open set wrt metric topology, what does it mean ?? what is the difference between an open set ( where one can simply note whether or not it includes its boundary points) and a set which is open with respect to a metric topology, Or how would we notice it like the way we notice inclusion of boundary points usually ?
My teacher never explained this to us but has given us a question concerning it .
 A: I'll try to give an intuitive explanation:
So an open set is the same as a set that is open.
The most direct concept of an open set could come from sets like $(a,b) \subset \mathbb R$, which as you described that it does not include its boundary.
Then if we apply some abstraction, we find that what is essential to the concept of open could be established purely by distance - that is, given any set, if we have a well-defined distance function for any of two elements in that set, we get a "metric space". Then we could define open set as a set that all its elements are "interior points" of the set. So here, a set is open or not depends on what "distance function" you define.
But we could apply the abstraction even further - we find "open" itself is a very interesting concept, and could be related to lots of other interesting concepts (like continuous, separation, etc.). Thus for an arbitrary set if we define a collection of subsets, and call them open - that is a "topological space". So here, a set is open or not depends on if you define it as open or not (i.e. how you define your topology).
A: A subset $ U$ of a top'l space $S$ is open  iff $U\cap \partial U=\emptyset,$ where $\partial U$ is the boundary of $U.$ That is, $\partial U=\overline U\cap \overline {(S \setminus U)}.$ The notation Fr$(U)$ is also used for $\partial U.$ ("Fr" is for "Frontier".)
(i).Suppose $U$ is open. Then  $S$ \ $U =\overline {S \setminus U}$ so $$U\cap \partial U=U\cap (\;\overline U\cap \overline {S \setminus U}\;)=$$ $$=U\cap (\;\overline U\cap (S \setminus U)\;)=$$ $$=(U\cap \overline U)\cap (S \setminus U)=$$ $$=U\cap (S \setminus U)= \emptyset.$$
(ii). Suppose $U\cap \partial U=\emptyset.$ Then $U\subset S\setminus (\overline {S\setminus U})$ because $$\emptyset=U\cap (\;\overline U \cap \overline {S \setminus U}\;)=$$ $$=(U\cap \overline U)\cap (\overline {S\setminus U})=$$ $$=U\cap (\overline {S\setminus U}).$$  .Also $\overline {S\setminus U}\supset S\setminus U$. So we have  $$U\subset S \setminus (\overline {S\setminus U}))\subset S \setminus (S\setminus U)) =U.$$ Therefore $U=S \setminus \overline {(S\setminus U)}.$ That is, $S$ is the complement of the closed set $\overline {S\setminus U},$ so $U$ is open.
