In a metric space is a dense subset of a dense subspace dense in the space itself? In a metric space is a dense subset of a dense subspace dense in the space itself?
I think that must be false, but I couldn't think of any examples of the contrary.
 A: It is true. Suppose $(X,d)$ is a metric space and we have inclusions $A \subseteq B \subseteq X$ where all the sets get the induced topology/metric. Suppose $A$ is dense in $B$ and $B$ is dense in $X$.
We show that $\operatorname{cl}_X(A) = X$, which will show denseness of $A$ in $X$.
Let $x \in X$. Let $\epsilon > 0$. Then since $B$ is dense in $X$ there is $b \in B$ with $d(b,x) < \epsilon/2$. Since $A$ is dense in $B$, there is $a \in A$ with $d(a,b) < \epsilon/2$. But then $d(a,x) \leq d(a,b) + d(b,x) < \epsilon$ so we have shown that $x \in \operatorname{cl}_X(A)$.
A: To add to the main answer. It's true for any topological space (no metric required). Suppose $X$ is a topological space, and $A\subseteq B\subseteq X$ with $A$ dense in $B$ and $B$ dense in $X$. Here: "$A$ dense in $B$" means $A$ is dense in the subspace topology on $B$.
Let $U\subseteq X$ be a nonempty open set. Then $U\cap B$ is nonempty since $B$ is dense in $X$. So $U\cap B$ is a nonempty open set in the induced topology on $B$. So $U\cap B\cap A$ is nonempty since $A$ is dense in $B$. So every nonempty open subset of $X$ intersects $A$ nontrivially, i.e., $A$ is dense in $X$.
