Intuition of open and closed sets? A set $S$ is said to be open in a topological space $X$. If it doesn't contain its boundary and closed if it contains its boundary.is this intuition right,then how can open and closed set be depicted in this manner.  
 A: Let $X$ be a topological space and $A\subseteq X$
$\partial{A}=\varnothing$ $\iff$ $A$ is open and closed.
Where $\partial{A}$ denotes the boundary of $A$.
Forward direction:
Since $\partial{A} \cap A$ $=$ $\varnothing$, it means that $A$ does not contain its boundary and thus, it is closed.
Since $\partial{A} = \varnothing$, it follows that $\overline{A}\subseteq A$, hence $A$ is closed.
Backwards direction:
If $A$ is both, open and closed, then $A \cap \partial{A}= \varnothing$. But, $A$ contains its boundary, $\partial{A}$. So $\partial{A} = \varnothing$.
In fact, if you consider any discrete topological space, $X$. Then every subset of $X$ is both, open and closed and consequently, has no boundary.
Furthermore, given any topological space, $X$. $X$ and $\varnothing$ are simultaneously, closed and open in any topology (since they are complements of each other), so their boundary is always empty.
Geometrically, it is easy to see that if a set does not contain its boundary, then it is equal to everything "inside it". This can be made formal through the following proposition:
Let $X$ be a topological space and $A\subseteq X$.
$A$ does not contain its boundary if and only if $A$ is open.
The proof is as follows:
If $A$ does not contain its boundary then $A\cap \overline{A^c} =\varnothing$ and so the complement of $A$ is closed which is equivalent to saying that $A$ is open. On the other hand, if $A$ is open, then as mentioned, it is equal to its interior. So $\partial{A}\cap A=\varnothing$.
Take a circle in the plane, centred anywhere you like, remove its boundary points, then you have an open disk, which is an open set. If you include the boundary points, then the set is closed. 
(Note: For any subset $B$ of a topological space $X$, $\overline{B}$ is the closure of $B$)
Example:
Consider $\{$ $(x,y)\in \mathbb{R}^2$: $x^2+y^2<1$ $\}$  where $\mathbb{R}^2$ is equipped with the standard metric topology.
Observe, that this set is equal to its interior, since it is an open disk. Hence it does not contain its boundary. I claim that the closure of $A$, is the set $\overline{A}=$ $\{$ $(x,y)\in \mathbb{R}^2$ $:$ $x^2+y^2\leq 1$ $\}$. (Prove this. I will post a proof later), and thus its boundary is the set $\partial{A}=$ $\{$ $(x,y)\in \mathbb{R}^2$ $:$ $x^2+y^2=1$ $\}$
