# Neighbourhood, Open Sets and Closed Set in R

## Neighbourhood of a point

A set A âŠ‚ R is called a neighbourhood (nbd) of a point aâˆˆR if there exists an open interval ( a- Îµ, a +Îµ) for some Îµ> 0 such that

a âˆˆ(a - Îµ,a + Îµ) âŠ‚ A

Equivalently A is nbd of a if âˆƒ an open interval I such that a âˆˆ I âŠ‚ A

### Open sets

A set A âŠ‚ R is called an open set if it is neighbourhood of each of its points. Alternatively, a set A âŠ‚ R is called an open set if for each a âˆˆ A there exists some Îµ > 0 such that a âˆˆ( a - Îµ,a + Îµ)âŠ‚ A

A is not an open set if A is not a neighbourhood of even one of its points.

The set R of real numbers is a neighbourhood of all its points. For any real a, we have

a âˆˆ (a - Îµ,a + Îµ)âŠ‚ R âˆ€ Îµ > 0

Thus the set R of real numbers is an open set.

The set Q of rational numbers is not a neighbourhood of any of its points because

a âˆˆ (a - Îµ, a + Îµ) âŠ‚ Q âˆ€ Îµ > 0

and any such interval contains rational as well as irrational points.

So set Q of rational numbers is not an open set.

The set of irrational numbers Qâ€™ = R â€“ Q is not a neighbourhood of any of its points as many interval around an irrational point will also contain rational points. So the set of irrational numbers Qâ€™ is not an open set.

The open interval (a,b) is a neighborhood of all its points since

x âˆˆ (a, b) âŠ‚ (a, b) âˆ€ x âˆˆ (a, b)

So every open interval (a,b) is an open set. (-âˆž,a)and (b,âˆž)are also open sets.

The interval (a,b] is a neighbourhood of all its points except b since

b âˆˆ (b - Îµ, b + Îµ) âŠ‚ (a, b] âˆ€ Îµ > 0 (since b + Îµ > b)

Similarly [a,b) is neighbourhood of all its points except a and [a,b] is a nbd of all its points except a and b. Thus (a,b]; [a,b) and [a,b] are not open sets.

(-1/n,1/n) is a nbd of 0 âˆ€ n âˆˆ N

(-1/n,1/n) is an open set âˆ€ n âˆˆ N

The set of natural numbers N, the set of whole numbers W, and, the set of integers Z are not neighbourhood of any of their points since N, W, Z cannot contain an interval. Thus the set of natural numbers (N), set of whole numbers (W) and the set of integers (Z) are not open sets.

Any non-empty finite set cannot be neighbourhood of any of its points as it cannot contain an interval which has infinite number of points. So, a finite set is not an open set.

The null set âˆ… is open in the sense that there is no point in the null set âˆ… of which it is not a neighbourhood.

### Some Theorems and their Proofs about Open Set

#### Theorem: The union of an arbitrary family of open sets is an open set.

Proof : Let {GÎ»: Î» âˆˆ Î›} be an arbitrary family of open sets.

Let G = â‹ƒ(Î»âˆˆÎ›) GÎ». We have to show that G is an open set.

Let a be any element of G = â‹ƒ(Î»âˆˆÎ›) GÎ» Then a âˆˆ GÎ» for some Î» âˆˆ Î›

Since GÎ» is given to be open set so it is neighbourhood of each of its points and hence GÎ» is neighbourhood of a.

Thus there exists some Îµ > 0 such that

a âˆˆ (a - Îµ, a + Îµ) âŠ‚ GÎ»

As GÎ» âŠ‚ â‹ƒ(Î» âˆˆ Î›) GÎ» = G

Therefore, a âˆˆ (a - Îµ, a + Îµ) âŠ‚ G

â‡’ G is nbd of a

As a is any point of G therefore G is neighbourhood of each of its points and hence G is open set.

#### Theorem : The intersection of a finite number of open sets is an open set.

Proof : We first prove the intersection of two open sets G1 and G2 is an open set.

