Proving the existence of a subset

Question

I am studying Discrete Maths on my own, and I need help proving that there exists a set $S \subseteq P(\mathbb{R})$ - the powerset of $\mathbb{R}$ with the following 3 conditions:

1. $S \sim \mathbb{R}$

2. if $X, Y \in S$ and $X \neq Y$, then $X \cap Y = \emptyset$

3. if $X \in S$, then $X \sim \mathbb{R}$

Recall that "$A\sim$ B" means A has the same amount of elements as B

My attempt: I understand that $P(\mathbb{R})$ is a set of all possible combinations of points arrangement on a line. Hence, if I take $\mathbb{R} = S$ the first two conditions are met. The third, however, is not, as any single point from $\mathbb{R} \nsim \mathbb{R}$.

How do I proceed from here?

Answer 1

First, notice that $|\mathbb{R}^2|=|\mathbb{R}|$ (for example, see this).

Then, there exists a bijection $f: \mathbb{R}^2\to \mathbb{R}$.

Define $S=\{f(\{x\}\times \mathbb{R})\space |\space x\in\mathbb{R}\}$.

As $f$ is a bijection, condition $2$ follows immediately.

Moreover, take $f(\{c\}\times \mathbb{R})\in S$. Then, it is clear $g:\mathbb{R}\to f(\{c\}\times \mathbb{R})$ given by $g(x)=f(c,x)$ is a bijection, so $|f(\{c\}\times \mathbb{R})|=|\mathbb{R}|$, that is, condition $3$ also holds.

Lastly, defining $h:\mathbb{R}\to S$ as $h(x)=f(\{x\}\times \mathbb{R})$ we find $h$ is a bijection and $|S|=|\mathbb{R}|$, as desired.