```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-domainattaches 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 toa function.  They say that "in set theory",a function is a special type of relation, so along theline of thought in Arturo Magidin's way of thinkingor my way of thinking.The "no co-domain included" definition is convenient,because for example we can sayf: C -> C is the polynomial function definedby 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 equivalentto 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'uhwilling.
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