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?