Re: Solving for an operator

On Tue, 15 Apr 2008, m7ossny wrote:

We have a metric space ((X, +), q) defined with a distance function
'q:X^2->Real' on an algebraic structure (mainly abelian group) defined
by a set 'X' and an operator '+:X^2->X'. The algebraic structure
features an identity element '0', and an inverse element '-x' for every
x \in X.

The question now is if for every x \in X

x+'0'=x implies q(x+'0', x)=0
x+(-x)=0 implies q(x+(-x), '0')=0

I don't look closely at equationswithout spaces. It causes too much
eye strain. Easier to read is = with a space before and after. Then it's
easier to read and I can give better thought to the problem. So without
going into the details, which is what happens when I don't look closely at
equationswithout spaces, I think you want to understand that when
d:X^2 -> R

is a metric, then for all x in X, d(x,x) = 0.

Are you acquainted with topological groups, or
in your case metric space topological groups?

Does that mean we can solve these three equations for '+' (basic
operator), 'q' (distance function), and '-' (inverse operator)!?

I don't know what you mean by solve in this context.

Of course now we have to set one of them arbitrary. Is this system
solvable!?

I think you're asking too much.

Thanks for help!!!

.

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