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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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You haven't defined what either $(x, f(x))$ or $f(x) = y$ means. –  Qiaochu Yuan Nov 24 '12 at 7:14
    
How bout this: $\{(x,f(x)) \in A \times B |\forall x \in A, \, \exists ! y\in B \mathrm{\,such \,that \,}f(x) = y \}$ –  fhyve Nov 24 '12 at 7:19
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You are mixing things up. In the "set builder notation" $\{x\mid \phi(x)\}$, the property on the right is a property of the object on the left ($x$, in this case). In your example, the statement on the right does not mention anything about the object on the left. Yes, you use $x$ on both sides, but they mean different things: On the left it is a free variable, on the right it is quantified; to say $\forall x \dots f(x)=y$ and to say $\forall z \dots f(z)=y$ is the same thing. –  Andres Caicedo Nov 24 '12 at 7:29
    
Since the statement on the right, appearances notwithstanding, does not depend on $x$, whether it holds or not depends only on $f$. But you still have not defined what $f$ is! –  Andres Caicedo Nov 24 '12 at 7:29
    
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? $$\{(x,f(x))| x \in A \, f(x) \in B \mathrm{\,and \,} \forall x \in A, \, \exists ! y\in B \mathrm{\,such \,that \,}f(x) = y \}$$ –  fhyve Nov 24 '12 at 7:48
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2 Answers

up vote 3 down vote accepted

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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This is something that I would put on a handout. But for some reason the prof wanted the set S in set builder notation and I want a more concise way to write it on the board. Also, it would be useful to know how exactly to write this stuff out using set builder and quantifiers and stuff. –  fhyve Nov 24 '12 at 8:47
    
@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. –  Henning Makholm Nov 24 '12 at 13:02
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