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Is it possible to find a function which maps the infinite 2D plane, ie, every single point on the infinite 2D plane to a finite 2D shape(eg-circle) while maintaining: 1. Continuity 2. One to one correspondence

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    The reason for asking this question is that i feel i know a good "proof" of this. – 2017-02-21

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Yes, very much so! An example of this mapping might be to write each (non-origin) point in $\mathbb{R}^2$ in polar coordinates $(r; \theta)$, and map that to $$\left(\frac{1}{1+e^{-r}}; \theta\right)$$ for instance, because if $f(t) = \frac{1}{1+e^{-r}}$ then $f$ takes $(0, \infty)$ to $(0, 1)$ continuously. (Also map the origin to itself). This map takes $\mathbb{R}^2$ to the open unit disc $D$. Now, a continuous map from $\mathbb{R}^2$ to a finite subset $U$ of $\mathbb{R}^2$ can only exist if $U$ is open. In fact the Riemann mapping theorem says that if you have a non-empty, simply connected, open subset $U$ of $\mathbb{R}^2$, then there is a continuous (in fact holomorphic if you view $\mathbb{R}^2$ as $\mathbb{C}$), bijective mapping between $U$ and $D$, and you can compose this with the map above to get a continuous bijection from $U$ to $\mathbb{R}^2$.

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    Firstly thank you b mehta for the wonderful answer. But doesnt this feel non intuitive if we consider the stereographic projection of the 2d plane and then consider the borsuk ulam theorem – 2017-02-21
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    It's important to remember that the stereographic projection has to miss out a point - typically the "north pole" of the sphere isn't mapped to anything on the plane so it doesn't give a function from $S^2$ to $\mathbb{R}^2$. – 2017-02-21
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    I agree. But you yourself said that "that" point maps to itself. – 2017-02-21
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    I said that the origin of $\mathbb{R}^2$ maps to itself under the map I specified above, but the north pole of $S^2$ isn't mapped to anything in stereographic projection. – 2017-02-21
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    One more thing, how can the seemingly "infinite" plane be mapped to a "finite" one. – 2017-02-21