Timeline for Is there an associative metric on the non-negative reals?
Current License: CC BY-SA 2.5
7 events
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| Feb 24, 2010 at 20:07 | comment | added | Joe Fitzsimons | The point I was trying to make was that given such a function h, any function of the form $g(h(g^{-1}(x),g^{-1}(y)))$ preserves definiteness, symmetry and associativity, but not necessarily the triangle inequality. So an approach is to pick some h of this form, and then look for a g which will allow you to satisfy the triangle inequality. Handily, g lets us change between, say, multiplication and addition, simply by picking g(x)=exp(x). | |
| Feb 24, 2010 at 19:57 | comment | added | aorq | You can find a definite, symmetric, and associative function h by just picking an arbitrary bijection between the non-negative reals and a countable product of (Z mod 2)s. The latter has a natural structure of a group where every element is an involution, so the non-negative reals do as well. The triangle inequality is the hard part here. | |
| Feb 24, 2010 at 19:32 | history | edited | Joe Fitzsimons | CC BY-SA 2.5 |
Generalized approach, and fixed an error.
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| Feb 24, 2010 at 13:21 | comment | added | Joel David Hamkins | By the way, welcome to MO! It is possible to edit your answer, by clicking on 'edit', if you wanted to clarify your idea along the lines of your comment. | |
| Feb 24, 2010 at 5:44 | comment | added | Joe Fitzsimons | Good point. Requiring g to be invertible is then too strong a condition. Actually, I hadn't really intended what I wrote. It isn't necessary that g be invertible, but only that for and given x>0, the g(x)=g(y) only for y=x or y=-x. Then g^{-1}(g(x)) is taken to be abs(x), but then its not clear the last bit holds. | |
| Feb 24, 2010 at 5:21 | comment | added | Joel David Hamkins | If g is even, how can it be invertible? | |
| Feb 24, 2010 at 5:01 | history | answered | Joe Fitzsimons | CC BY-SA 2.5 |