Cardinality and Countably Infinite

Question

Find the cardinality of the set of all subsets $A \subset R$ for which $R \setminus A$ is countably infinite.

I honestly have no idea how to approach this. I have no background in Cardinality, other than the 3 pages of reading we had that go with this question set. Based on it, it seems like you determine Cardinality by finding a bijection to a set with known Cardinality, but how would you even know what set to choose?

Answer 1

Instead of explicitly finding a bijection to a set of known cardinality, you can use cardinal arithmetic to find the cardinality of the set $S$ of all countably infinite subsets of $R$.

First, there are $2^{\aleph_0}$ real numbers and this gives a lower bound to $|S|$.

On the other hand, $|S|$ is certainly less than the cardinality of the set $X$ of all countable sequences of real numbers. The cardinality of this set can be computed as $$ |X| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0}, $$ where $\aleph_0 \cdot \aleph_0 = \aleph_0$ can be shown using a pairing function (see the comments). You can find these computation rules on Wikipedia.

Putting both bounds together, we have $2^{\aleph_0} \leq |S| \leq 2^{\aleph_0}$, hence $|S| = 2^{\aleph_0}$.

Answer 2

Consider the collection $C$ of countably infinite subset of $R$. For any $A\subset R$ as defined above,

$$R\setminus A = A^c \in C$$

And for any $c \in C$,

$$R/(c^c) = c$$

So the cardinality of this collection is the same as the cardinality of the collection of countable subsets of $R$.

Answer 3

Here is a way to injectively identify the set of countably infinite real $sequences$ with a subset of $\Bbb R$. There is a bijection $f:\Bbb R\to (0,1)$ so it suffices to consider sequences of members of $(0,1)$. For a sequence $x=(x_n)_{n\in\Bbb N}$of members of $(0,1)$, for each $n$ let $0.x_{n,1},x_{n,2},x_{n,3},...$ be the decimal representation of $x_n$ in which infinitely many of the digits are $not$ $9$. Let $g(x)$ be represented by the decimal sequence $0.x_{1,1},\;x_{1,2},x_{2,1},\;\;x_{1,3},x_{2,2},x_{3,1},\;\;$ $x_{1,4},x_{2,3},x_{3,2},x_{4,1},\;x_{1,5},x_{2,4},...$ et cetera. Then $g$ is injective.