$A$ is a subset such that $A$ and its complement will divide $\mathbb{Z}^+$ into two partitions (using your hint). $M$ is any subset of $\mathscr{P}(\mathbb{Z}^+)$ which can be the whole set, some set that span across the two partitions, or a subset of only one partition.

In all cases, $M=A_i \cup B_i$ where $A_i$ and $B_i$ are subsets of the partitions $A$ and $A^c$ respectively to give the required $M$ ( I think I can do away with the index $i$ completely (which runs from $0$ to $|\mathscr{P}(\mathbb{Z}^+)|$ anyway, by instead calling $C=A_i$ and $D=B_i$ so that $C$ and $D$ will depend o…

Now I am wondering about the case $P(\mathbb{Z}^+)\to \{0,...,n\}$ :
If $n$ is of the form $2^m$ for $m \in \mathbb{Z}^+$ (The case when $n=\aleph_0$ is trivial because you end up having the same set, which is obviously in bijection correspondance), then we can achieve the required parition by partitioning the entries in the sequence into m-1 blocks (because the last two case is taken cared by $x\in A$ and $x \not\in A$. But what if $n$ is odd, how should I partition $A$ so that $h^{-1}$ can map nicely?

Uh, is the following proof legal?

Given $D={\mathbb{Z}^+}^{\mathbb{Z}^+}$ and $E=\{0,1\}^{\mathbb{Z}^+}$, we can deduce by inspection that $E \subset D$ since $D$ contains all sequences with countable entries including binary sequences, thus a injective $g: E\to D$ can be defined as the (not surjective) fixed point function. $f(x)=x$ for all $x \in E$

Now it remains to get an injective $h: D \to E$ to invoke the theorem in Q6a to show $|D|=|E|$
Take any subset $A \subset D$, partition $A$ into countable parts. For all $x \in D$, define

For that what I am trying to say is given a sequence $(x_0,x_1,x_2,...)$ while $x_i$ is just one number for each sequence, there are a maximum of |D| number of choices for $x_i$. and the sequence itself is countably long. Perhaps I am trying to say the number of choices of each entry is countable, not the entry itself i countable (something I am not very sure how to handle)

$h$ acts on $A$ and start labelling the binary sequence based on whether the ith element is present or absent in the sequence. Maybe the domain of h should be the set of all $A$ instead

both G and F are uncountable sets that might be larger than the reals, say $F=2^\mathbb{R}$ and $G=\mathbb{R}^{\mathbb{R}}$. I want to check whether G and F bijects each other, then I ran into problem when I tried to write a sequence of it to figure how to partition it and define my bijective function (if any). Because attempt to writing down the sequence we will get (where x is some number from the domain)

(xxx... xxx... xxx... x...)

But those ... may not be countable and I don't know their uncountable size until I can find a bijective function, but these ... is what cause me to struggle…

the blocks appear when you put all possible f s together. e.g. for $\mathbb{N}\to \{0,1\}$ all possible fs are

{0,0,0,0,0,...}
{1,0,0,0,0,...}
{0,1,0,0,0,...}
{0,0,1,0,0,...}
{0,0,0,1,0,...}
{0,0,0,0,1,...}
and so on
{1,1,0,0,0,...}
{0,1,1,0,0,...}
{0,0,1,1,0,...}
{0,0,0,1,1,...}
and so on
{1,1,1,0,0,...}
{0,1,1,1,0,...}
{0,0,1,1,1,...}
and so on
{1,1,1,1,0,...}
{0,1,1,1,1,...}
and so on
{1,1,1,1,1...}
and so on
{1,0,1,1,1,...}
and so on
{1,0,0,1,0,...}
and so on
{0,1,0,0,0,1,...}
and so on
...