cardinality of some subsets of the power sets of $\mathbb Z$ and $\mathbb R$ What is the cardinal number of these sets :

*

*$X:= \left\{ A \in P(\mathbb Z) \mid \sum_{a \in A} |a| \text{ is finite} \right\}$


*$Y:= \left\{ B \in P(\mathbb R) \mid(\mathbb {R}\setminus B) \space \text {is countable} \right\}$


*$Z:= \left\{ C \in P(\mathbb R) \mid|C| = \mathfrak{c}  \right\}$
My solutions:
$1.$ Define $ f : \left\{x\in P(\mathbb Z) | \text {x is finite subset}\right\}  \to \mathbb N$
$\space \text{defined by :} \begin{cases}  
  P_{lastdigit7}  \space ,  & \text{if $y\in x>0$} \\
  P_{lastdigit3},  & \text{if $y\in x<0$} \\\end{cases} $
so $f(X) = P_{y1}*P_{y2}* \cdots * P_{yn}$
$Py:=$ the prime number in the y place  , $P_1 = 2 , P_2=3$ and so on..
$P_{lastdigit7}:=$ the prime number in the y place  when last digit is 7 $P_1 = 7 , P_2=17$...
$P_{lastdigit3}:=$ the prime number in the y place  when last digit is 3 $P_1 = 3 
  , P_2=13$...
I think its a bijection and its prove that is countable , $\aleph_0$.
2.We want $(\mathbb {R}\setminus B)$ to be countable so $B$ must be
uncountable .
so lets find cardinality of countable sets.
High limit : $|\mathbb R ^{\mathbb N}| = \mathfrak{c}.$
Low limit: $| \left\{0,1 \right\} ^{\mathbb N}| = \mathfrak{c}$.
So $|Y|=\mathfrak{c}$.
$P(\mathbb R)=\left\{ Countable \space subsets \right\} \bigcup  \left\{ 
  Uncountable \space subsets \right\} \implies $
$2^\mathfrak{c}=|P(\mathbb R)|=|\left\{ Countable \space subsets 
  \right\}| +  |\left\{ Uncountable \space subsets \right\}| = $
$\mathfrak{c} + |\left\{ Uncountable \space subsets \right\}| \implies 
  |\left\{ Uncountable \space subsets \right\}|=2^\mathfrak{c}. $
$|Y|=2^\mathfrak{c}.$
$3.$ In excercise 2 we show that $|C|=2^\mathfrak{c}.$
Is my proofs correct ?
 A: *

*First, let $g$ be any bijection $\mathbb Z \to \mathbb N$, for example
$$g(n) = \begin{cases} 2 \cdot n,\ n \geq 0\\ -2 \cdot n - 1,\ n < 0\end{cases}$$
Now, say $f(X) = \sum_{x \in X} 2^{g(x)}$. Check that $f$ is bijection.


*Let $Y' = \{B \in P(\mathbb R) | B \text{ is countable}\}$. Do you know that $Y'$ has cardinality $\mathfrak{c}$? If yes, then there is simple bijection $Y \leftrightarrow Y'$.


*Consider set $D = \{[0, 1] \cup T | T \subseteq [3, 4]\}$. What is cardinality of $D$? What is relation between $C$ and $D$?
A: Your answer for (1) works, modulo some more careful wording, but your $f$ isn’t a bijection.
But $f$ is one-to-one. It is easy to prove that if $W$ is infinite, and there is a one-to-one function $f:W\to\mathbb N,$ then $W$ is countably infinite.
So you don’t need your $f$ to be a bijection to get the cardinality of $X$ is countably infinite.

An explicit bijection for (1).
If $x=\{a_1<a_2<\cdots <a_k\},$
then define $$f(x)=\begin{cases}0&x=\emptyset\\\sum_{i=1}^k2^{a_i+1}&a_1\geq 0\\
-1+\sum_{i=1}^k 2^{a_i-2a_1}&a_1<0
\end{cases}$$
This is slightly easier to understand by seeing that $g(x)=f(x)+1$ is a bijection with $\mathbb N^+,$ the set of non-zero natural numbers.
Given $n\in\mathbb N,$ write $n+1$ in binary:
$$n+1=2^{b_1}+2^{b_2}+\cdots+2^{b_k}.$$
With $0\leq b_1<b_2<\cdots<b_k.$
Then if $b_1=0,$ we get:
$$f^{-1}(n)=\{b_2-1,b_3-1,\dots,b_k-1\}$$
When $b_1>0,$ you get: $$f^{-1}(n)=\{b_1-2b_1,b_2-2b_1,b_3-2b_1,\dots b_k-2b_1\}.$$
I’ll leave it to you to prove this is a bijection.

This is a tricky way of doing something that is easy for the set $$X’=\{x\in P(\mathbb N)\mid x\text{ finite}\}$$ There is an easy binary encoding of $X’$:
$$f’(x)=\sum_{i\in x} 2^{i}$$
We could just use a simple bijection $g:X\to X’,$ of course, and define $f(x)=f’(g(x)),$ but I wanted a more direct numeric encoding.
