Point Set Topology - II | Mathematics for Competitive Exams PDF Download

Limit Points

Neighbourhood of a Point

Definition : A set S ⊂ R is called a neighbourhood (nbd) of a point p ∈ R if there exists an open interval (p - ε, p + ε) for some ε > 0,
such that

p ∈ ( p - ε, p + ε) ⊂ S.

Examples

1. The set R of real numbers is nbd. of all its points.

For any real number x, we have

x ∈ (x - ε  x + ε) ⊂ R, ε > 0.

2. The open interval (a, b) is a nbd. of all its points, since

x ∈ (a + ε, b - ε) ⊂ (a,b)

3. The half open interval (a, b] is a nbd, of all its points except b, since for any ε > 0

b ∈ [b - ε, b + ε ) <z ( a, b].    (∵ b + ε > b)

Similarly, [a, b) is a nbd. of all its points except a and [a, b] is a nbd. of all its points except a and b.

4. The set Z of integers is not a nbd. of any of its points, since for any integer p and any

ε > 0,

p ∈ (p - ε , p + ε ) ⊄ Z.

5. Each of the following open intervals.

6. For any p ∈ R, N = {p} is not a a nbd. of p, since for any ε  > 0, (p - ε , p + ε ) ⊂ N.

7. The set Q of rational numbers is not a nbd. of any of its points, for if (a, b) is any open interval around a point p ∈ Q, then

p ∈ (a, b) ⊂ Z.

since between any two real numbers a, b there is always an irrational number which does not belong to Z.

8. The set R  ~ Q of irrational numbers is not a nbd. of any of its points.

9. A non-empty finite set S is not a nbd. of any of its points, if (a, b) is any open interval around any point x ∈ S, then

x ∈ (a, b) ⊄ S,

since (a, b) necessarily contains infinite number of points and S is a finite set.

10. The empty set Ф is a nbd. of each of its points, since there is no point in Ф which is not a nbd.

Example 16: If M is a nbd. of a point p and N ⊃ M, then N is also a nbd. of the point p.

Since M is a nbd. of the point p therefore, there exists an open interval (p - ε, p + ε),

ε > 0 such that

p ∈ (p - ε, p + ε) ∈ M

⇒ P ∈ (p - ε, P + ε) ⊂ M ⊂ N    [∵ N ⊃ M ⇒ M ⊂ N]

⇒ P ∈ (p - ε, p + ε) ⊂ N.

Hence N is a nbd. of p.

Example 17: If M and N are nbds. of a point p, then M ∩ N is also a nbd. of p.

Since M and N are nbds. of p, therefore, there exist ε₁ > 0 and ε₂ > 0 such that

p ∈ (p - ε₁, + ε₂) ⊂ M,
and p ∈ (ε₂ - c2, p + ε₂) ⊂ N.

Let ε = min {ε₁, ε₂} so that ε < ε₁, and ε < ε₂. Thus

....(1)

∴ p ∈ (p - ε, p + ε) ⊂ M ∩ N.

Hence M ∩ N is a nbd of p.

Note. Obviously, p ∈ (p - ε, p + ε) ⊂ M υ N and so M υ N is also a nbd. of p.

Example 18: Show that intersection of a finite number of neighbourhoods of a point is also a neighbourhood of the point.

Let M₁, M₂,M_n be a finite number of neighbourhoods of a point p. Then there exists ε_i > 0 such that

p ∈ ] p - ε_i, p + ε_i [ ⊂ M_i for i = 1, 2,...., n.

Let ε = min {ε₁, ε₂, .... ε_n. Then

p ∈ ] p - ε, p + ε [ ⊂ M_i for i = 1, 2, n

⇒ p ∈ ] p - ε, p + ε [ c    M₁, ∩ M₂ ∩ ... ∩ M_n.

Hence M₁ ∩ M₂ ∩ ... ∩ M_n is a nbd. of p.

Example 19: Define a neighbourhood of a point. Show that the intersection of the family of all neighbourhoods of a point x is {x}.

A subset S ⊂ R is called a neighbourhood of a point p ∈ R if there exists some ε > 0 such that ( p - ε, p + ε ) ⊂ S,

Let {N_x} be a family of all neighbourhoods of a point x. We have to show that ∩ N_x = {x}.

Consider any point y ≠ x and Let ε = I x - y |

then, (x - ε, x + ε) is a nbd. of x and y ∉ (x - ε, x + ε),

since y ∈ (x - ε, x + ε) ⇒ x- ε < y < x + ε ⇒ lx-y| < ε, a contradiction.

Since y ∉ (x - ε, x + e), y ∉ n N_x. Hence n Nx = {x}.

