Thursday 28 March 2013

elementary set theory - 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?


Answer



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


  • 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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