# Cardinality of $P_\mathfrak{c}(\mathbb R)$

Calculate the cardinality of $\mathcal P_\mathfrak{c}(\mathbb R)=\{\mathcal R \in \mathcal P(\mathbb R) \, /\, \#(\mathcal R)=\mathfrak{c}\}$.

I thought that instead of $\mathcal P_\mathfrak{c}(\mathbb R)$ I should think in $\mathcal P_c([0,1))$ so i can work with the representation of $x \in [0,1)$ in the binary numeral system but I don't know how to do a bijection with that set. Of course, I am thinking that $\mathcal P_c(\mathbb R)$ has the cardinality of $\mathbb R$. (Aclaration: $\mathfrak{c}=\#\mathbb R)$

• If by $\mathfrak{c}$ you mean the cardinality of the continuum, then the cardinality will be $2^{\mathfrak{c}}$. – André Nicolas Jul 12 '15 at 1:59
• Not a great notation. Usually in set theory $\mathcal P_\kappa(X)$ denotes the set of subsets of $X$ of cardinality strictly less than $\kappa$. – Asaf Karagila Jul 12 '15 at 2:20

The cardinality of $\mathcal{P}_\mathfrak{c}(\mathbb{R})$ is $2^\mathfrak{c}$.
Clearly we have $\mathcal{P}_\mathfrak{c}(\mathbb{R}) \subseteq \mathcal{P}(\mathbb{R})$ so that $\left|\mathcal{P}_\mathfrak{c}(\mathbb{R})\right|\le 2^\mathfrak{c}$.
On the other hand, we have $X\cup (0,\infty) \in \mathcal{P}_\mathfrak{c}(\mathbb{R})$, where $X$ is an arbitrary subset of $(-\infty,0)$. But the number of such subsets is $2^\mathfrak{c}$ since $|(-\infty,0)| = \mathfrak{c}$. Hence $\left|\mathcal{P}_\mathfrak{c}(\mathbb{R})\right| \ge 2^\mathfrak{c}$.