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Here is the interval (0,1). Step 1 : Divide the interval by half. Then we get cut 1/2. Step 2 : Divede the intervals of Step 1 by half. Then we get cuts 1/4, 2/4, 3/4. ................... In this way, we get infinite cuts contained in (0,1). And we can correspond each cut to one of binary number. Like 1/2 -> 0.1 1/4 -> 0.01 3/4 -> 0.11 ...... In this way, the binary numbers in the interval (0,1) are in 1:1 correspondence with the cuts constructed in above manner. And we know the real numbers in (0,1) are in 1:1 correspondence with the binary numbers in the same interval. Thus, the cuts are in 1:1 correspondance with the real numbers. And the number of cuts is countable, because it's the countable sum of countable numbers. Thus, the set of real numbers in the interval (0,1) is countable set. Is it right?

PHilo
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  • I don't think that it is countable! – user Dec 28 '17 at 14:30
  • Wrong. See Cantor’s iconic result. The set of real numbers in the interval $(0,1)$ is not countable. See Theorem 3.1. –  Dec 28 '17 at 14:30
  • Can you state the first number you could count? – For the love of maths Dec 28 '17 at 14:31
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    Is it right? NO – Batman Dec 28 '17 at 14:31
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    And to what cut do you map $1/3$? In binary: $0.010101\ldots$ – ajotatxe Dec 28 '17 at 14:32
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    This method only produces real numbers with terminating binary expansions. Most real numbers have non terminating expansions. – kimchi lover Dec 28 '17 at 14:34
  • We can map 1/3 to some binary number with infinite sequence. – PHilo Dec 28 '17 at 14:34
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    @PHilo: at each step, you only add numbers which, when written as a fraction in lowest terms, can be written with a power of 2 in the denominator. The number $1/3$ cannot be written that way, so it is never added as as cut. The problem with your proof is that you only get numbers with a finite binary expansion. – Carl Mummert Dec 28 '17 at 14:36
  • See Rohan's comment – For the love of maths Dec 28 '17 at 14:36
  • Why does it generate only terminating binary expansions? Steps are infinite, so infinite expension is possible. – PHilo Dec 28 '17 at 14:36
  • @PHilo: there are an infinite number of steps, but at each step you only add particular numbers as cuts, and the numbers you add always have terminating binary expansions. If you argue that $1/3$ is added, for example, you should be able to say the number of the step at which it is added... – Carl Mummert Dec 28 '17 at 14:37
  • @Mummert: Even though I can not say the number of step, is it not possible there 'exist' such infinite steps? – PHilo Dec 28 '17 at 14:42
  • @PHilo: every number that is added to your countable set is added at a particular step. (That is, a number ends up in the overall union of sets if and only if it is one of the sets that you are taking the union of). Because the number $1/3$ is never added at any step, it is not in the overall countable set of numbers that you generate. That is where the argument goes wrong - it assumes that more things are in the union than are actually there. In the end, the only numbers actually in the countable set are the ones that have finite binary expansions. – Carl Mummert Dec 28 '17 at 14:45
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    Once you allow "infinitely many steps" you are no longer concerned with finite binary strings, but with infinite strings. And there are uncountably many infinite binary strings. – Asaf Karagila Dec 28 '17 at 14:45

2 Answers2

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The binary numbers are not in one-to-one correspondence with the cuts you describe; you only obtain the binary numbers which have a finite binary expansion (which, you correctly prove is a countable set). However, binary numbers with an infinite expansion, for example $$\frac{1}{3} = 0.0101010101\ldots$$ or, importantly, any irrational number, will be missed by this enumeration.

preferred_anon
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The problem is that you can not build a one to one correspondence from binary numbers with infinite expressions to that set of cuts.

ajotatxe
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