Prove: Any open interval has the same cardinality of $\Bbb R$ (without using trigonometric functions) I want to prove that every open interval has the same cardinality of $\Bbb R$.
The question is:
Is it enough to prove that any open interval is uncountable? If I prove it, can I say that this interval has the same cardinality of reals?
I know that I can get a bijection using the tangent function, but I am not allowed to use trigonometric functions.
I've proved that $|(a,b)|=|(0,1)|$ so I may prove that $|(0,1)|=|\Bbb R|$ but I did not find the bijection without using trigonometric functions.
 A: Consider the function
$$g(x)=\frac{x}{1+|x|}$$
Verify that $g$ is a bijection from real numbers to $(-1,1)$.
A: "Uncountable" just means "not countable", where countable is the smallest infinity. If you show that $(a,b)$ is uncountable, and that $\Bbb{R}$ is uncountable, you haven't shown that they have the same cardinality.
You need to exhibit a bijection. That is the very definition of "same cardinality". Any way you prove that the two have the same cardinality will, at least implicitly, exhibit a bijection.
"Same cardinality" means that there is a bijection. So you're asking if you can show that a bijection between these two sets, but without showing that there is a bijection between these two sets. You have to use a bijection. Any theorem, lemma, etc. MUST use bijections, as that is the very definition. What you're asking is like asking "can you show that 2 is even, without showing that 2 is even".
Edit, for an explicit bijection I leave the full construction to you (as it is somewhat tedious), but you could have something like:
$$f(x)=\begin{cases} \frac{1}{x-a} \text{ for } x\in (a,a+\mu/4)\\
\text{ linear connecting term for } x\in [a+\mu/4,a+3\mu/4]\\
\frac{1}{x-b} \text{ for } x\in (a+3\mu/4,b)
\end{cases}$$
where $\mu$ is the length of the interval.
A: This is an old question, but there is one very simple bijection without trigonometric functions. Consider $f:(0,1) \to \mathbb{R}$ given by
$$f(x) =
\begin{cases}
 \frac{1}{2x}  & 0<x<\frac 12 \\
\frac 1{2x-2} + 2, & \frac 12 \leq x < 1
\end{cases}$$
If you want $(a,b)$, it is, of course, simply a matter of scaling/shifting. 
A: I'll give another bijection, to add to the list (it has a geometric interpretation) : 
$x\to \ln(\frac{1}{x-a} -\frac{1}{b-a})$ for $x\in ]a,b[$ is easily seen to be a bijection $]a,b[\to \Bbb{R}$
A: Suppose $m$ is odd and $n$ is even. Then
$$ \frac{x^m}{1-x^n} $$
is a bijection $(-1,1) \to \mathbb{R}$. (It suffices to check it is increasing and unbounded in both directions: the former implies injectivity, the latter surjectivity.) The same applies to, for example,
$$ \log{\left(\frac{1+x}{1-x}\right)} = 2\arg\tanh{x} $$
A: I posted the following answer to your post over here which was later marked as a duplicate of this question. As I take a different (and arguably more elementary) approach than the previous answers, I thought it might be a good idea to repost it for future reference:

You already know that there is a bijection $(c,d) \to (-1,1)$ and it thus suffices to find a bijection $f \colon (-1,1) \to \mathbb R$. The following will do:
For all $n \in \mathbb N_0$ we let $f \restriction_{[1- 2^{-n},1-2^{-n-1}]}$ be the linear function from $(1- 2^{-n} ; 2^n)$ to $(1-2^{-n-1} ; 2^{n+1})$ and likewise $f \restriction_{[-1+ 2^{-n-1},-1+2^{-n}]}$ is the linear function from $(-1+ 2^{-n-1} ; -2^{n+1})$ to $(-1+ 2^{-n} ; -2^{n})$.
The picture below demonstrates what's going on:

