cardinality of some subsets of the power sets of $\mathbb Z$ and $\mathbb R$

Question

What is the cardinal number of these sets :

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

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

3. $Z:= \left\{ C \in P(\mathbb R) \mid|C| = \mathfrak{c} \right\}$

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 ?

Answer 1

1. 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.

2. 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 $Y \leftrightarrow Y'$.

3. Consider set $D = \{[0, 1] \cup T | T \subseteq [3, 4]\}$. What is cardinality of $D$? What is relation between $C$ and $D$?

Answer 2

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.

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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.

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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.