Example 20: (i) Can a set N be a neighbourhood of any point of R ~ N ?

(ii) Can a non-empty finite set be a neighbourhood of any of its points ?

(i) We shall show that N cannot be a neighbourhood of any point of R ~ N.

Let x ∈ R ~ N then, x ∉ N.

Suppose N is a nbd. of x then there exists some c > 0 such that
x ∈ (x - ε, x + ε) ⊂ N ⇒ x ∉ N, which is a contradiction.

Hence the assertion made above is not TRUE.

(ii) The answer of the second part is also No.

Let x ∈ N, where N is a non-empty finite set. If N is nbd. of x, then
x ∈ (x - ε, x + ε) ⊂ N for some ε > 0.

This is impossible, since N is a finite set and (x - ε, x + ε) contains infinite number of elements.

Limit Points

Definition 1 : A real number p is said to be a limit point of a set S ⊂ R if for each ε > 0 there exists at least one point p₁ ∈ S such that

P₁ ∈ (p - ε, p + ε) and p₁ ≠ p.

Definition 2 : The set of all limit points of a set S ⊂ R is called the derived set of S and is denoted as S’.

x ∈ S' ⇔ x is a limit point of S.

Example

1. 0 is the only limit point of the set
2. 1 and -1 are the only limit points of the setNotice that 1, -1 both belong to S and S’ = {1, -1}.

Remarks

• A set may or may not have a limit point.
  For Example, R has limit points but Z has no limit point.
• Limit point of a set may or may not belong to the set.
  For Example, every limit point of R belongs to R and the only limit point 0 of the set
  does not belong to the set i.e., 0 ∉ S’.

Examples :

1. The set I has no limit point for a nbd of m ∈ 1 contains no Point I other than m. Thus the derived set of I is the null set ∅.2. Every point of R is a limit point for every nbd of any of its points contains an infinity of members of R. Therefore, R’ = R.

Examples: Find the limit points of each of the following sets :
(a)(b)(c)(d) D = {x : 0 < x < 1}
(e)(f) F = {3^-n + 5^-n : n ∈ N}
(a) The only limit point of A is 0 i.e., A’ = {0}. Notice that 0 ∉ A.
(b) We have B = The only limit point of B is 1 i.e., B’ = {1}. Notice that 1 ∉ B.(c) The set C = has only two limit points 1,-1. Notice that 1 e C and -1 e C. We have C' = {1, -1}.

(d) We observe that D = [0, 1) and so D' = [0, 1].

(e) First we write all the elements of the set
For m = 1, the elements of E are :...(1)For m = 2, the elements of E are :...(2)For m = 3, the elements of E are...(3)and so on.It is clear that1 is the limit point of the elements in row (1),
1/2 s the limit point of the elements is row (2),
1/3 is the limit point of the elements in row (3),
and so on.
Thus ... are all limit points of E.

Since E contains an infinite number of fractions lying between 0 and 1, so each nbd. of 0 surely contains a positive fraction and as such 0 is also a limit point of E.
Hence (f) We have As discussed in part (e), we observe that

Example. Give an example of a set whose derived set is
(i) void
Ans. If Z is the set of integers, then Z' = Ф

(ii) subset of the given set
Ans. Let S = Then S’ = { 1 , - 1 } ⊂ S.(iii) superset of the given set
Ans. Let S =Then S’ = {0}.S’ is neither a subset nor a superset of S.

(iv) same as the given set
Ans. R' = R or [a, b]’ = [a, b].

Example : Give an example of a set which has
(i) no limit point

Ans. The set Z of integers has no limit point.

(ii) unique limit point
Ans. has 0 as the unique limit point.(iii) two limit points
Ans.has two limit points viz 1 and -1.(iv) three limit points
Ans. has three limit points viz. 1, 2, 3

(v) infinite number of limit points

Ans. The set R of real numbers has infinite number of limit points. In fact every real number is a limit point of R.

(vi) a set whose every point is a limit points of the set
Ans. The set R of real numbers.

(vii) a set none of whose points is a limit point of the set
Ans. The set Z of integers.

(viii) A set with only √3 as its limit point.
Ans. S = Example 21: Define limit point of a set. Show that the set N of natural numbers has no limit point.

Let m be any natural number.
Then is a nbd. of m which does not contain any point of N different from m.Let m ∈ R ~ N i.e., m is any real number but not a natural number. Then we can find some integer j such that j < m < j + 1. Consequently (j, j + 1) is a nbd. of m, which does not contain any point of N. Hence N has no limit point i.e., N' = Ф.