Let a âˆˆ G1 âˆ© G2 â‡’ a âˆˆ G1 and a âˆˆ G2

Since G1 and G2 are open sets therefore they are neighbourhoods of each of their points, in particular G1 and G2 are nbds of a. therefore there exists Îµ1 >0 and Îµ2 >0 such that

a âˆˆ (a-Îµ1,a + Îµ1 ) âŠ‚ G1

a âˆˆ (a - Îµ2,a + Îµ2 ) âŠ‚ G2

Let = Min{Îµ1,Îµ2 } so that Îµ â‰¤ Îµ1 and Îµ â‰¤ Îµ2. Thus

a âˆˆ (a - Îµ, a + Îµ) âŠ‚ (a-Îµ1, a + Îµ1 ) âŠ‚ G1

a âˆˆ (a - Îµ, a + Îµ) âŠ‚ (a - Îµ2, a + Îµ2 ) âŠ‚ G2

Therefore, a âˆˆ (a - Îµ, a + Îµ) âŠ‚ G1 âˆ© G2

This implies that G_1âˆ©G_2 is nbd of a

Since a was any point of G1 âˆ© G2 therefore G1 âˆ© G2 is nbd of each of its points and hence G1 âˆ© G2 is open.

Now consider three open sets G1, G2, G3. Then,

Now RHS is an intersection of two open sets.

G1 âˆ© G2 and G3 is an open set and so G1 âˆ© G2 âˆ© G3 is an open set.

Proceeding like this if G1,G2,G3,â€¦, Gn are finite number of open sets,then,

#### Theorem: The intersection of the family of all neighbourhoods of a point a is {a}.

Proof: Let {Na} be family of all neighbourhoods of a point a.

We have to show â‹‚ Na = {a}

Let b â‰ a and Îµ =|a - b| then,

(a - Îµ, a + Îµ) is nbd of a

and b âˆˆ (a - Îµ, a + Îµ) because if b âˆˆ (a - Îµ, a + Îµ) then a - Îµ < b < a + Îµ

or â€“ Îµ < b - a < Îµ or |b-a| < Îµ which is a contradiction.

Since b âˆˆ (a - Îµ,a + Îµ) therefore b âˆˆ âˆ© Na

So any b â‰  a results in b âˆˆ âˆ© Na.

Hence â‹‚ Na = {a}

#### Theorem: Every open set is a union of open intervals.

Proof: Let G be any open set. By definition for each x âˆˆ G,âˆƒ some Îµx > 0 such that

x âˆˆ (x - Îµx ,x + Îµx ) âŠ‚ G

Thus G âŠ‚ â‹ƒ(xâˆˆG) (x - Îµx , x + Îµx ) âŠ‚ G

Hence G = â‹ƒ(xâˆˆG) (x - Îµx , x + Îµx )

### Closed Set

A set A âŠ‚ R is called a closed set if and only if its complement Aâ€™ = R â€“ A is an open set.

Thus A is closed â‡” R-A is open

The set of real numbers R is closed set as R'= âˆ… is an open set. Therefore the set of real numbers R is both open set and closed set.

The set of rational numbers Q is not closed set as Qâ€™ the set of all irrational numbers is not an open set. Also the set of irrational numbers Qâ€™ is not a closed set as (Qâ€™)â€™ = Q (the set of rational numbers) is not an open set. So the set of all rational numbers Q and the set of all irrational numbers Qâ€™ are both neither open set nor closed set.

Every closed interval [a,b] is a closed set as R - [a,b] = (-âˆž,a) âˆª (b,âˆž) being union of two open sets is an open set.

[a,b) is not a closed set as R - [a,b) = (-âˆž,a) âˆª [b,âˆž) is not open set as it is neighbourhood of all its points except b.

Similarly (a,b] and (a,b) are not closed sets.

The empty set Î¦ is closed set as Î¦â€™ = R is open set. Therefore the null set Î¦ is both open set and closed set.

### Some Theorems and their Proofs on Closed Set

#### Theorem: The intersection of an arbitrary family of closed sets is closed.

Proof: Let {AÎ» âˆ¶ Î» âˆˆ Î›} be an arbitrary closed family of closed sets.

We have to show that

A= â‹‚(Î»âˆˆÎ›) AÎ» is a closed set.

By De Morganâ€™s law

Since each AÎ» is closed therefore each R - AÎ» is open set. Therefore Aâ€™ being arbitrary union of open sets is open set. Hence A is closed set.

#### Theorem: The union of a finite number of closed sets is a closed set.

Proof: Let A1, A2,â€¦,An be n closed sets. Then R - A1,R - A2,â€¦,R - An are open sets.

Therefore (R - A1) âˆ© (R - A2)â€¦âˆ© (R - An is an open set. (Since finite intersection of open sets is open set)

By De Morganâ€™s law

( R - A1 ) âˆ© (R - A2 )â€¦âˆ© (R - An ) = R - {A1 âˆª A2â€¦âˆª An }

Which implies R - {A1 âˆª A2â€¦âˆª An } is an open set.

Hence A1 âˆª A2â€¦âˆª An is closed.

##### Note: The union of an arbitrary family of closed sets may not be closed set.

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