How can I show that any two arbitrary intervals in the set of reals have the same cardinality? 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 $f(x)=1/x$.
 A: 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.
A: 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.
A: 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 $(0,1)$.
