Dense subsets in Metric spaces. Let $(X, d)$ be a metric space. $A \subset X$ is said to be dense if the closure of the set $A$ is X. Recall that the closure of the set $A$ means that the set itself along union with all of its limit points i.e. $\bar{A}=A \cup A^d$, where $A^d=$ all limit points of $A$ in $X$.
If look at some examples like $\mathbb{R}$ with usual metric spaces has infinitely many dense subsets. Let $X$ be an uncountable set, can we put a metric on $X$ such that $X$ has only finitely many dense subsets? Or does there exist an uncountable metric space having only finitely many dense subsets?
Or, for any natural number $n$ can we always produce a metric(Countable and uncountable in both cases) space with exactly  $n$ dense subsets?
 A: By taking the disjoint union of $n$ copies of $\{0\} \cup \{1/k\}$ we get a metric space with exactly $2^n$ dense subsets.
To make the space as large as you want, add in that many isolated points, which have to all be included in any dense subset. This recovers the discrete metric space example for $n = 0$, as pointed out by geetha in the comment.
On the other hand, if $A$ is dense, any superset of $A$ is also dense. So if $X \setminus A$ is infinite then there are infinitely many dense subsets of $X$ (by taking the union of $A$ with any subset of $X \setminus A$).
So for there to be finitely many dense subsets, $X \setminus A$ has to be finite for any dense $A$.
Now, consider all $B$ such that $X \setminus B$ is dense.
Each $B$ is finite and there are finitely many of them, so their union $B_*$ is finite as well.
We'll show that $X \setminus B_*$ is dense. Then the number of dense subsets is the same as the choices of $B \subset B_*$, which is $2^{|B_*|}$.
Let $b \in B_*$ and $U$ be any neighborhood of $b$.
Then there is some $B \subset B_*$ with $b \in B$ and $X \setminus B$ dense. So $U \cap (X \setminus B)$ is infinite. Since $B_*$ is finite, $U \cap (X \setminus B_*)$ must be infinite as well, so $b$ is an accumulation point of $B_*$.
