Closed set and set, closed in $\mathfrak M$ I've read in my textbook that a set $A$ is called closed if it contains its limit points, i.e. $A'\subseteq A$. But then, coming to next chapter, I came across a term of set $B$, closed in metric space $\mathfrak M$ (or in its set? this was not quite clear in the book).
I'm not sure how I should understand it. After some thinking it seems that the first definition of a "just closed" set without any reference to where it is closed implied some set which would contain the limit points we're interested in. From this, $B$ is closed in $\mathfrak M$ would mean that 
$$( x\in\mathfrak M)\wedge(x\in B')\implies x\in B.$$
Thus, the set $B$ may be closed in $\mathfrak M$, but not closed in $\mathfrak N$, e.g. for the case $B=\mathbb R\setminus\mathbb N$, $\mathfrak M=\mathbb R\setminus\mathbb N$, $\mathfrak N=\mathbb Q$, where $\mathfrak M$ and $\mathfrak N$ have the standard metric $\rho(x,y)=|x-y|$.
Is my understanding right?

It's true that when we talk about closed sets, we need to specify in what metric space it's closed in. In fact, this generalizes to not just metric spaces, but topological spaces.
In general, if $X$ is a topological space and $Y \subseteq X$ is given the subspace topology, then we say that a subset $A \subseteq Y$ is closed in $Y$ iff there is some set $C \subseteq X$ such that $A = C \cap Y$ and $C$ is closed in $X$.
For example, consider $X = \mathbb R$ and $Y = (2, 7) \subseteq X$, each with the standard topology (that is, view each of them as a metric space with the standard metric $\rho(x, y) = |x - y|$). Notice that $Y$ is not closed in $X$. Now consider $A = [4, 7)$. Notice that $A$ is not closed in $X$, but we do have that $A$ is closed in $Y$ (since $A = [4, 8] \cap Y$ and $[4, 8] \subseteq X$ is closed in $X$).
