What is the cardinality of the set of all infinite sequences? The set is defined as {$(n_1, n_2,...n_k ..) | n_k \in \mathbb{N}$}. What are some approaches to finding and proving the cardinality of this set? 
 A: You have $|\mathbb{N}|$ options for as many slots, so $|\mathbb{N}|^{|\mathbb{N}|}$ different sequences.
There are several ways to prove that this is $|\mathbb{R}|$. I'll give one, but try to come up with another (hint, power sets).
Continued Fractions. Given a sequence $a_i$ we can map it to a continued fraction by $f(\{a_i\})=[0;a_1,a_2,\ldots]$. When the sequences are taken over the naturals, this is a bijection with the $(\mathbb{R}-\mathbb{Q})\cap[0,1]$. Since there are only countably many rationals, the irrationals in $[0,1]$ and $[0,1]$ itself have the same cardinality by the properties of cardinal arithmetic. It's well-known that $|[0,1]|=|\mathbb{R}|$, and several proofs are given here.
A: We call the set of all infinite sequences of natural numbers $E$.
Notice that for each infinite sequence in $E$, each term ($n_k$) in the  sequence 
can be mapped to $k\in\mathbb{N}$, hence the cardinality of 
each infinite sequence is $\aleph_0$.   Because $E$ is
an infinite set of infinite sequences, the cardinality must 
be bigger than $\aleph_0$. In fact, $\#E$ can be expressed as
$$\#E = \aleph_0^{\aleph_0}.$$
Because $\aleph_0 > 2$, we follow the law of exponents and get 
$2^{\aleph_0} \leq \aleph_0^{\aleph_0} \leq \left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0 *\aleph_0} = 2^{\aleph_0}.$
Therefore the cardinality of E is: $$\#E = \aleph_0^{\aleph_0} = 2^{\aleph_0}.$$
QED
Note: The above assertion that the exponents $\aleph_0 \times \aleph_0=\aleph_0$  is to recognize the Cartesian product of two countably infinite sets are still countably infinite.  The sketch of the proof is as follows: Suppose $A$ and $B$ are both countably infinite, so we
can write $A$ and $B$ in list form as
$$A = \{a_1, a_2, a_3, \dots \},$$
$$B = \{b_1, b_2, b_3, \dots \},$$
and form a rectangular table with elements
$(a_1, b_1), (a_1, b_2), (a_1, b_3),\dots$ in the table. We  can process
each element in the table
diagonally and see each element of $A\times B$ will be listed, therefore
$A\times B$ can be expressed in list form, so $A\times B$ is also countably infinite. In other words, we have proved that
$$\aleph_0 \times \aleph_0=\aleph_0.$$
A: An infinite sequence in N can be veiwed (is defined as) a function from N -> N (with domain all of N, i.e. a: N -> N and notate a1 = a(1), a2 = a(2), etc.). 
How many different functions are there ? 
Each function is a subset of N x N, i.e. an element of P(N x N), so S the set of all functions (sequences) is a subset of P(N x N) and so |S| <= |P(N x N)|.
As there is a bijection between N and N x N then there is a bijection between P(N) and P(N x N) so that |S| <= |P(N x N)| = |P(N)|.
Permutations of N are a subset of possible functions denoted as sym(N), so that |sym(N)| <= |S|. How many permutations are there ? There is a proof (below) that shows an injection from P(N) into sym(N) so that |P(N)| <= |sym(N)| <= |S| <= |P(N)|
So the answer is that |S| = |P(N)|.
Proof P(N) injects into sym(N).
Split N into two infinite subsets N1 and N2 (say even and odd). There is a bijection between them , say f: N1 -> N2.
We can define a mapping g: P(N1) -> sym(N) as follows:
For A ∈ P(N1), p ∈ sym(X) is the permutation p: N -> N as n exchanges with f(n) if n ∈ A, else n is unchanged.
Then g is an injection from P(N1) into sym(N), so that | P(N1)| ≤ |sym(X)|. 
But then |N1| = |N|, so that | P(N1)| = | P(N)| and thus |P(N)| ≤ |sym(X)|.