Example 22: Given an example each of :

(i) a bounded set of real numbers

(ii) an unbounded set of real numbers
Also find the set of limit points of both the sets.

(i)is a bounded set of real numbers. Its limit points are 1 and -1.(ii)is an unbounded set of real numbers. 0 is the only limit point of S.

Example 23: Find the derived sets of the following :
(i) (ii)(iii)(iv)

A’ = {1}, B’ = {-1, 1}, C’ = {0}, D’ = ϕ

Example 24: If S and T are sets of real numbers, then show that

(a) S ⊂ T ⇒ S’ ⊂ T’
(b) (S ∪ T)’ = S’ ∪ T’

(c) (S ∩ T)’ ⊂ S’ ∩ T’

Given an example to show that (S ∩ T)’ and S’ ∩ T’ may not be equal.

(a) Let S ⊂ T. We shall prove that S’ ⊂ T’.

Let x ∈ S’ so that x is a limit point of S. Thus for each ε > 0, (x - ε, x + ε) contains a point x, ∈ S such that x₁ ≠ x. Since S ⊂ T, x₁ ∈ S ⇒ x₁ ∈ T.

Consequently, (x - ε, x + ε) contains a point x₁ ∈ T, x₁ ≠ x, and so x is a limit point of T
i.e., x ∈ T’.
∴ x ∈ S' ⇒ x ∈ T’.

Hence S ’ ⊂ T
(b) We know S ⊂ S ∪ T S’ ⊂ (S ∪ T)^,_, by part (a).

Similarly, T ⊂ S ∪ T ⇒ T ’ ⊂ (S ∪ T)’.

S’ u T ' ⊂ (S ∪ T) ’ ...(1)
∴ Now we prove (S ∪ T)’ ⊂ S’ υ T’. ...(2)
Let x ∈ (S ∪ T)' ⇒ x is a limit point of S ∪ T
⇒ every nbd. of x contains a point y ∈ S ∪ T, y ≠ x
⇒ every nbd. of x contains a point (y ∈ S or y ∈ T), y ≠ x
⇒ every nbd. of x contains a point y ∈ S, y ≠ x,
or every nbd. of x contains a point y ∈ T, y ≠ x.
⇒ x is a limit point of S or x is a limit point of T
⇒ x ∈ S’ or x ∈ T’

⇒ x ∈ S’ ∪ T'.

Thus (2) is proved.

From (1) and (2), (S ∪ T)' = S’ υ T.

(c) We know S ∩ T ⊂ S    ⇒    (S ∩ T)' c S’, by part (a)

and    S ∩ T ⊂ T    ⇒    (S ∩ T)’ ⊂ T.

(S ∩ T)’ ≠ S’ ∩ T^,.

To prove that the equality does not hold, we take

S = ]1, 2[ and T = ]2, 3[ so that S ∩ T = ϕ.

⇒ S’ = [1, 2], T = [2, 3] and

(S ∩ T)’ = ϕ' = ϕ).

Now S’ ∩ T' = {2} and (S ∩ T)’ = ϕ.

Thus (S ∩ T)' ≠ S’ ∩ T.

Some Theorems of Limit Points

Theorem : A point p is a limit point of a set A if and only if every neighbourhood of p contains infinitely many points of A.

Proof. The condition is necessary.

Let p be a limit point of A. We shall show that every nbd. of p contains infinitely many points of A. Suppose this is false then, there exists some ε > 0 such that (p - c, p + e) contains only a finite number of points, say pt, p2, pn ; which are different from p.

Let δ = min {Ip - p₁ I , I p - p₂ I ,..., I p — p_n I } (δ > 0)

⇒ δ < I p - P_i I for i = 1, 2 ..., n

⇒ (p - δ, p + δ) contains no point of A other than p

⇒ p is not a limit point of A, which is a contradiction.

Hence every nbd. of p must contain infinitely many points of A.

The condition is sufficient.

Suppose every nbd. of p contains infinitely many points of A. Surely then, every nbd. of p contains a point of A different from p and so p is a limit point of A.

Corollary. Show that a finite set has no limit points.

Proof. Let S be a finite set. Let if possible p be a limit point of S.

By Theorem for every ε > 0, (p - c, p + c) must contain an infinite number of elements of S, which is impossible, as S is a finite set. Hence p cannot be a limit point of S.

Theorem : (Bolzano-Weierstrass Theorem)

Every infinite bounded set of real numbers has a limit point.

Proof. Let S be any infinite bounded set of real numbers with bounds

m = inf S and M = sup S i.e., m < x < M ∀ x∈S.    ...(1)

We define a set of real numbers as follows :

T = {x : x Exceeds only a finite number of elements of S}.    ...(2)

From (2), it follows that

⇒ x g S’ u T'.

Thus (2) is proved.

From (1) and (2), (S ∪ T)' = S’ ∪ T'

(c) We know S ∩ T ⊂ S ⇒ (S ∩ T)' ⊂ S’, by part (a)

and S ∩ T ⊂ T ⇒ (S ∩ T)’ ⊂ T.

(S ∩ T)’ ⊂ S’ ∩ T^,.

To prove that the equality does not hold, we take

S = ]1, 2[ and T = ]2, 3[ so that S ∩ T = ϕ.

⇒ S’ = [1, 2], T = [2, 3] and

(S ∩ T)’ = ϕ^, = ϕ.

Now S’ ∩ T' = {2} and (S ∩ T)’ = ϕ.

Thus    (S ∩ T)' ≠ S’ ∩ T.

Some Theorems of Limit Points

Theorem : A point p is a limit point of a set A if and only if every neighbourhood of p contains infinitely many points of A.

Proof. The condition is necessary.

Let p be a limit point of A. We shall show that every nbd. of p contains infinitely many points of A. Suppose this is false then, there exists some ε > 0 such that (p - ε, p + ε) contains only a finite number of points, say p₁, p₂, p_n ; which are different from p.

Let δ = min { I p - p₁ I , I p - p₂ I,... , I p - p_n I } (δ > 0)

⇒ δ < I p - P_i I for i = 1, 2 ...., n

⇒ (p - δ, p + δ) contains no point of A other than p

⇒ p is not a limit point of A, which is a contradiction.

Hence every nbd. of p must contain infinitely many points of A.

The condition is sufficient.

Suppose every nbd. of p contains infinitely many points of A. Surely then, every nbd. of p contains a point of A different from p and so p is a limit point of A.

Corollary. Show that a finite set has no limit points.

Proof. Let S be a finite set. Let if possible p be a limit point of S.

By Theorem for every ε > 0, (p - ε, p + ε) must contain an infinite number of elements of S, which is impossible, as S is a finite set. Hence p cannot be a limit point of S.

Theorem : (Bolzano-Weierstrass Theorem)

Every infinite bounded set of real numbers has a limit point.

Proof. Let S be any infinite bounded set of real numbers with bounds

m = inf S and M = sup S i.e., m < x < M ∀ x∈S.    ...(1)

We define a set of real numbers as follows :

T = {x : x Exceeds only a finite number of elements of S}.    ...(2)

From (2), it follows that
From (1) and (4), we observe that m ∈ T and so T is non-empty.

Again from (1), M > x ∀ x ∈ S (and S is infinite).

From (3), ∀ y ∈ T
⇒ y > only a finite number of elements of S.

Thus M > y ∀ y ∈ T and so M is an upper bound of T i.e., T is bounded above. Since T is a nonempty and bounded above subset of R, so by order-completeness property of R, T must have the supremum, say p i.e.,

sup T = p.

We shall show that p is a limit point of S. Let ε > 0 be arbitrary.

Since p + ε > p and p is an upper bound to T, p + ε ∉ T

⇒ p + ε > infinite number of elements of S, using    (3)

⇒ infinite number of elements of S < p + ε.    ...(5)

Since p - ε < p and p is the l.u.b. of T, so p - ε cannot be an upper bound of T. Consequently, there exists some q ∈ T such that

q > p - ε, i.e., p - ε < q, where q ∈ T.

Now q e T ⇒ q < infinite number of elements of S, by (4).

Thus  p - ε < q < infinite number of element s of S.    ...(6)

From (5) and (6), it follows that any nbd. (p - ε, p + ε) of p contains infinite number of elements of S. Hence p must be a limit point of S.

Remark : The conditions of the Bolzano-Weierstrass Theorem cannot be relaxed.

We have seen in the corollary of Theorem that a finite set (which is of course bounded) can never have a limit point. Further there exist infinite unbounded sets having no limit point.

For Example,

(i)    The set Z of integers is an infinite unbounded set, which has no limit point.

(ii)    The set S = {1², 2², 3², ... n², ...}

is an infinite unbounded set, which has no limit point.

Remark : It may, however, be noticed that

is an infinite unbounded set, which has the limit point viz. 0.

Theorem 1 : Every finite set is bounded.

Proof : Let A = {a₁, a₂, .... a_n} be a finite set and
let h = min(a₁, a₂, .... a_n) and k = max(a₁, a₂, ..., an)
then h < x < k, ∀ x ∈ A
Therefore A is bounded.

Theorem 2 : Every infinite bounded set of real numbers has a limit point.

Proof : Every infinite and bounded subset of R has a limit point i.e. atleast one limit point.

The Archimedean Property Of Real Numbers

If x, y are two positive real numbers, then there exists a positive integer n such that ny > x.

Proof : Suppose there does not exist any positive integer n such that ny > x. Then ny < x ∀ n ∈ N. Consequently, the set

S = {y, 2y, 3y    ny, (n + 1) y, ...}

is non-empty and bounded above by x.

By order-completeness property, S has the supremum, say p.

∴ (n + 1) y < p    ∀ n ∈ N

⇒ ny < p - y    ∀ n ∈ N, (p - y < p, as y > 0)

⇒ p - y (< p) is an upper bound of S, which is a contradiction to the fact that p = sup S.

Corollary 1 : If x be any positive real number, then there exists a positive integer n such that n>x.
Corollary 2 : Let x by any real number and y be a positive real number, then there exists a positive integer n such that ny > x.

Corollary 3 : For any real number x, there exists a positive integer n such that n > x.

Corollary 4 : For any real number x, there exists an integer m such that m < x.

Example 25: If x be any positive real number, then there exists a positive integer n such that 1/x < x.

Applying the Archimedean property, for y = x > 0 and x = 1, there exists a positive integer n such that ny > x nx > 1 ⇒ n > 1/x. Hence 1/n < x.

Remark : The integer n satisfying n < x < n + 1 is called an integral part of x and is denoted by [x].

Example 26: If x and y are real numbers and x > 1, prove that there exists a positive n such that xⁿ > y.

Suppose that conclusion of the problem is false.

Then xⁿ < y ∀ n ∈ N. Consequently, the set
S = {x, x², x³, ..., xⁿ, ...}

is non-empty subset of R, which is bounded above by y. By order-completeness property of R, S has the l.u.b. Let p = l.u.b. S.    ...(1)

Since x > 1, px > p ⇒ p > p/x ⇒ p/x < p

⇒ p/x is not an upper bound of S, by (1) and so there exists some xⁿ ∈ S such that
xⁿ > p/x ⇒ x^{n +1} > p, where x^{n +1} ∈ S.

Thus shows that p is not an upper bound of S, which is a contradiction.

Hence xⁿ > y for some positive integer n.

Intervals

The representation of real numbers as points on a straight line helps us to define four types of intervals.

Let a and b be any two real numbers such that a < b.

1. Open Interval, denoted by ]a, b[, is defined as

(a, b) = {x : a < x < b}.

The end points a and b do not belong to (a, b).

2. Closed Interval, denoted by [a, b], is defined as
[a, b] = {x : a < x < b}.

The end points a and b belong to [a, b].

Semi-Open or Semi-Closed Intervals are defined as follow :

3. (a, b] = {x : a < x < b}.

Here a ∉ (a, b] and b ∈ (a, b].

The interval ]a, b] is said to be open from the left and closed from the right.

4. [a, b) = {x : a < x < b}.

Here a ∈ [a, b) and b ∉ [a, b).

The interval [a, b) is said to be open from the right and closed from the left. The intervals (a, b] and [a, b) are also known as half-open or half-closed intervals.

Open And Closed Sets

Open Sets
Definition : A set S ⊂ R is called an open set if it is a nbd of each of its points.
In Other Words : a set S ⊂ R is called an open set if for each p ∈ S, there exists some ε >
0 such that
p ∈ (p - ε, p + ε) ⊂ S ∀ p ∈ S.
Examples
1. The set R of real numbers is an open set.
2. Every open interval (a, b) is an open set.
3. [a, b) is not an open set, since [a, b) is a nbd. of all is points except a.

Similarly, (a, b], [a, b] are not open sets.
4. The set Z is integers is not an open set,
since Z is not a nbd. of each of its points.
5. G = (1, 2) ∪ (3, 4) is an open set, as G is a nbd. of each of its points.
6. is an open set for each n e N.7. The empty set ϕ is an open set.

8. The singleton set {x} is not an open set, since for any ε > 0, (x - ε, x + ε) ⊄ {x}

9. The set Q of rational numbers is not an open set, since Q is not a nbd. of each of its points.

10. The set R - Q of all irrational numbers is not an open set.
11. The set S = is not an open set, as S is not a nbd. of each of its points.
For Example, (1 - ε, 1 + ε) is not an open set, as S is not a nbd. of each of its points.

For Example, (1 - ε, 1 + ε) contains an infinite number of irrational numbers and so (1 - e, 1 + ε) ⊂ S for each ε > 0.

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

Proof. Let {G_I : λ ∈ A} be an arbitrary family of open set.

Let  . We have to show that G is an open set.

Let x be any element of G = ∪ G_λ
Then x ∈ G_λ for some X ∈ A.

