Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to show that $(a,b)$ and $(c,\infty)$ have the same cardinality, yet I am not sure what to do. The only example I have to go by that is similar is $(0,1)$ and $(1,\infty)$... And that one used the function $f(x)=1/x$.

share|cite|improve this question
up vote 4 down vote accepted

You have done the hard part. Can you find a bijecton between $(a,b)$ and $(0,1)$? There is a natural one. Then can you find one between $(1,\infty)$ and $(c,\infty)?$. Then you can just compose the bijections $(a,b)\leftrightarrow (0,1) \leftrightarrow (1,\infty) \leftrightarrow (c,\infty)$. Normally you wouldn't be asked to do it explicitly, you have shown that you could and that is enough.

share|cite|improve this answer

Show $(a,b)$ and $(c,d)$ have the same cardinality, with $a,b,c,d\in\mathbb{R}. a<b,c<d$
Define $\Phi:(a,b)\to(c,d)$ as
$$\Phi(x) = \frac{(x-a)}{(b-a)}*(d-c)+c$$ The intuitive idea is that the relative length of segment a-x with respect to a-b is preserved and projected on c-d. Show this is bijective.

To show $(0,\frac{\pi}{2})$ has same cardinality with $\mathbb{R}$, look at $$\arctan:(0,\infty)\to(0,\frac{\pi}{2})$$ Show it is bijecitive.

Then we show $(a,b)$ is bijective with $(0,\frac{\pi}{2})$ by method 1, which is bijective with $(0,\infty)$ by arctan, which is bijective to $(c,\infty)$ by shifting by c.

share|cite|improve this answer

You could try to transform the interval $(a,b)$ to the interval $(0,1)$ like so:

$(a,b) - a \rightarrow (0,b-a)$

$(0,b-a) * \frac{1}{b-a} \rightarrow (0,1)$

Then you can use your existing argument for the cardinality of $(0,1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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