What is the relation between Neighbourhood of a point,Interior point and open set? I want to know what is the relation between of Neighbourhood of a point,Interior point and open set?
Definition: A set $N \subset \mathbb{R}$ is called the $\textbf{neighbourhood}$ of a point a, if there  exists an interval I containing a and contained in N, i.e ,$$a\in I \subset N $$.
Definition:     A point x is an interior point of a set S if S is a nbd of x. In other words, x is an interior point of S if $\exists $ an open interval $(a,b)$ containing x and contained in S , i.e., $$x\in(a,b)\subseteq S$$.
Definition: A set S is said to be open if it is a nbd of each of its point, i.e, $x\in S$, there exists an open interval $I_x$ such that $$mx \in I_x \subseteq S $$.
 A: Intuitively:


*

*A neighbourhood of a point is a set that surrounds that point. That is, if you move a sufficiently small (but non-zero) amount away from that point, you won't leave the set.

*An interior point of a set is a point that is surrounded by the set. Note that this is really the same relation, only the subject has changed.

*An open set is one which surrounds all its points. That is, wherever you are in that set, a sufficiently small move will not get you out.
A: General definitions:


*

*Neighborhood: a neighborhood of a point $x$ is a set that contain some open set that contains $x$. For the case of the standard topology of $\Bbb R$ an open set in composed of arbitrary union of open intervals, and the minimal expression of an open set is an open interval. Then if a set $S$ contain some open interval $I$ that contain $x$ then $S$ is a neighborhood of $x$, i.e.


$$x\in I\text{ and } I\subseteq S\iff S\text{ is a neighborhoof of } x $$
(for the standard topology of $\Bbb R$)


*Interior point: an interior point of a set $S$ is a point of an open set $U$ contained in $S$. For the case of the standard topology on $\Bbb R$ because every open set is composed of the union of open intervals (that are the minimal expression of open sets in the standard topology of $\Bbb R$) then this mean that $x$ is a interior point of $S$ if exists some open interval $I$ that contains $x$ and $I$ is contained in $S$, i.e.


$$x\in I\text{ and }I\subseteq S\iff x\text{ is an interior point of } S$$
(for the standard topology of $\Bbb R$)


*Open set: the open sets are the definition of a topology in some space, i.e. the collection of subsets $\mathcal T$ of some space $G$ is a topology of $G$ if and only if


*

*the empty set and $G$ belongs to $\mathcal T$

*the arbitrary union of elements of $\mathcal T$ belongs to $\mathcal T$

*the finite intersection of elements of $\mathcal T$ belongs to $\mathcal T$ 
Any element of $\mathcal T$ is named an open set. The standard topology of $\Bbb R$ contains all open intervals, so it contains too the arbitrary union of open intervals, the empty set, $\Bbb R$ itself and any finite intersection of all these elements.
A: It might help to look at a simple example.  Let $S = [0,1]$.  This is a neighbourhood of each point of $(0,1)$, but not a neighbourhood of $0$ or $1$.
Since it is not a neighbourhood of $0$ or $1$, and those points are in $S$, it is not an open set.  The interior of $S$ is $(0,1)$.
A: Okay, I'm going to give an analysis metric space definition.
Preliminary:  First we need to consider what "universe" we are defining these points and sets to belong to.  The OP seems to be assuming this universe is $\mathbb R$ which is fine.  But it could be $\mathbb R^k$ as well.  Or it could be more abstract and simply be any "metric space".  I am not going to explain what a metric space is here but simply say it is a "universe" of points where somehow a distance between two points is somehow defined.  In $\mathbb R$ the distance between $x$ and $y$ is $|x - y|$.  In $\mathbb R^k$ it is $||x, y|| = \sqrt{x^2 - y^2}$.  It doesn't matter how distance is defined just that it is.
First:  If we have a point $x$ we need somehow to describe the idea of "a bunch of points that immediately surround $x$ in all directions".
If our universe is $\mathbb R$ any interval $I = (a,b)$ where $a < x < b$ will do.  So will $[a,b]$ where $a < x <b$.  But $[x,b]$ or $[x,b)$ will not as it doesn't include any points immediately to the left of $x$.
In $\mathbb R^2$ for $(x,y)$ is could be a rectangle $[a,b] \times [c,d]$ where $a < x < b$ and $c < y < d$.  or it could be the disc centered at $(a,c)$ with radius $r$ st $(a - x)^2+ (c -y)^2 < r^2$.
Or if it's any metric space in general, I like to define a "open ball" as set $N = $ "all the points, $z$ so that $||x,z|| < d$ for some distance $d$.
But it doesn't matter which of these methods we use to find "a bunch of points immediately surrounding $x$ in all directions".  Any one of them will do
As the OP wants to use an interval we will.  But it needs to be an open interval.  $x \in [x, b]$ simply won't do as it doesn't include any points immediately to the left of $x$.
.....
Okay now we can start!
A neighborhood of $x$ is any set that contains ""a bunch of points that immediately surround $x$ in all directions".
In other words: Definition:  A set $S$ is a neighborhood of $x$ if there is an open interval $I_x$ such that $x \in I_x$ (in other words that $I_x = (a,b)$ and $a < x < b$) and $I_x \subset S$.
It actually wouldn't change things very much to define that a nieghborhood is an open interval.  It wouldn't hurt things but it wouldn't help things either.  I'm personally very used to defining a neighborhood as being an "open ball" as I described above.  (In $\mathbb R$ an open ball is the exact same thing as the open interval $(x - d, x + d)$).
But we will use the open interval criterion as a definition.
So.... a neighborhood is any set around x that contains points that "completely surround"  x  i.e.  a set that contains an open interval containing x. 
Now interior points:  An interior point of a set $S$, is a point in $S$ that is completely surrounded by points of $S$.  If for example $S = [a,b]$ then if $a < x < b$ then$x$ is "smack dab in the middle" and absolutely surrounded by points of $S$.  But the two points $a$ and $b$ are not.  All points except $a$ and $b$ are interior points of $S$.
Definition: A point $x \in S$ is an interior point of $S$ is a neighborhood of $x$.  In other words $x \in S$ is an interior point of $S$ if there exists an open interval $I_x$ so that $x \in I_x \subset S$.
Definition: A set $S$ is open if every point in $S$ is an interior point of $S$.  In other words, for all $x \in S$ there are open intervals (maybe very small) such that $x \in I_x \subset S$.
Example: consider $S = (-\infty, 3] \cup (5, 7]$.  $S$ is not open because $3$ and $7$ are not interior points.  There is no $(a,b)$ so that $a < 3 < b$ and $(a,b) \subset S$.  Same thing for $7$.
But all other points are interior points.  If $x < 3$ then $(x-1, 3) \subset S$ so $x$ is interior point.  If $5 < y < 7$ then $(5,7)\subset S$ so $y$ is an interior point.
So $W = (-\infty, 3) \cup (5,7)$ would be open.
...
So the relationship...
A neighborhood of a point surrounds the point completely (but maybe only for a very small distance).
An interior point of a set is a point in the set that is completely surrounded by the set.
A set if open if every point in the set is completely surrounded by other points in the set.  (In other words no point is on the "edge" of the set where "right next to it" the set "ends".)
