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Not sure if this is possible, but I want to make a continuous function that serves as a bijection between any finite subspace of $\mathbb R^3$, and a predefined subspace of $\mathbb R^2$. I'm looking for a practical/realizable (in the sense of software implementation) method to fit a continuous function approximator (like a neural network) as opposed to some really crazy mathematical trick that is super ugly and hardly possible to implement in software. I found some good comments and answers in this following question that make it seem like in the more general case ($\mathbb R^3$ and $\mathbb R^2$) is very ugly and probably not going to have a very implementable solution in real-time in software, but was hoping that a finite subspace of $\mathbb R^3$ might be constrained enough that there is a nice solution Does same cardinality imply a bijection?

user3496060
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    When you say "finite subspace of $\Bbb{R}^3$", what precisely do you mean? – Theo Bendit Mar 29 '19 at 21:36
  • The question is vague. Terms need to be defined more precisely. The first sentence, for example, makes my head explode in a bad way. – zhw. Mar 29 '19 at 22:07
  • @TheoBendit well could be a hypercube, could be a complex surface. I don't know how the feasibility of a solution changes by the type of finite subspace so I'm trying to be general. I know how to do this pragmatically from an $\Bbb{R}^N$ to $\Bbb{R}^N$ space, but in this case i need to "flatten" and condense finite information in $\Bbb{R}^3$ into a bounded pre-defined geometry in $\Bbb{R}^2$. I'm not a mathematician, so my ability to type questions like one is going to reflect that. Help is appreciated. – user3496060 Mar 30 '19 at 04:57
  • @zhw. Maybe i should be asking this on the paper mache stack exchange :-) – user3496060 Mar 30 '19 at 05:14
  • I still don't really know what a "finite subspace" is. By what I'm hearing, it sounds like a bounded set, I'm not really sure. I can say that the solution changes *heavily* depending on the underlying set. If you're mapping from, say, an open two-dimensional disc, then it's easy. If you're mapping from the Sierpinski cube, it's hard (probably; I'm not going to try it). – Theo Bendit Mar 31 '19 at 01:26

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