Is there any way to reorder R so that 3 < 2? And a similar question, which probably can be answered in the same breath: Is d(2, 3) < d(2, 100) for all metrics? Is there a nice theorem that talks about this?
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2Can't reorder in a way that "plays nice" with the algebra. – André Nicolas Oct 19 '13 at 01:09
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You can wellorder it, but you won't be able to say exactly what such an order looks like. – Francis Adams Oct 19 '13 at 03:51
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Thanks everybody for the helpful remarks, it is much clearer now. – Julian Cienfuegos Oct 20 '13 at 03:32
3 Answers
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Yes: define $x\prec y$ if and only if $x>y$ in the usual ordering.
No, not even for metrics generating the usual topology: define $d(x,y)=\min\{|x-y|,1\}$.
Brian M. Scott
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1Brian problem is that does not order f as a field. You end up having -1>0 but then (-1)^2>0 so 1>0 and hence 0>(0-1) 0>-1 this is a contradiction. If you just want to order it as a set there are of course infinitely many ordering I mean we can use your thing but rather then inverting all of R all we need to do is invert interval [2,3]. – Oct 19 '13 at 03:13
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3@Anthony: The OP did not impose any algebraic requirements on the order. – Brian M. Scott Oct 19 '13 at 03:56
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Given a ring $R$ let $f(a,b,c)$ be a value in the set $\{-1, 1\}$. Then define the ordering so that $f(a,b,c) c\cdot a \lt f(a,b,c) c\cdot b$ whenever $a \lt b$. Do a similar thing for addition.
Daniel Donnelly
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Put $3$ at the front, $2$ at the back, all the rest in between, ordered among themselves in the usual way (or any way you like).
An easy way to get a metric on $\mathbb R$ satisfying $d(2,3)\gt d(2,100)$ is to identify $\mathbb R$ with a suitably bent curve in the plane and use the planar distance. E.g., define $$z(t)=e^{(\pi i-1)t}$$ and then define $$d(s,t)=|z(s)-z(t)|=|e^{(\pi i-1)s}-e^{(\pi i-1)t}|;$$ then $$d(2,3)=e^{-2}+e^{-3}\gt e^{-2}$$ while $$d(2,100)=e^{-2}-e^{-100}\lt e^{-2}\lt d(2,3).$$
bof
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