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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 ???

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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 $\le 2^{2^{\aleph_0}}$.

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