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We can find a bijection from $(0,1)$ to $\mathbb R$. For example, we can use $f(x)=\frac{2x-1}{1+|2x-1|}$ composed of parts of two hyperbolas, see the graph here. Or we could appropriately scale the tangent function to get $g(x)=\tan\pi\left(x-\frac12\right)$, see the graph here. Several such bijections are suggested in the answers to this post: Is there a bijective map from $(0,1)$ to $\mathbb{R}$?

But does there exist a bijection from $[0,1]$ to $\mathbb R$? If yes, then what is it?

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Fawkes4494d3
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  • Yes. What do you think? – Did Sep 30 '15 at 10:20
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    Use the ideas in this post. – David Mitra Sep 30 '15 at 10:20
  • I don't understand the downvote... For cardinality reason, the answer is obviously yes. The interesting part is "If yes, then what is it?" – mathcounterexamples.net Sep 30 '15 at 10:22
  • So you're considering that it might be possible for $[0,1]$ to be "larger" than $\mathbb R$? – Git Gud Sep 30 '15 at 10:23
  • @mathcounterexamples.net some guys here are very fast with downvotes, but the interesting part is not that interesting, because Cantor-Schröder-Bernstein gives you a bijection, and it is a very familar example of how to construct a bijection from $[0,1)$ to $(0,1)$ and the rest is made via compositions – Dominic Michaelis Sep 30 '15 at 10:26
  • @Domini Michaelis I agree, but if your don't have this background, which is probably the case of the OP writter, it is interesting to brainstorm on the topic. – mathcounterexamples.net Sep 30 '15 at 10:29
  • @mathcounterexamples.net yeah for sure, I think a comment that this is a duplicate and linking to an answered question would have been much better – Dominic Michaelis Sep 30 '15 at 10:31
  • @GitGud Maybe he's considering that they might be incomparable. It's not immediately obvious that you don't need AC for CBS. – DRF Sep 30 '15 at 10:40
  • @DRF As $[0,1]$ is a subset of $\mathbb{R}$ I highly doubt they could be incomparable – Dominic Michaelis Sep 30 '15 at 11:00
  • @DominicMichaelis They can't, assuming you use classical logic (Cantor-Schroeder-Bernstein). And they can't even in general I think since the bijection is actually strictly constructible. But it certainly is not right away obvious to a layman. – DRF Sep 30 '15 at 11:14

1 Answers1

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Let’s fix $f:(0,1)\to\mathbb{R}$.

Define $g:[0,1]\to\mathbb{R}$ as follows:

  • $g(0) = -1$
  • $g(1) = 1$

and for $0<x<1$,

  • if $f(x)\in\mathbb{N}^*$, then $g(x) = f(x)+1$
  • if $-f(x)\in\mathbb{N}^*$, then $g(x) = f(x)-1$
  • otherwise, $g(x) = f(x)$

Then, if $f$ is a bijection, so is $g$.

Guillaume Brunerie
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