1
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Another cute function to play with is $\tan(\pi x/2)$. $\endgroup$ – André Nicolas Oct 24 '15 at 23:11
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.