Since G_λ is given to be an open set and x ∈ G_λ so G_λ is a nbd. of x. Thus there exist some ε > 0 such that

x ∈ (x - ε, x + ε) ⊂ G_λ
⇒ x ∈ (x - ε, x + ε) ⊂ G_λ ⊂  G_λ = G

⇒ x ∈ (x - ε, x + ε) ⊂ G ∀ x ∈ G
⇒ G is a nbd. of x, ∀ x ∈ G

It follows that G is a nbd. of each of its points and consequently G = G_λ. is an open set.

Theorem : The intersection of a finite number of open sets is an open set.
Proof. First of all we show that the intersection of two open sets G₁ and G₂ is an open set. Let x be any point of Gt n G₂. Then x ∈ G₁ ∩ G₂
⇒ x ∈ G₁ and x ∈ G₂

⇒ G₁ and G₂ are nbds. of x (∵ G₁ and G₂ are open sets)

⇒ G₁ ∩ G₂ is nbd. of x

⇒ G₁ ∩ G₂ is a nbd. of x for each x ∈ G₁ ∈ G₂.

Thus G₁ ∩ G₂ is an open set.

Now we consider three open sets G₁,G₂, G3. Then
G₁ ∩ G₂ n G₃ = (G₁ ∩ G₂) ∩ G₃.

R.H.S. being the intersection of two open sets viz. G₁ ∩ G₂ and G₃ is an open set and so G₁ ∩ G₂ ∩....∩ G₃ is an open set.

Proceeding in a similar way, if G₁, G_2......, G_n are a finite number of open sets, then Gt n G₂ ∩ ... ∩ Gn is an open set.

Remark : The intersection of an arbitrary family of open sets may not be an open set.

Let us consider G_n = ( - 1/n, 1/n ) ∀ n ∈ N. Then

G₁ = ( - 1, 1), G₂ = (- 1/2, 1/2 ), G₃ = (- 1/3, 1/3), ...

Since every open interval is an open set, therefore {G_n : n ∈ N} is a family of infinite number of open sets.

Clearly, G₁ ∩ G₂ = (- 1, 1) ∩ (- 1/2, 1/2) = (- 1/2, 1/2),

G₁ ∩ G₂ ∩ G₃ = (- 1/2, 1/2) ∩ ( - 1/3, 1/3) = (- 1/3, 1/3)
and so on. Thus at every step, the size of the open interval is reducing and getting closer to 0.
Indeed  G_n = {0} which is not an open set,
}since for any ε > 0, (0 - ε, 0 + ε) = (- ε, ε) ⊄ {0}.

Example : Let 

Prove that l_n is an open set for each positive integer n and 

Hint. Since each open interval is an open set, (x - 1/n, x + 1/n) is an open set for n = 1, 2, 3,
... As shown above Similarly,Example : Is every infinite set open? Justify your answer.

Hint. Every infinite set need not be an open set.

For Example,

(i) the set Z of integers is an infinite set, which is not an open set.
(ii)  is an infinite set, which is not an open set.Definition : A set S ⊂ R is called a closed set if and only if its complement R ~ S is an open set.

S is closed ⇔ R ~ S is open.

Examples

1. The set R of real numbers is a closed set,
as R ~ R = ϕ is an open set.

2. The empty set ϕ is a closed set,
as R ~ ϕ = R is open set.

3. The set R of real numbers is both open and closed.

The empty set ϕ is both open and closed.

4. Every closed interval [a, b] is a closed set,

∵ R ~ [a, b] = (- ∞, a) ∪ (b, + ∞)
and R.H.S. being the union of two open sets is an open set.

5.    [a, b) is not a closed set,

R ~ [a, b) = ( - ∞, a) ∪ [b, + ∞)

is not an open set as the R.H.S. is a nbd. of each of its points except b.

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

6.    [0, 3] ∩ [1, 2] is a closed set,

since [0, 3] ∩ [1, 2] = [1, 2], which is a closed set.

7. F = [1, 2] ∪ [2, 4] is a closed set, since F = [1, 4] is a closed set.

8. Q is not a closed set, since R ~ Q being the set of all irrational numbers is not an open

set. Thus the set Q of rational numbers is neither a closed set nor an open set.

Theorem : (a) The union of a finite number of closed set is a closed set.

(b) The intersection of an arbitrary family of closed sets is a closed set.

(c) Given an Example to show that an arbitrary union of closed sets may not be a closed set.
Proof, (a) Let F₁, F₂, ..., Fn be a finite number of closed sets so that R ~ F₁, R ~ F₂, ..., R ~ F_n are open sets.

