How exactly do I go about finding a bijection between (0,1) → N \ {0}

so $(0,1) → (1, \infty)$. I figured I could look at this as finding a function from $(0,1) → (0, \infty)$ and just adding 1.

I've seen examples where f(x) = $\frac{1}{x} -1$ then $f(0) = \infty$ and $f(1) = 0$ (but these were on closed sets)

I couldn't find an example of a function such that $\lim_{x\to 1} = \infty$ or $\lim_{x\to 0} = \infty$ which is what it looks like I need here.

Can someone give me an example, or a way to find such a function?

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    $\begingroup$ What is $N$? If it is the natural number, you'll have a hard time finding the bijection since the two sets don't have the same cardinality. $\endgroup$ – Nitrogen Oct 24 '15 at 23:04
  • $\begingroup$ Right, is this a question that ask you to prove or disprove? $\endgroup$ – More water plz Oct 24 '15 at 23:09
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    $\begingroup$ If you are looking for the interval $(0,1)$ of reals to the interval $(1,\infty)$ of reals, simple is $f(x)=\frac{1}{1-x}$. $\endgroup$ – André Nicolas Oct 24 '15 at 23:09
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    $\begingroup$ Another cute function to play with is $\tan(\pi x/2)$. $\endgroup$ – André Nicolas Oct 24 '15 at 23:11
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    $\begingroup$ N isn't (1, infinity). N is {1,2,3,4,....}. No such bijection exists. Ask for (0,1) to (0, infinity) instead. $\endgroup$ – fleablood Oct 24 '15 at 23:31

Here are two bijections $f$ from $(0,1)$ to $(1,\infty)$:

1) Let $f(x)=\frac{1}{1-x}$;

2) Let $f(x)=1+\tan\left(\frac{\pi x}{2}\right)$.


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