How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$?

Question

How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$?

I realize that any of the class of functions $f:x\to (n\cdot x)$ gives a bijection between $\mathbb{N}$ and the subset of $\mathbb{N}$ whose members equal multiples of n. So, we have at least a countable infinity of sets which have the same cardinality of $\mathbb{N}$. But, we could remove a single element from any countably infinity subset of the natural numbers and we still will end up with a countably infinite subset of $\mathbb{N}$. So (the reasoning here doesn't seem quite right to me), do there exist uncountably many infinite subsets of $\mathbb{N}$ with the same cardinality of $\mathbb{N}$?

Also, is the class of all bijections $f: \mathbb{N} \to \mathbb{N}$ and a given countably infinite subset uncountably infinite or countably infinite?

Answer 1

One can write a bijection between $\mathrm{Fin}(\mathbb N)$ and $\mathbb N$, i.e. between the set of finite subsets of $\mathbb N$ and $\mathbb N$. For example: $$f(A)=\sum_{i\in A}2^i$$

Note that $A$ is finite so this is a well-defined function. It turns out that this is a bijection as well.

Then one can show that $\mathcal P(\mathbb N)\setminus\mathrm{Fin}(\mathbb N)$ is uncountable, and in fact equipotent to $\mathcal P(\mathbb N)$. This is because $k+\aleph_0=2^{\aleph_0}$ implies that $k=2^{\aleph_0}$. So if $k=|\mathcal P(\mathbb N)\setminus\mathrm{Fin}(\mathbb N)|$ then $k=2^{\aleph_0}=|\mathcal P(\mathbb N)|$.

---

To the second question, one can show that there are $2^{\aleph_0}$ permutations of $\mathbb N$. Therefore if $f\colon A\to\mathbb N$ is a bijection composing it with a permutation of $\mathbb N$ will result in another bijection. It is not hard to show that if we compose different permutations we have different bijections (i.e. they disagree on the value of some point in $A$). Therefore there are $2^{\aleph_0}$ many bijections between any two countable sets.

(I should also remarked that none of the points in this answer requires the axiom of choice.)

Answer 2

$\Bbb N$ has only countably many finite subsets, but it has uncountably many subsets altogether, so it must have uncountably many countably infinite subsets (i.e., subsets with the same cardinality as $\Bbb N$ itself).

If $A$ is any countably infinite set, there are uncountably many bijections between $\Bbb N$ and $A$.

Answer 3

Some more answers:

There are uncountably many sets of odd numbers, and adjoining the even numbers to each of these yields a set with the same cardinality as $\mathbb N$.

Given any real number $x$, we can form the set $S(x)=\{\lfloor x \rfloor,\lfloor 10x \rfloor,\lfloor 100x\rfloor,\dots\}$ (e.g., $S(\pi)=\{3,31,314,3141,31415,\dots\}$), which clearly has the same cardinality as $\mathbb N$, and the map $S$ is a bijection, so this process yields uncountably many sets.

(Really the same as the last one) Given any infinite sequence $(a_n)$ of natural numbers, we can form the set $\{2^{a_1},3^{a_2},5^{a_3},\dots\}$ which has the same cardinality as $\mathbb N$.

Answer 4

Here is a simple explicit bijection from $\mathcal P(\mathbb N)$ to $\mathcal P(\mathbb N)\setminus \operatorname{Fin}(\mathbb N)$:

$$ f(A) = \begin{cases} \{ n+1 \mid n\in A \} & \text{if $A$ is cofinite} \\ \mathbb N \setminus \{ n+1 \mid n\in A \} & \text{if $A$ is finite} \\ A & \text{if $A$ is neither finite nor cofinite} \end{cases} $$

($A$ is said to be cofinite iff $\mathbb N\setminus A$ is finite).

Answer 5

As great answers have been given already, I'd merely like to add an easy way to show that the set of finite subsets of $\mathbb{N}$ is countable:

Observe that $$\operatorname{Fin}(\mathbb{N}) = \bigcup_{n\in\mathbb{N}}\left\{ A\subseteq\mathbb{N}: \max(A) = n \right\},$$ which is a countable union of finite sets as for each $n\in\mathbb{N}$ there certainly are less than $2^n = \left|\mathcal{P}(\{1,\ldots,n \})\right|$ subsets of $\mathbb{N}$ whose largest element is $n$. Hence, $\operatorname{Fin}(\mathbb{N})$ is countable itself.

From here on, you can use Asaf's argument to show that the set of infinite subsets of $\mathbb{N}$ (which are precisely the sets with the same cardinality as $\mathbb{N}$) must be uncountable.

Answer 6

Let

$\quad \mathcal I(\mathbb N) = \{A \in \mathcal P(\mathbb N) \, | \, A \text{ is infinite} \} $

By the Schröder–Bernstein theorem, if we can construct an injection $\Phi$ from $\mathcal P(\mathbb N)$ into $\mathcal I(\mathbb N)$, then we can claim that $|\mathcal I(\mathbb N)| = \mathcal |\mathcal P(\mathbb N)|$.

Observe that by using this technique it is not necessary to determine beforehand the cardinality of all the finite subsets of $\Bbb N$.

Set $K = \{k \in \mathbb N \mid k \text{ is odd}\}$ and define

$$ \Phi(A) = \begin{cases} \{ 2n \mid n\in A \} & \text{if $A$ is infinite} \\ \{ 2n \mid n\in A \} \cup K& \text{if $A$ is finite} \end{cases} $$

Answer 7

Let

$\quad \mathcal I(\mathbb N) = \{A \in \mathcal P(\mathbb N) \, | \, A \text{ is infinite} \} $

Let $K = \{n \in \mathbb N \mid n \text{ is an odd integer}\}$.

For each $F \subset \mathbb N$ where $|F|$ is finite consider the sequence

$ F,2F \cup K, 4F \cup 2K, 8F \cup 4K, \dots = (2^nF \cup 2^{n-1}K)_{\,n\ge0}$

where we agree that $2^{-1}K = \emptyset$.

Since $2^nF \cup 2^{n-1}K = 2^mG \cup 2^{m-1}K$ implies $m = n$ and $G = F$ we can define a bijective correspondence $\Psi$ from $\mathcal I(\mathbb N)$ to $\mathcal P(\mathbb N)$,

$$ \Psi(A) = \begin{cases} 2^{n-1}F \cup 2^{n-2}K & \text{where } A = 2^nF \cup 2^{n-1}K \text{ and } n \gt 0 \\ A & \text{otherwise} \end{cases} $$