(R ~ F₁) n (R ~ F₂) ∩ ... ∩ (R ~ F_n) is an open set.

By De Morgan’s law,

R~(F₁u F₂ ∪ ... ∪ F_n) = (R ~ F₁) n (R ~ F₂) ∩ ... ∩ (R ~ F_n)

⇒    R ~ (F₁ ∪ F₂ ∪ ... ∪ F_n) is an open set.

Hence F₁ u F₂ ∪ ... ∪ F_n a closed set.

(b) Let {F_λ : X ∈ ∧} be an arbitrary family of closed sets.

We have to prove that  is a closed set.

Equivalently, we have to prove that R ~ F is an open set.

By De Morgan's law,

R ~ F = R ~ (∩ F_λ) = ∪ (R ~ F_λ).

Since each F_λ is a closed set, so R ~ F_λ is an open set.

Now ∪ (R ~ F_λ) being an arbitrary union of open sets is an open set and so R ~ F is an open set. Hence F = ∩ F_λ is a closed set.
(c) ThenSince every closed interval is a closed set, therefore {F_n : n ∈ N} is a family of infinite number of closed sets. We observe that
Indeed,  which is not a closed set.

Example : Given an example of each of the following :

(a) an interval which is an open set.,

Ans. (a, b).
(b) an interval which is not an open set,

Ans. [a, b].
(c) an open set which is not an interval,

Ans. R or (0, 1) u (2, 3).

(d) a set which is neither an interval nor an open set.

Ans. The set Z of integers.

Example : Give an example of each of the following :

(a) an interval which is a closed set,

Ans. [a, b].

(b) an interval which is not a closed set,

Ans. [a, b).

(c) a closed set which is not an interval,

Ans. R or [0, 1] υ [2, 3].

(d) a set which is neither an interval nor a closed set.

Ans. The set Z of integers.

(e) A set which is neither an open set nor a closed set.

Ans. The set Q of all rational numbers is neither an open set nor a closed set. Example : Comment on the following statements :

(a) Can a finite set be open?

Ans. Yes. An empty set is a finite set, which is also an open set.

(b) Can a finite non-empty set be open?

Ans. No, Let S is a non-empty finite open set, then for any x ∈ S and any ε > 0
(x - ε, x + ε) ⊄ S,
since the interval (x - ε, x + ε) necessarily contains an infinite number of points and S is a finite set.

(c) Is every infinite set open?
Ans. No. The set Z of integers is an infinite set, which is not an open set.
(d) Is the intersection of any arbitrary family of open sets an open set?
Ans. No. G_n = ( - 1/n, 1/n) ∀ n ∈ N is a family of infinite number of open sets and which is not an open set,

Theorem : A set S is closed if and only if it contains all its limit points.

Equivalently, S is closed o S' ⊂ S.

Proof : By definition of a closed set,

S is closed ⇔ R ~ S is an open set

⇔ R ~ S is a nbd. of each of its points
⇔ for each x ∈ R ~ S, ∃ some c > 0, such that
(x - ε, x + ε) ⊂ R ~ S or (x - ε, x + ε) ∩ S = ϕ

(∵ y ∈ (x - ε, x + ε) ⇒ y ∈ R ~ S ⇒ y ∉ S)
⇔ for each x ∈ R ~ S, ∃ a nbd. of x which contains no point of S
⇔ x is not a limit point of S ∀ x ∈ R ~ S.
⇔ x ∉ S’ ∀ x ∈ R ~ S
⇔ x ∉ S' ∀ x ∉ S
⇔ S' ⊂ S.

Remark. The above Theorem gives us a second definition of a closed set.

In order to show that a set S is closed, we must prove that if p is any limit point of S. then p ∈ S.
Corollary 1. Every finite set S is a closed set.

Proof. Since S is a finite set, S has not limit points. .\ S’ = ϕ.

Since ϕ ⊂ S. so S’ ⊂ S. Hence S is a closed set.

Corollary 2. Every singleton is a closed set.

Proof. Any singleton set {x} is a finite set and hence by Corollary 1, it is a closed set. Corollary 3. The set Z of integers is a closed set.

Proof. We know Z‘ = ϕ and so Z’ ⊂ Z. Hence Z must be a closed set.

Corollary 4. The set N of natural numbers is a closed set.

Proof. We know N' = ϕ and so N' ⊂ N. Hence N is a closed set.

Theorem. Prove that the supremum of a non-empty bounded set is either the greatest member of the set or is a limit point of the set.

Or

Prove that the supremum of a non-empty bounded set S of real numbers when not a member of S must be a limit point of S.

