# How is this a function? - Analysis.

Let $$X = \{1, 2, 3\}, Y = \{4, 5, 6\}$$. Define $$F \subseteq X \times Y$$ as $$F = \{(1, 4),(2, 5),(3, 5)\}$$. Then $$F$$ is a function.

I simply do not see how this could be a function, as there is nothing that it is mapping to, if anyone can explain how this is a function, that would be lovely.

Yes, under $$F$$ the mapping is as follows: $$1 \mapsto 4, 2 \mapsto 5, 3 \mapsto 5$$.

• Ohh, I understand. Thank you. I did not see that, how silly. – Sam May 13 at 19:36

This is the set-theoretic definition of "function".

Once you have defined ordered pairs, so that $$(x,y)=(a,b)$$ if and only if $$x=a$$ and $$y=b$$, given sets $$A$$ and $$B$$ we define the set $$A\times B$$ to be $$A\times B = \{(x,y)\mid x\in A,y\in B\}.$$

Given sets $$A$$ and $$B$$, a "function from $$A$$ to $$B$$" is defined to be a subset $$f\subseteq A\times B$$ with the following properties:

1. For each $$a\in A$$ there exists $$b\in B$$ such that $$(a,b)\in f$$;
2. For each $$a\in A$$, if $$b,b'\in B$$ are such that $$(a,b)\in f$$ and $$(a,b')\in f$$, then $$b=b'$$.

When this happens, we write $$f\colon A\to B$$, and we say that the value of $$f$$ at $$a$$ is $$b$$ if and only if $$(a,b)\in f$$; so we write $$f(a)=b\iff (a,b)\in f.$$

The set you describe is a subset of $$X\times Y$$ that satisfies conditions 1 and 2, hence it is a function from $$X$$ to $$Y$$.

• A descriptive answer. (+1) – VIVID May 13 at 20:28
• Maybe worth adding that in set theory, a function and its graph are the same thing (contrary to precalculus or calculus courses). – Taladris May 14 at 5:49
• And thinking of a function like this makes later (when defining functions on equivalence classes/quotient spaces) the phrase '... is well defined..' clear. One defines the subset f first and then shows that condition 2 holds. Otherwise it is really a mind twister what is actually shown in such situation. – lalala May 14 at 6:36

Basically, $$F = \{(a,b),...\}$$ stands for $$F(a) = b$$.

Recall that a function is comprised of three parts:

1. Domain
2. Codomain
3. The way to uniquely associate every element in its domain to an element in its codomain

So it's perfectly OK to denote a function $$f$$ as the set of all pairs $$(x, f(x))$$, since it encodes all the necessary information to represent a function.

• The graph of a function does not determine the codomain. Indeed, from this definition of $f$ we can't say for sure if $Y = \left\{4,5,6\right\}$ or $Y = \left\{4,5,6,\text{Paris Texas}\right\}$. That being said, in some contexts the codomain is not considered to be part of the data which defines a function. – user1892304 May 17 at 10:25
• In a positive way, a set $$S$$ is a function if and only if : $$S$$ is a binary relation ( that is, a set such that all its elements are ordered pairs) AND for all objects $$x$$, $$y$$ and $$z$$ the pairs $$(x,y)$$ and $$(x,z)$$ belong to $$S$$ only if $$y=z$$ ( meaning, only if these " two" pairs are , in fact, identical). In short, quoting Enderton : " a function is a single valued relation" .

• This test / definition can be turned negatively using DeMorgan's law. Let $$S$$ be a set ; there are only $$2$$ reasons you can give in order to assert that $$S$$ is not a function ( and each reason is sufficient by itself) :

(1) $$S$$ is not a set of ordered pairs , that is $$S$$ is not a binary relation

OR

(2) there are at least, in $$S$$ , two different ordered pairs with the same first element (meaning : two pairs with the same first element, but with non-identical second elements).

• Applying this to the case you present :

(1) is there an element of $$F$$ that is not an ordered pair?

(2) are there in $$F$$ two pairs with the same first element, but with different second elements?

• By reason (1) the set $$\{(a, 1), c, (b,2)\}$$ is not a function.

• By reason (2) the set $$\{(a, 1), (b,3), (b,2)\}$$ is not a function.