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I found a lot of problems of bijections like $[0,1) \rightarrow[0, \infty)$. But I don't find ant of these:

$$\text{Is it possible to make a bijection } [0,1] \rightarrow[0, \infty)? \text{ If yes, find at least one bijection.} $$

Can anyone help?

My work: I found one function: $f(x)=\frac{1}{\left( x-\frac{1}{2} \right) ^2}-4$. What about $f(x)=\frac{1}{\left( x-1 \right) ^2}-1$?

Karagum
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  • Your function is not injective, because $f(0) = f(1)$. Do you want the bijection to be continuous? – Brazilian Cérebro Feb 11 '18 at 10:51
  • I guess it doesn't matter if the bijection is continuous or not. But in this case the bijection cannot be continuous, right? @BrazilianCérebro – Karagum Feb 11 '18 at 10:53

1 Answers1

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Let $$f(x):=\begin{cases} x&\text{if $x\in[0,1)\setminus\{1/n: n\in\mathbb{N}^+\}$}\\ \frac{1}{n+1}&\text{if $x=\frac{1}{n}$ for some $n\in\mathbb{N}^+$}\\ \end{cases}$$ The function $f$ is bijection from $[0,1]\to [0,1)$ which acts as a left-shift in the set $\{1,1/2,1/3,\dots\}$ and it is the identity elsewhere. Now take your bijection $g$ from $[0,1)$ to $[0,+\infty)$ (for example $g(x)=\frac{x}{1-x}$) and consider the composition $g\circ f: [0,1]\to [0,+\infty)$.

P.S. Note that there is no continuous bijection from $[0,1]$ to $[0, \infty)$ because $[0,1]$ is a compact set and $[0, \infty)$ is not a compact set.

Robert Z
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  • Can you give me more explanation about the function $f(x)$, @Robert Z? – Karagum Feb 11 '18 at 11:14
  • @Karagum $f$ is a bijection from the set $A={1,1/2,1/3,\dots}$ to ${1/2,1/3,\dots}=A\setminus {1}$. Is it clear now? Any further doubt? – Robert Z Feb 11 '18 at 19:06
  • @Karagum: this is a common technique for finding bijections. You find one that does most of what you want, then take care of the problem points. There are many cases here of bijections between $[0,1]$ and $(0,1)$ which do basically the same thing. You can even take care of a countable infinity of points this way. Once you know that, you stop worrying about the problem children. – Ross Millikan Feb 11 '18 at 19:43