# how to prove that any two circles and any two disks have the same cardinality

1.I am trying to prove that two circles has the same cardinality I declared two intervals $[0,2\pi R]$ and $[0,2\pi \widetilde{R}]$ I build I bijection between the two interval It's the correct proof ? $2$.The equation for a closed disk $(x-a)^{2}+(y-b)^{2}\leqslant R^{2}$. Can I prove it by doing in the same way that I prove the circles cardinality but now with the area $\pi R^{2}$ instead of using the circumference. Thanks

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Notice that any continuous curve has the cardinality of $\mathbb{R}$ being an injective image of $[0, 1]$ and also every set containing an open set can be shown to contain a continuous curve and thus has the same cardinality. – Karolis Juodelė Aug 20 '12 at 11:15
$(1)$. Need to be careful, you want to prove a geometric fact. Note want to use half-open intervals, like $[0,2\pi R)$. One cannot assess what you did without being given some detail. $(2)$. Again, need explicit geometric bijection. – André Nicolas Aug 20 '12 at 11:32
Hello Nicolas:Why did you write half open interval $[0,2\pi R)$ if the point $2\pi R$ is a memeber it is the last point of the circle. – Hernan Aug 20 '12 at 12:18
@Hernan: Circles don't have endpoints. – Hurkyl Oct 29 '12 at 8:13

For discs, the map $$f(r,\theta )=(\frac{r R_2}{R_1},\theta)$$ will give a bijection from the disc of radius $R_1$ to the disc of radius $R_2.$
Yeah, your proof is also correct, but I found it more easy to visualize. Here $(r,\theta )$ is the polar coordinates of a point in the plane – pritam Aug 20 '12 at 10:34