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I always thought that the points on a circle would be uncountable, like the real numbers between 0 and 1. But I read or saw something that made me wonder. It said you could form a one to one pairing between the integers and the points on a circle of radius one. Start at any point and pair that to the integer 1. Move along the circumference a distance of one unit and pair that point with the integer 2. Keep doing this. Since the length of the circumference is irrational, it was claimed, you will not pick the same point ever again and you have a one to one pairing. Can this be right? It used to be thought that rational numbers were uncountable until someone came up with the clever idea to zip back and forth through a grid of all the fractions. Is this the clever idea that shows that the points on a circle are countable or is there something wrong with the claim? And if it turns out to be correct, aren't then the real numbers between 0 and 1 countable by wrapping that line segment into a circle and moving along that circle by a distance of some irrational number?

bazbsg
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    No; you would end up with a countable dense subset of the unit circle. You can also find a countable dense subset of $[0,1]$ --- for example, $\Bbb Q\cap [0,1]$. In other words, you get arbitrarily close to every point on the target set, but you don't actually obtain the target set. It's weird, I know. – pancini May 21 '17 at 08:39
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    For example, you will never reach the point opposite from the starting point – Hagen von Eitzen May 21 '17 at 08:40
  • Have you ever looked up Stereographic Projection? If not, I highly suggest you do because it maps all the points on the circle to the real line. Which is uncountable – Malcolm May 21 '17 at 09:23
  • Also, the fact that the rationals are countable with the argument you mentioned is not mathematically rigorous per say. The fact that it is follows from the argument that any rational can be thought as a pair of elements $(p,q) $ which is an element of $\mathbb {N} \times \mathbb {N} $ (more or less) where $ \mathbb {N}$ are the natural numbers. You can prove that if you take the cartesian product of countable sets, then the result is also countable. – Malcolm May 21 '17 at 09:32
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    Thanks everyone for your insights. After reading that I will never reach the point opposite the starting point and thinking about it for a while, I realized that the procedure I described does not prove that the points on a circle are countable, because some (one or more) points on the circle do not map to an integer. And even if I did eventually reach that point opposite the starting point, it means I would after twice as many moves come back to the starting point and that's well before reaching the end of the integers which also means I couldn't have hit every point. – bazbsg May 22 '17 at 11:21
  • Now I hope I can remember where I saw or read that so I can go back and post a comment that it's not correct. – bazbsg May 22 '17 at 11:22

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