# Does there exist a bijection from $[0,1]$ to $\mathbb R$?

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?

• Yes. What do you think?
– Did
Sep 30, 2015 at 10:20
• Use the ideas in this post. Sep 30, 2015 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?" Sep 30, 2015 at 10:22
• So you're considering that it might be possible for $[0,1]$ to be "larger" than $\mathbb R$? Sep 30, 2015 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 Sep 30, 2015 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 $g$.

• What's $\mathbb{N}^{*}$?
– AJY
Jul 6, 2016 at 19:29