Cardinality of real numbers Given an open interval, say $(a,b]$, how do we show that it has the same cardinality as the set of real numbers? Or is there a bijective mapping in the 1st place?
 A: The simplest proof sketch, in my opinion is:


*

*$|(a,b)| = \mathfrak{c}$

*$|(a,b]| = |(a,b)| + 1$

*$\mathfrak{c} + 1 = \mathfrak{c}$


where $\mathfrak{c}$ is the cardinality of the reals.
A: Take a countable subset of $(a,b)$. and index the elements as $K=\{c_i\}$.
Take the map $f:(a,b] \rightarrow (a,c)$ as:
$f(b) = c_1$
$f(c_i) = c_{i+1}$
$f(x)_{x\ne b; x\not \in K} = x$.
That is easily shown to be a bijection so the cardinality of $(a,b]$ is the same cardinality as $(a,b)$.
If you are like me, the blase command "Take a countable subset of $(a,b)$ " will leave you cold.  But it is legitimate via the inductive property of the natural numbers.  I don't think (I could be wrong) that it would be valid to say "take an uncountable subset" but a countable subset is fine.
Oh, and $(a,b)$ has the same cardinality as $\mathbb R$ is a standard exercise.  Your tangent function with proper shifting and scaling will do. 
A: You can use the tangent function to construct a bijection $f$ from $(a,b)$ to $\mathbb{R}$. (That's an idea from your comment.)
Now you have to find a way to map the extra point $b$. The trick is to give up on looking for a nice formula and reuse a trick you may have seen when "counting" the natural numbers. Use your bijection $f$ to find the points $x_n$ such that $f(x_n) = n$ for $n= 0, 1, \ldots$. Now redefine $f$ so that $f(x_{n+1}) = n$. Then $f$ is is injective on $(a,b)$ and its range is all of $\mathbb{R}$ except $0$. So define $f(b) = 0$ and you're done.
