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Can you give me an example of a bijective function $f:\mathbb{C}\rightarrow\mathbb{R}$? Can you parameterize a continuous plane with a continuous line?

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  • $\begingroup$ For the bijection, look here: mathoverflow.net/questions/126069/… $\endgroup$ – Lee Mosher Jul 11 '17 at 18:11
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    $\begingroup$ For the parameterization, do you mean a continuous surjection? or a continuous bijection? or what do you mean? $\endgroup$ – Lee Mosher Jul 11 '17 at 18:11
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Regarding the "continuous" part of the question, it's unclear what you want, but there isn't much good news.

There does not exist a continuous injection $\mathbb R^2\to\mathbb R$. Any such function would restrict to a continuous injection from the circle $S^1\to\mathbb R$, which is already impossible by the intermediate value theorem.

Also, there does not exist a continuous bijection $\mathbb R\to\mathbb R^2$. This is less obvious. See this question: Is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$

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  • $\begingroup$ I suppose I could be more clear: Say I have a Vectorspace $\mathbb{R^2}$ over the field $\mathbb{R}$ and I want to find a bijective function $g:\mathbb{R^2}\rightarrow\mathbb{R}$ where for all $x_0\in \mathbb{R^2}\quad \lim_{x\rightarrow x_0} f(x)=f(x_0)$. $\endgroup$ – ty. Jul 11 '17 at 18:36
  • $\begingroup$ Thanks for the link $\endgroup$ – ty. Jul 11 '17 at 18:42
  • $\begingroup$ @ty. Ah, so you want a continuous bijection $\mathbb R^2\to\mathbb R$. Unfortunately, that's impossible; see my second paragraph. $\endgroup$ – Chris Culter Jul 11 '17 at 18:50
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Since you tagged this as linear algebra, let's proceed as follows:

Let $\{\alpha_j\}_{j\in J}$ be a basis of the $\Bbb Q$-vector space $\Bbb R$, then $\{u\alpha_j\}_{(j,u)\in J\times\{1,i\}}$ is a basis of $\Bbb C$. Since $J$ is infinite, it has the same cardinality as $J\times\{1,i\}$. Any bijection $J\to J\times\{1,i\}$ then gives rise to a $\Bbb Q$-linear map $\Bbb R\to \Bbb C$.

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