Date: Oct 6, 2013 5:51 PM
Author: David Bernier
Subject: Re: Is (t^2-9)/(t-3) defined at t=3?
On 10/06/2013 04:42 PM, Peter Percival wrote:

> Arturo Magidin wrote:

>

>>

>> Technically, two functions are equal if and only if they have the

>> same domain, and they take the same value at each point on the

>> domain.

>

> I wonder why you don't require the codomains to be the same as well.

> Two functions might be equal according to that definition but one could

> be invertible and the other not.

>

In the way I think of a function 'f', no co-domain

attaches to 'f' in a natural way. For example,

for the function f with domain R (= the set of real numbers)

and such that, for any x in R,

f(x) = sin(x),

the domain of f is R, but I would write:

"f with co-domain R", "f with co-domain [-1, 1]",

or by abuse of language,

"f(x) with co-domain R", "f(x) with co-domain [-1,1]".

Nevertheless, the default, primary, definition of "function"

at PlanetMath provides for a co-domain being attached to

a function. They say that "in set theory",

a function is a special type of relation, so along the

line of thought in Arturo Magidin's way of thinking

or my way of thinking.

The "no co-domain included" definition is convenient,

because for example we can say

f: C -> C is the polynomial function defined

by f(z):= z^137 + 202 z^23 - 45z^4 + 32 z^3 - 1001 z^2 + 7z ,

all this without bothering to specify a co-domain.

And in any case, if we say "f maps C onto C", it's equivalent

to saying f(C) = C.

cf.:

http://planetmath.org/node/30360 (def. of function)

David Bernier

--

Let us all be paranoid. More so than no such agence, Bolon Yokte K'uh

willing.