Bijection between the reals and the set of permutations of the natural numbers?

In analysis today we talked about re-arrangements of sequences, and one student asked how many re-arrangements there are of a given sequence. We were able to very quickly create a one-to-one function from the reals to the set of permutations on $\mathbb{N}$ by simply noting that for any real number, there is a re-arrangement of a conditionally convergent series that converges to that number.

What we were not easily able to do was either prove that function was onto, or create an injection from the permutations on $\mathbb{N}$ back to the reals. So we know the number of re-arrangements is at least the cardinality of the reals, can we show it is exactly the same as the cardinality of the reals?

Here you’ll find a proof that the infinite continued fractions with $0$ integer part are precisely the irrationals in $(0,1)$. The map

$$\left(\Bbb Z^+\right)^{\Bbb Z^+}\to(0,1):a\mapsto[0;a_1,a_2,a_3,\dots]$$

is therefore an injection.

I have just done this as an exercise in my introductory Set Theory text.

Let $( a_1, a_2, \dots )$ be a permutation of $\mathbb N$.

For each $a_i$, form a string of $(a_i-1)$ zeros followed by a $1$. Call this string $s_i$. For example, if $a_i = 5$, then $s_i = 00001$.

Define a mapping $f : (a_1, a_2, \dots ) \mapsto 0.s_1s_2s_3\dots$, where $0.s_1s_2\dots$ is the concatenation of the strings $s_i$.

Then $f$ gives the desired injection when $0.s_1s_2\dots$ is read as the binary expansion of $x \in [0,1]$. The injection going the other way can be obtained as per your OP.

Edit : Come to think of it, you don't need to view the image as a binary expansion. It works equally well as a decimal expansion.

The number of permutations of $\Bbb{N}$ is at most the number of functions from $\Bbb N$ to $\Bbb N$, which is $\aleph_0^{\aleph_0}$. But $\aleph_0^{\aleph_0} \le (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0 \cdot \aleph_0}=2^{\aleph_0}$.