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?

  • $\begingroup$ Yes. What do you think? $\endgroup$ – Did Sep 30 '15 at 10:20
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    $\begingroup$ Use the ideas in this post. $\endgroup$ – David Mitra Sep 30 '15 at 10:20
  • $\begingroup$ I don't understand the downvote... For cardinality reason, the answer is obviously yes. The interesting part is "If yes, then what is it?" $\endgroup$ – mathcounterexamples.net Sep 30 '15 at 10:22
  • $\begingroup$ So you're considering that it might be possible for $[0,1]$ to be "larger" than $\mathbb R$? $\endgroup$ – Git Gud Sep 30 '15 at 10:23
  • $\begingroup$ @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 $\endgroup$ – Dominic Michaelis Sep 30 '15 at 10:26

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 $g$.

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    $\begingroup$ What's $\mathbb{N}^{*}$? $\endgroup$ – AJY Jul 6 '16 at 19:29

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