So it's easy to show that the rationals and the integers have the same size, using everyone's favorite spiral-around-the-grid.
Can the approach be extended to say that the set of complex numbers has the same cardinality as the reals?
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Yes. $$|\mathbb R|=2^{\aleph_0}; |\mathbb C|=|\mathbb{R\times R}|=|\mathbb R|^2.$$ We have if so: $$|\mathbb C|=|\mathbb R|^2 =(2^{\aleph_0})^2 = 2^{\aleph_0\cdot 2}=2^{\aleph_0}=|\mathbb R|$$ If one wishes to write down an explicit function, one can use a function of $\mathbb{N\times 2\to N}$, and combine it with a bijection between $2^\mathbb N$ and $\mathbb R$. |
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Of course. I will show it on numbers in $[0,1)$ and $[0,1)\times[0,1)$. Consider $z=x+iy$ with $x=0.x_1x_2x_3\ldots$ and $y=0.y_1y_2y_3\ldots$ their decimal expansions (the standard, greedy ones with no $9^\omega$ as a suffix). Then the number $f(z)=0.x_1y_1x_2y_2x_3y_3\ldots$ is real and this map is clearly injective on the above mentioned sets. Generalization to the whole $\mathbb C$ is straighforward. This gives $\#\mathbb C\leq\#\mathbb R$. the other way around is obvious. |
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One particularly nice class of bijections from $\Bbb R$ to $\Bbb C = \Bbb R^2$, which is in my opinion a little bit similar to the spiral around the grid, is given by the space-filling curves. |
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