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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.

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

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    $\begingroup$ Ohh, I understand. Thank you. I did not see that, how silly. $\endgroup$ – Sam May 13 at 19:36
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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 $Y$.

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  • $\begingroup$ A descriptive answer. (+1) $\endgroup$ – VIVID May 13 at 20:28
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    $\begingroup$ Maybe worth adding that in set theory, a function and its graph are the same thing (contrary to precalculus or calculus courses). $\endgroup$ – Taladris May 14 at 5:49
  • $\begingroup$ 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. $\endgroup$ – lalala May 14 at 6:36
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Basically, $F = \{(a,b),...\}$ stands for $F(a) = b$.

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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.

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    $\begingroup$ 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. $\endgroup$ – user1892304 May 17 at 10:25
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  • 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.

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