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
> 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.
http://planetmath.org/node/30360 (def. of function)
Let us all be paranoid. More so than no such agence, Bolon Yokte K'uh