Bijection between $3^{\mathbb N}$ and $\mathcal P(\mathbb N)$?

Since $\mathcal P(\mathbb N)$ is the set of all subsets of $\mathbb N$,

For each element $a \in \mathbb N$ and every subset $S\subset N$, we can define a function $f$:

$f(a) = 0$ if $a \notin S$

$f(a) = 1$ if $a \in S$

This proves a bijection between $2^{\mathbb N}$ and $\mathcal P(\mathbb N)$ exists and therefore, $|2^{\mathbb N}| = \mathcal P(\mathbb N)$.

Is there a bijection between $|3^{\mathbb N}|$ and $\mathcal P(\mathbb N)$?

Every infinite ternary sequence would need to correspond with one and only one subset $S \subset N$.

An approach similar to a powerset operation is as follows:

$f(a) = 0$ if $a \notin S$

$f(a) = 1$ if $a \in S \oplus a+1 \in S$

$f(a) = 2$ if $a \in S$ ^ $a+1 \in S$ (^ denotes AND operator)

It is fairly easy to see that this function is a bijection for all ternary sequences with only $0$ or $2$, since we know exactly what elements will be a member of $S$ and which ones won't (i.e. if $a$ is a member, so is $a+1$, and if $a$ isn't a member, then neither is $a+1$).

The only problem is when $f(a)=1$. We know that either $a$ or $a+1$ will be in the subset, but not both, so there are exactly two subsets for each $1$ in the sequence. Being an infinite ternary sequences, infinitely many $1$'s may occur, thus we may end up with an uncountable number of subsets for a single ternary sequence.

This approach also fails for finite sets instead of $\mathbb N$.

A more general set theoretical approach towards even defining $3^A$ is as follows, but it doesn't provide a solution to my initial question:

If we have two sets, $A$ and $B$, with $|A| = |B|$ and $A \cap B = {\emptyset}$ ($A$ and $B$ are disjoint to avoid multiset theory), then we can define a non communitive binary operation $ϛ$ as follows

$A ψ B = C$

$C$ is the set of all subsets $S$ with the following properties:

For every element $a \in S$:

$a \in A$

or

If $a \in B$, then some corresponding (via a bijection) $a' \in A$.

From this definition, it is clear that for any $T \subset B$, $T \not\subset C$, thus the order of $A$ and $B$ matters.

Therefore, $C = 3^A$.

This can clearly be seen with finite sets for example:

$A = $ {$0,1$}

$B = $ {$2,3$}

Assume a bijection $a\in A$:

$f(a) = a+2$, $a+2 \in B$.

$C = $ {{},{$0$},{$1$},{$0,1$},{$0,2$},{$1,3$},{$0,1,2$},{$0,1,3$},{$0,1,2,3$}}

The following example could be instead written using multisets, with $2$ being a repetition of $0$, and $3$ being a repetition of $1$, but by definition, sets cannot contain multiple elements, so the need for a disjoint set with same cardinality is necessary.

At the very least, is possible to define $|3^{A}|$ using a non-communative binary operation for sets $A$ and $B$, but can this be done for unary operations on $A$ (which is necessary to define a bijection between $3^\mathbb{N}$ and the powerset $\mathcal{P}(\mathbb N)$? Any thoughts?

All answers are overkill IMO. Its easy to define a bijection between $2^\mathbb{N} $ and $3^\mathbb{N}$ which then answers your question;for this map $0-1,1-01,2-00$ so for instance $0121..$ gets mapped to $1010001...$ its easy to see this is a bijection.

$2^{\mathbb{N}}$ is in a bijective correspondence with the real numbers between $0$ and $1$ (think of their binary representation). Similarly, $3^{\mathbb{N}}$ is in a bijective correspondence with the same interval of real numbers (ternary representation). Therefore, they have the same cardinality, that of the Continuum.