# Cardinality of the family of all possible dense subsets(dense in $\mathbb R^{n}$) of $\mathbb R^{n}$

Actually, for each irrational number $$i$$, the set $$\{m+ni | m,n \in \mathbb N\}$$ , is dense in $$\mathbb R$$, as there are continuum many irrationals, hence obviously there are at least continuum many dense sets .

And, for a fixed integer $$k$$, $$\{m/k^{n} | m,n \in \mathbb N\}$$ is dense in $$\mathbb R$$, but as this family is countable, hence they do not add up extra cardinality.

And, it can be proved that if $$X$$ is a metric space with exactly $$n$$ non-isolated points , then there are exactly $$2^{n}$$ dense subsets of $$X$$ . But I figure out this problem ???

Define a set $$\mathcal{D}$$ of subsets of $$\mathbb{R}^n$$ by $$\mathcal{D} = \{ \mathbb{Q}^n \cup U \mid U \subseteq \mathbb{R}^n \setminus \mathbb{Q}^n \}$$ Then $$\mathcal{D}$$ is in bijection with $$\mathcal{P}(\mathbb{R}^n \setminus \mathbb{Q}^n)$$, which has cardinality $$2^{2^{\aleph_0}}$$.

Every set in $$\mathcal{D}$$ is dense in $$\mathbb{R}^n$$ since $$\mathbb{Q}^n$$ is dense in $$\mathbb{R}^n$$, so the set of dense subsets of $$\mathbb{R}^n$$ has cardinality $$\ge 2^{2^{\aleph_0}}$$.

But $$\mathcal{P}(\mathbb{R}^n)$$ has cardinality $$2^{2^{\aleph_0}}$$, so the set of dense subsets of $$\mathbb{R}^n$$ has cardinality $$\le 2^{2^{\aleph_0}}$$.