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Initially on [0,1], all elements should have a one-one relation with ℝ. I thought that the deleted open middle-third intervals of the Cantor Set when deleted lose out on their images in ℝ so essentially there will now be some images on ℝ which have no pre-images on the Cantor Set and thus there won't be a one-one relation with ℝ. But I am not very sure of this and even if I'm right a better explanation by someone else would really help a lot

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    $\begingroup$ What did you try? $\endgroup$ – Parcly Taxel Aug 27 '17 at 14:39
  • $\begingroup$ Initially on [0,1], all elements should have a one-one relation with ℝ. I thought that the deleted open middle-third intervals of the Cantor Set when deleted lose out on their images in ℝ so essentially there will now be some images on ℝ which have no pre-images on the Cantor Set and thus there won't be a one-one relation with ℝ. But I am not very sure of this and even if I'm right a better explanation by someone else would really help a lot $\endgroup$ – Ritojeet Basu Aug 27 '17 at 14:54
  • $\begingroup$ Welcome to Math.SE! I think you misunderstood a key aspect of this site. This is more about helping you find the right answer then giving you answers to arbitrary problems. For example, if you were to give us what you've tried, where you've looked, what you've read to answer the problem, and describe why you aren't satisfied/doubt what you've come across, I'm sure others would help. Also, choosing a descriptive title that described the content would help. For more, I recommend reading How to ask a good question? $\endgroup$ – davidlowryduda Aug 27 '17 at 14:55
  • $\begingroup$ You have made the correct observation that the inclusion map from the Cantor set to $\mathbb{R}$ is not a bijection. Also, the inclusion map from the Cantor set to $[0,1]$ is not a bijection. However, there are many, many, many, many other functions from the Cantor set to $\mathbb{R}$. Just because one of them is not a bijection, does not mean there are no bijections. $\endgroup$ – Lee Mosher Aug 27 '17 at 17:28
  • $\begingroup$ Yes, I get that my logic had a flaw there but I have another solution now. Since the Cantor set is broken up into uncountably infinite points in [0,1], isn't it true that if it is "uncountable" then we cannot have a one-one relation with ℝ. For a one-one relation, it needs to be countable. $\endgroup$ – Ritojeet Basu Aug 28 '17 at 13:34
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Each element of the Cantor set C, can be expressed as a decimal base
three without any 1's. Take that expression and change every 2 to
a 1 and read the result as a decimal base two. Thus a bijection
between C and [0,1].

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