Existence of a sequence that has every element of $\mathbb N$ infinite number of times I was wondering if a sequence that has every element of $\mathbb N$ infinite number of times exists ($\mathbb N$ includes $0$). It feels like it should, but I just have a few doubts.
Like, assume that $(a_n)$ is such a sequence. Find the first $a_i \not = 1$ and the next $a_j, \ \ j > i, \ \ a_j = 1$. Swap $a_j$ and $a_i$. This can be done infinitely many times and the resulting sequence has $1$ at position $M \in \mathbb N$, no matter how large $M$ is. Thus $(a_n) = (1_n)$. 
Also, consider powerset of $\mathbb N, \ \ 2^{\mathbb N}$. Then form any permutation of these sets, $(b_{n_{\{N\}}})$. Now simply flatten this, take first element $b_1$ and make it first element of $a$ and continue till you come to end of the first set and then the next element in $a$ is the first element of $b_2$... But again, if you start with set $\{1 | k \in \mathbb N\}$ (infinite sequence of ones), you will get the same as in the first case...
You would not be able to form a bijection between this sequence and a sequence that lists every natural number once, for if you started $(a_n)$ with listing every natural number once, you would be able to map only the first $\aleph_0$ elements of $(a_n)$ to $\mathbb N$.
But consider decimal expansion of $\pi$. Does it contain every natural number infinitely many times?
What am I not getting here?
What would the cardinality of such a sequence, if interpreted as a set, be?

$$\langle \underbrace{0}_{1\text{ term}},\underbrace{0,1}_{2\text{ terms}},\underbrace{0,1,2}_{3\text{ terms}},\underbrace{0,1,2,3}_{4\text{ terms}},\underbrace{0,1,2,3,4}_{5\text{ terms}},\dots\rangle$$
Each $n\in\Bbb N$ appears in all but the first $n$ blocks, hence infinitely often.
