Does there exist a bijective mapping $f: \Delta\to\mathbb{C} $?
Yet I have not found such example. Is it false (why?),please help me. $ \Delta $ is the unit disk in $\mathbb {C}$.
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2How are you defining $\Delta$ – MITjanitor Mar 19 '13 at 04:03
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$\Delta $ means unit circle in $\mathbb{C}$ – Andy Mar 19 '13 at 04:09
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Unit circle or unit disk? – Lubin Mar 19 '13 at 04:10
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1Do you require the map to be biholomorphic, or just bijective? – Potato Mar 19 '13 at 04:10
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just bijective mapping – Andy Mar 19 '13 at 04:15
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($\Delta$ is standard notation for the unit disc. http://planetmath.org/UnitDisk) – Andrés E. Caicedo Mar 19 '13 at 04:29
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A continuous $f$ cannot exist, since $\Delta$ is compact and $\mathbb{C}$ is not. Surely a biholomorphic cannot exist. – Teddy Mar 19 '13 at 04:39
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I would do this radius by radius. So in case we were talking about the open disk, all we’d need to do is get a nice map from $[0,1\rangle$ onto $[0,\infty\rangle$, like $x/(1-x)$. In case we were talking about the closed disk, you need a (discontinuous) map from $[0,1]$ onto $[0,1\rangle$, and follow it by the map you chose before. For this discontinuous map, one might send $1/n$ to $1/(n+1)$ for all $n\ge1$.
Lubin
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1) I'll assume $\Delta$ is the unit circle first, since that's what you said in a comment above. Remove $1$ from the unit circle, and $(0,0)$ from the complex plane. Then take the polar coordinates. You see your question amounts to finding a bijection between $(0,2\pi)$ and $(0,+\infty)\times [0,2\pi)$. I will take for granted that for every interval which is not a singleton, there exists a bijection with $(0,1)$ (see here if you want to construct such bijections). Now the problem is equivalent to finding a bijection between $(0,1)$ and $(0,1)^2$.
Recall that every real number $x$ has a unique proper decimal expansion $x=x_0,x_1x_2\ldots$ By proper, I mean that the sequence of decimals is not eventually equal to $9$. Now the function $$ (0.x_1x_2\ldots,0.y_1y_2\ldots)\longmapsto 0.x_1y_1x_2y_2\ldots $$ defines an injection of $(0,1)^2$ into $(0,1)$. It is easy to get an injection of $(0,1)$ into $(0,1)^2$. By Cantor-Schroeder-Bernstein, there exists a bijection between $(0,1)$ and $(0,1)^2$.
2) If $\Delta$ is the open unit disk now (which would make more sense notationwise), you can find a homeomorphism between $\Delta$ and $\mathbb{C}$. Note that you can't find a biholomorphism, due to Liouville.
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You might want to include that such a biholomorhpic map cannot exist, by Liouville's theorem. – Potato Mar 19 '13 at 04:43
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I do not think that $g$ is a bijection, since nothing is mapped to 0.0909090909.... Luckily we have Cantor-Bernstein theorem we only need injective maps in both directions. This is almost the same mistake as the one made famously by Cantor (and noticed by Dedekind), see, for example, Dauben, p.55-57. – Martin Sleziak Mar 19 '13 at 16:12
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