# Best bijection between $\mathbb{N}^\mathbb{N}$ and $[0,1] \subset \mathbb{R}$.

Cantor's diagonal argument proves as a special case that the set of functions from $$\mathbb{N}$$ to $$\{0,1\}$$ has the same cardinality as the interval $$[0,1].$$ There is an 'obvious' mapping from functions to real numbers that almost provides a bijection, except very few pairs of functions happen to map to the same reals. This can then be dealt with by using the Cantor-Bernstein-Schröder Theorem to prove the existence of a bijection. For an application in number theory I would like to apply similar bijections between $$\mathbb{N}^\mathbb{N}$$ and $$[0,1].$$ The construction which is 'obvious' is to use dove-tailing, as follows. Let $$M$$ be a matrix with rows and columns indexed by $$\mathbb{N}$$ so that if $$f$$ is a function, then the binary representation of $$f(n)$$ is contained in the $$n$$'th row of $$M.$$ Now let $$f$$ correspond to the real number which has as its binary digits the entries (0,0),(0,1),(1,1),(0,2),(1,1),(2,0),(0,3) etc. of $$M.$$ Clearly each function produces a unique real number between $$0$$ and $$1.$$ Again it may be necessary to apply the Cantor-Bernstein-Schröder Theorem to eliminate double occurrences of real numbers. In my particular application I do not worry about this, since all my functions map only a finite number of natural numbers to zero. But my question is whether other bijections are also known, other than this 'obvious' one.

Assuming you're talking about $$\Bbb{N}$$ starting at $$1$$ rather than $$0$$, the bijection I like best is $$(n_1, n_2, n_3, ...) \mapsto \frac{1}{2^{n_1}} + \frac{1}{2^{n_1 + n_2}} + \frac{1}{2^{n_1 + n_2 + n_3}} + ...$$ The first number of the sequence tells me how many digits after the binary point to count to see the first $$1$$ in the binary expansion of $$x$$, and every number after that counts the number of binary digits to the next $$1$$. Although $$0$$ is not in the image of this map as a result, it is a true bijection of $$\Bbb{N}^{\Bbb{N}}$$ with $$(0, 1]$$--nothing gets double counted because this function can never output an infinite tail of zeroes.
The function provided in the answer by Rivers McForge can be slightly modified to include $$0$$: $$$$f(n_1,n_2,n_3,...) = \begin{cases} 0, & n_i = 1\\ g(n_1 -1, n_2, n_3,...),&n_1 >1, ~ n_{i>1}=1\\ g(n_1, n_2, n_3, ...), & \text{otherwise} \end{cases}$$$$ where $$$$g(n_1,n_2,n_3,...) = \frac1{2^{n_1}}+\frac1{2^{n_1+n_2}}+\frac1{2^{n_1+n_2+n_3}}+...$$$$
This works by "making some space" when $$n_i = 1$$ for $$i>1$$ by shifting the function in $$n_1$$ in the sense of Hilbert's paradox of the Grand Hotel, which leaves an "empty space" at $$n_i=1$$ for all $$i \in \mathbb N$$. This "empty space" is then used to assign the value $$0$$ for $$f(1,1,1,...)$$, which makes $$f(n_1,n_2,n_3,...)$$ the bijection between $$\mathbb N^{\mathbb N}$$ and $$[0,1]$$, as requested in the question.
In general, a bijection $$h: S \to S \cup \{a\}$$, where $$S$$ is an infinite set, can be constructed by defining a sequence $$(s_n)$$ in $$S$$ with no repeating members to define the function $$h(x)$$ in the following way: $$$$h(x) = \begin{cases} a, & x= s_1 \\ s_{n-1}, & x=s_n, ~n>1 \\ x, & x \notin (s_n) \end{cases}$$$$ In the above example $$S=(0, 1]$$, $$a=0$$ and $$s_n = 2^{-(n-1)}$$ in order to have $$f(n_1,n_2,n_3,...) = h(g(n_1,n_2,n_3,...))$$.