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Is there a way to map from $\mathbb{R}^2$ to $\mathbb{R}^1$, such that every point in $\mathbb{R}^2$ has a unique point in $\mathbb{R}^1$ and you preserve the distance (isometry) relations of $\mathbb{R}^2$?

Injective map from $\mathbb{R}^2$ to $\mathbb{R}$ gives an example on how to map $\mathbb{R}^2$ to $\mathbb{R}^1$. I'm curious if there is a way to do this and while maintaining $\mathbb{R}^2$'s triangle inequality, which is for $x,y,z\in\Bbb R^2$ we have $|f(x)-f(z)|\leq|f(x)-f(y)|+|f(y)-f(z)|$.

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VenomFangs
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1 Answers1

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Such a map can't exist. Let $p$ be the image of the origin. Picture the circle $x^2 + y^2 = 1$. That whole circle would have to map to a set of points of distance $1$ from $p$. But the set of points of distance 1 from a fixed point in $\mathbb{R}$ is finite.

hunter
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  • You're assuming the map is supposed to be an isometry. In fact the map $f(x,y)=0$ satisfies the (current version of) the problem... – David C. Ullrich Apr 30 '16 at 15:31
  • @David, I apologize for the confusion. The link I provided I think implied the isometry, but I didn't explicitly state that in the question. I updated the question to help reflect that. – VenomFangs Apr 30 '16 at 15:44
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    @VenomFangs Aargh. Are you now saying you want an isometry? The updated version of the question does not say so. – David C. Ullrich Apr 30 '16 at 15:47
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    @VenomFangs The updated updated version does. Sort of. If you mean you're llooking for an isometry why not say so? "Preserve the diistance relations" is not clear. So now the answers showing that there is no such isometry answer the question. – David C. Ullrich Apr 30 '16 at 15:49
  • @DavidC.Ullrich, my formal math background is weaker than I wish it was. I have an engineering (computer science) degree and know I miss some of the fine points of math. I've tried to update the question again. If it is not clear, if you want to update it, I would appreciate to see how to do the proper wording. – VenomFangs Apr 30 '16 at 15:50
  • @VenomFangs And now the updated updated uupdated version says isometry. You know people have answered about ten different questions, all trying to answer the actual question you had in mind. – David C. Ullrich Apr 30 '16 at 15:50
  • @DavidC.Ullrich, I've tried to update the question to add clarification as I saw people had issues. I promise I'm learning, so my future questions will be of better quality. – VenomFangs Apr 30 '16 at 15:52
  • @DavidC.Ullrich at the time i posted the answer the OP asked for it to preserve the metric space structure... since then the question has changed – hunter Apr 30 '16 at 16:08