# Neighborhood of a Point

Let $$\left( {X,\tau } \right)$$ be a topological space. A subset $$N$$ of $$X$$ containing $$x \in X$$ is said to be the neighborhood of $$x$$ if there exists an open set $$U$$ containing $$x$$ such that $$N$$ contains $$U$$, i.e.

\[x \in U \subseteq X\]

A neigborhood of a point is not necessarily an open set. However, if a neighborhood of a point is an open set, we call it an open neighborhood of that point.

If $$X = \left\{ {a,b} \right\}$$ with topology $$\tau = \left\{ {\phi ,\left\{ a \right\},X} \right\}$$ (known as a Sierpinski space), then $$\left\{ a \right\}$$ and $$X$$ are neighborhoods of $$a$$ because we can find an open set $$\left\{ a \right\}$$ such that

On the other hand, $$X$$ is the only neighborhood of $$b$$ because we can find the open set $$X$$ such that

\[b \in X \subseteq X\]

As another example, let $$X = \left\{ {a,b,c,d} \right\}$$ with topology $$\tau = \left\{ {\phi ,\left\{ a \right\},\left\{ b \right\},\left\{ {a,b} \right\},X} \right\}$$ then $$\left\{ a \right\},\left\{ {a,b} \right\},\left\{ {a,c} \right\},\left\{ {a,d} \right\},\left\{ {a,b,c} \right\},\left\{ {a,c,d} \right\},X$$ are neighborhoods of $$a$$. Similarly, $$\left\{ b \right\},\left\{ {a,b} \right\},\left\{ {a,c} \right\},\left\{ {b,d} \right\},\left\{ {a,b,c} \right\},\left\{ {a,b,d} \right\},\left\{ {a,c,d} \right\},X$$ are neighborhoods of $$b$$, and $$X$$ is the only neighborhood of $$c$$ and $$d$$. It is clear from this illustration that a point $$x$$ may have more than one neighborhood.

**Neighborhood System**

Let $$\left( {X,\tau } \right)$$ be a topological space. The set of all neighborhoods of a point $$x \in X$$ is said to be a neighborhood system of $$x$$. It is denoted by $$N\left( x \right)$$. The above example shows this neighborhood system.

**Theorems**

• The topological space $$X$$ itself is a neighborhood of each of its points.

• A subset of a topological space is open if and only if it is the neighborhood of each of its own points.

• The intersection of two neighborhoods of a point is also its neighborhood in a topological space.

• The union of two neighborhoods of a point is also its neighborhood in a topological space.

• If $$A$$ is a neighborhood of $$x$$ and $$A \subset B$$, then show that $$B$$ is also a neighborhood of $$x$$.

• If $$A$$ is a neighborhood of $$x$$, then show that there exists an open set $$B$$ such that $$B$$ is also a neighborhood of $$x$$ and $$A$$ is a neighborhood of each point of $$B$$.

• The neighborhood system of a point is a non empty set.

• The intersection of a finite number of the neighborhoods of a point is also its neighborhood.

• Any subset $$M$$ of a topological space $$X$$ which contains a member of $$N(x)$$ also belongs to $$N(x)$$.

• Each neighborhood of a point of a cofinite topological space is open.

karim dad

February 14@ 5:13 pmi can’t understand that when topology is given of a non empty set X,(i.e T={{ },{a},{b},{a,b},X } then to find nbhd of a point ,we see those open sets of X,that are subsets of X or those subsets that are exists in topology.I.i.e, if we choose subsets those are nbhds of a point ,a, which are exist in topology,are {a},{a,b},X. But if we choose nbhds from all subsets of X,then all those which are given in above example,but if we choose nbhds of c,from all subsets of X,then {c},{a,c},{b,c},{c,d},{a,b,c},{a,c,d},x. but in given topology,nbhd of a number c is the set only X. so finally my question is that, please tell me,when we choose nbhd of a point (i.e in a topological space),either we choose all those subsets that contains that point from topology or all origional subsets of X.

Rashmi B.

August 18@ 6:12 pmFrom example 3, I don’t get it how to prove, can please explain it.

Shah mansoor

December 9@ 4:52 pmyou have a mistake in neighborhood of ‘b’

which is that; Nb:{a,c} which is not true the true neighborhood is {b,c}

Murerwa

December 31@ 12:09 amYou did well but about nbh of element d make it clear as well