Bijection from $\mathbb N^\mathbb N$ Let's say I'm trying to find a bijection from $\mathbb N^\mathbb N$, i.e., the set of all functions from $\mathbb N$ to $\mathbb N$, to some other set, say an open interval $(a,b)∈R$. What do I need to establish, to say there can exist a bijection? I do understand that I need to prove it's both injective and surjective, but how do you do it with a set of functions as the domain?
Let's say I'm trying to see if the injective part is true. So, I need to find a function that takes every possible sequence of natural numbers as input, and associates each of them with one real number in the interval $(a,b)$. But what does it mean to have a sequence of natural numbers as the input of a function? And even if that makes sense, what sort of functions can satisfy such a purpose? I hope I sound sane.
 A: Typically it's not so easy to explicitly write down a bijection.  Instead
you can use the Cantor-Bernstein-Schroeder theorem.
You can map an interval injectively into $\{0,1\}^{\mathbb N}$ using a base-$2$ representation.
You can map ${\mathbb N}^{\mathbb N}$ injectively into an interval using
continued fractions.
A: A function from $\mathbb N$ to $\mathbb N$ would be of the form 
$$f(1)=n_1, f(2)=n_2, f(3)=n_3, f(4)=n_4, f(5)=n_5, \ldots$$ for particular values of $n_1,n_2,n_3,n_4,n_5,\ldots$, so producing a sequence of natural numbers.  You can then regard the function and the sequence as equivalent.
A function from these to the real numbers is much as you might expect, so for example the continued fraction:
$$g(n_1,n_2,n_3,n_4,n_5,\ldots)= n_1 + \cfrac{1}{n_2 + \cfrac{1}{n_3 + \cfrac{1}{n_4 + \cfrac{1}{n_5+\cdots} } } }.$$
To show this is injective (which it is in this case) you need to show that each different sequence produces a different real number
To show this is surjective (which it is not quite in this case - try finding a rational)  you need to show that each real number is produced by some sequence 
