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I am giving a two part presentation to a class on cardinality. I have done the first part but the prof wasn't satisfied with my definition of a function. For this presentation I have limited time so I want to go through definitions as quickly as possible. For this reason I don't want to talk about relations.

Does this work as a definition of a function:

$\{(x,f(x))|\forall x \in A, \, \exists ! y\in B \mathrm{\,such \,that \,}f(x) = y \}$

where A is the domain and B is the codomain? Specifically, does this ensure that everything in A gets mapped to something? Do I have to say that x is in A and f(x) is in B and for all A there exists a unique etc etc?

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    f(x) is supposed to just be suggestive notation for an element of B, in particular, the unique element of B that$x$maps to. How do I say this? Like this? $$\{(x,f(x))|$x$\in A \, f(x) \in B \mathrm{\,and \,} \forall$x$\in A, \, \exists ! y\in B \mathrm{\,such \,that \,}f(x) = y \}$$2012-11-24

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As Andres wrote in the comments, you are using $f$ to define what a function is, but the notation $f(x)$ already implies (at least intuitively) that $f$ is a function.

You should write that $f\colon A\to B$ is a function from $A$ to $B$ if:

  1. $f\subseteq A\times B$.
  2. For every $a\in A$ there exists a unique $b$ such that $(a,b)\in f$.

Alternatively, you could define a function independently of $A$ and $B$ by saying that $f$ is a function if:

  1. Every element of $f$ is an ordered pair.
  2. If $(a,b)\in f$ and $(a,c)\in f$ then $b=c$.

Then you can add that $f\colon A\to B$ is a shorthand for the following:

  1. $f$ is a function.
  2. For every $a\in A$ there is some $b\in B$ such that $(a,b)\in f$.
  3. For every pair $(a,b)\in f$ we have that $b\in B$.
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I don't see anything wrong with the standard definition of a function: a subset $S$ of $A\times B$ with the property that for each $x\in A$, there exists a unique $y\in B$ such that $(x,y)\in S$. And given a function $S$, we define the notation $f(x)$ to mean exactly that unique such $y$.

This will probably be more digestible to the audience (in a short amount of time) anyway than trying to write it in symbols. Oversymboling mathematics rarely makes it more understandable to those who don't already understand it.

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    @fhyve: It doen't seem to make sense to demand a definition of what "$S$ is a function" means in set builder function. Set builders are for defining _one particular_ set, but for a definition you want something that you can apply to _any_ set $S$ you want, and it will then tell you whether that $S$ is a function or not. Set builders do not appear to be helpful for that.2012-11-24