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Say I have a 2 axis system $(x,y)\in \mathbb R^2$ and I want to transform it to a 1 axis system $(w)\in \mathbb R$ so order will be kept and there will be a bijection between them.

What would be the right way to do this?

$w=5x+7y$ won't work, neither will $w=17xy$ since those are real numbers.

Is it even possible to have a bijection between them, i.e to be able to go back to $(x,y)$ from $(w)$?

PS: order preservation is more important.

Furthermore, how can I use that transformation on distances? Say I want to know what is distance 1 from $(x,y)$ in $w$, so $5\sqrt{(a-x)^2}+7\sqrt{(b-y)^2} = 5u+7v$ and I know that $u+v=1$ so how can I find what is $5u+7v$?

Note: let's assume $\mathbb R^2$ is arbitrarily ordered by x, then by, i.e x tie breaker is y ordering.

shinzou
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    You can have a bijection, but not order preserving. – N74 Apr 24 '17 at 09:18
  • How is it a bijection? $0.55+70 == 50+7(2.5/7)$ and why is not order preserving? @N74 – shinzou Apr 24 '17 at 09:29
  • Never said the functions you listed were the bijection you are looking for. And the problem with ordering is that no order can be defined on $R^2$: i.e. is $(4,5)$ bigger or smaller than $(5,4)$? – N74 Apr 24 '17 at 09:56
  • Let's say I arbitrarily ordered by x then y. What is the bijection I was looking for then? @N74 – shinzou Apr 24 '17 at 09:59
  • Read the answer to this post: https://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr?lq=1 , if you could have used the single digit interleaving you had probably found the bijection that preserves your ordering... too bad it is not injective and the bijective ones cannot respect your ordering. – N74 Apr 24 '17 at 10:29
  • Seems like getting 0 included in the [0,1] interval is a little difficult to implement in code but other than that it seems it could work. Are the simple transformation in my questions order preserving though? @N74 – shinzou Apr 24 '17 at 10:48
  • I see now that they don't mention order preservation, which is not good for me @N74 – shinzou Apr 24 '17 at 11:05
  • Why is this tagged (well-orders)? I do not really see anything in this question actually related to well-orderings. – Martin Sleziak May 02 '17 at 14:46

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