How does metric decide if set is open or closed? I have a problem in which I am to decide if $A \subset (X,d)$ is open or closed.
There are two subproblems,  in which the set $A$ is the same but the metrics are different. My question is how does the metric "influence" if the set is open or closed? 
My notion of a metric is that is some kind of "distance" function inherent to the set but I'm having a hard time seeing how this affects open and closedness. I guess my question is very general. 
Thanks in advance.
 A: You can define openness (as your book almost certainly does) in terms of the metric:

A set $A \subseteq X$ is open if, for every $a \in A$, there is some real $r > 0$ such that the ball of radius $r$ (as measured by $d$), centered at $a$, is contained in $A$.

The general idea is that a set is open if no point in it is "right on the edge," i.e., around every point is at least little neighborhood of points still in the set, as measured by the metric. (A closed set is one whose complement (i.e., the set of everything outside it) is open.)
A: A set is open if for every point in the set there exists $\epsilon-ball$, with $\epsilon > 0$ such that all points lying in this ball belong to the set. Metric is used to determine which points lie in this $\epsilon-ball$. For example on real line with usual distance the set {1,2,3} is not an open set because these are just points and regions around each point contain points that are not included in the set. But if we define the metric(you are right this corresponds to distance with particular properties) as $d(x,y)=1$ if $x \neq y$ and $d(x,y)=0$ if $x = y$, then the same set becomes an open set, because for any $\epsilon <1$ the $\epsilon-ball$ around a point $x$ contains only the point $x$ so all points in the $\epsilon-ball$ are included in the set.
