are these closures equal in metric space

Question

I have a question. let $M$ be a metric space. $A, B$ are its subsets such that $A{\subset}B{\subset}M$.
To my understanding, $A$ is dense in $B$ if the closure of $A$ in B!!! equals $B$. and $B$ is dense in M, if the closure of $B$ in $M$!!! equals M.
To prove $A$ is dense in $M$, I am going to show the closure of $A$ in $M$!!! equals $M$.
My question however is this: Can I claim the closure of $A$ in $B$ equals the closure of $A$ in $M$? I think my claim is intuitively correct. however what do you think? if yes, how to prove this? ps. By !!! I just mean this is the point of my question, where I am kinda confused. clearly not a mathematical symbol.

Answer 1

Suppose $C\subseteq M$ is a closed subset of $M$ containing $A$. Then $C\cap B$ is closed in $B$ and contains $A$, hence contains the closure of $A$ in $B$. Since the closure of $A$ in $B$ equals $B$, we have $C\cap B\supseteq B$ which implies $C\supseteq B$. Since the closure of $B$ in $M$ is equal to $M$, we know the only closed subset containing $B$ is $M$. Therefore, $C=M$. Therefore, the closure of $A$ in $M$ is equal to $M$.