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Hi am am taking a course in measure- and integral theory. Here I have been asked to prove that there exists a bijection $ f : [0,1] \rightarrow \mathbb{R} $. Would it suffice to find a function that is a bijection and prove injectivity and surjectivity or would this be not be the point of such an exercise?

Function found:

$ g(x)=\ln\left(\frac{1}{x}-1\right) $

$ g^{-1}(x)=\frac {1}{1+e^x} $

comjens
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First of all, notice that $[0,1]$ is a closed interval, so you can't really expect to create a continuous bijective function between it and $\mathbb R$. There exist bijections between $(0,1)$ and $\mathbb R$, however, and it isn't hard to produce a discontinuous bijection between $[0,1]$ and $(0,1)$.

Second of all, your function most certainly is not a bijection, since $g(x)>0$ for all values of $x$. In fact, your function maps $[0,1]$ to $[\frac{1}{1+e}, \frac12]$.

5xum
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  • That makes sense, hadn't really noticed that obviously. I do think i made a small mistake in the post, switched the function and inverse, look at edit. But even with the edit the is a problem because it is the closed interval. Do you have any idea how to solve this? – comjens Mar 17 '16 at 10:14
  • @comjens Now, your function isn't even defined for $x\in{0,1}$, so it is not even a function from $[0,1]$ to $\mathbb R$ – 5xum Mar 17 '16 at 10:16
  • I see that too, I am too hasty sorry. – comjens Mar 17 '16 at 10:18