at 00:44
@WilliamElliot for example. If you want to nitpick, the graph itself is also just a representation of the curve in $\mathbb{R}^n$ and hence can be looked at as a function as well... – gt6989b Jun 15 '18 at 13:19
If I cannot even plot a particular function, does that also mean that it's graph 'exists' in theory? – StubbornAtom Jun 23 '18 at 08:02

Does every function have a graph?

2 Answers2