Proof. Since S is non-empty and bounded above, S must have the supremum, say s i.e., s = sup S (O-C Property).

⇒ x < s ∀ x ∈ S.

If s < S, then s is the greatest member of S. Let s ∉ S.

We shall show that s is a limit point of S.

For any ε > 0, we have s - ε < s.    ...(1)

Since s is the l.u.b. of S, so s - ε cannot be an upper bound of S. Consequently, there exists some t ∈ S such that

t < s < s + ε ⇒ t < s + ε.    ...(2)

From (1) and (2), it follows that any nbd. (s - ε, s + ε) of s contains a point t ∈ S, t ≠ s. Hence s is a limit point of S.
Theorem : Show that a non-empty, bounded and closed set S contains its supremum and infimum.

Proof : Since S is non-empty and bounded above S must have a supremum say s

i.e., s = sup S (O-C property). If we show that s ∈ S, then S has a maximum viz. s. We now proceed to show that s is a limit point of S.

Let ε > 0 be arbitrary. Since s = l.u.b.S. s - ε is not an upper bound of S and so there exists some t ∈ S such that

t > s - ε or s - ε < t    ...(1)

Clearly, t < s < s + ε    ...(2)

From (1) and (2), it follows that (s - ε, s + ε) contains a point t ∈ S, t ≠ s for all ε > 0.

Thus s is a limit point of S. Since S is a closed set, s ∈ S.

Hence s is a maximum of S. Similarly, s’ = inf S is a minimum of S.

Theorem : Prove that the derived set of any set is a closed set.

Proof : Let S’ be the derived set of a set S. We have to show that S’ is a closed set. Equivalently, we have to show that if p is limit point of S’, then p ⇐ S’.

Suppose p is a limit point of S’. The for any ε > 0, (p - ε, p + ε) contains a point q of S’, q ≠ p. Now q ∈ S’ ⇒ q is a limit point of S.

⇒ Every nbd. of q (in particular, (p - ε, p + ε) ) contains a point r of S, r ≠ q
(∵ r ≠ q and q ≠ p)

⇒ for each ε > 0, (p - ε, p + ε) contains a point r e S, where r ≠ p

⇒ p is a limit point os S p ∈ S’. Hence S’ is closed set.

Theorem : The derived set of a bounded set is bounded.

Proof : Let S be any bounded set whose bounds are
m = inf S and M = sup S

i.e., m < x < M ∀    x ∈ S    ...(1)

We shall prove that

m < x < M ∀ x ∈ S’

Let, if possible, there exist some y ∈ S' such that y > M.

Now,    y > M    ⇒    y - M > 0.

We choose ε > 0 such that

ε < y - M    ⇒    M < y - ε  ...(2)

From (1) and (2),

x < x ∀ x ∈ S.

In follows that there exists a nbd. ( y - ε, y + ε ) of y which does not contain any point of S and so y in not a limit point of S i.e., y ∉ S’, which is contradiction.

Thus x < M ∀ x ∈ S’.

Similarly, we can show that

m < x ∀ x ∈ S’

Hence m < x < M ∀ x ∈ S' and so S’ is bounded.

Example  27: Show that S =is closed but not open.

The only limit points of the set S are 1 and - 1 i.e., S’ = {1, - 1}.
Since S' ⊂ S, S is a closed set.

Since S is not a nbd. of each of its points, S is not an open set.
Notice the for each ε > 0, 1 ∈ (1 - ε, 1 + ε) ⊄ S.

Example : Comment on the following statements :

(a) Can a singleton be closed?

Ans. Yes. Every singleton is closed

(b) Can a finite set be closed?

Ans. Yes. Every finite set is closed

(c) Is every finite set closed?

Ans. Yes.

(d) Is every infinite set closed?
Ans. No. S = is an infinite set which is not closed, since its only limit point 0 does not belong to the set S.(e) Is the union of an arbitrary family of closed sets a closed set?
Ans. No. {F_n = [1/n, 2] : n ∈ N} is a family of infinite number of closed sets, and which is not a closed set.
(f) Is the intersection of an arbitrary family of closed sets a closed set?
Ans. Yes.

Example 28: Give Examples of sets S and S’ such that
(a) S ∩ S ’ = ϕ, ( b ) S ⊂ S ’ , (c) S ’ ⊂ S, (d) S = S ’ .

(a) Let S =  Then S’ = {0}. Clearly, S ∩ S’ = ϕ(b) Let S = ( 0, 1 ) so that S’ = [0, 1]. Obviously, S ⊂ S’.(c) Let S =  Then S’ = {1, - 1} c S.(d) Let S = [0,1] so that S’ = [0,1]. We have S = S’.

Example 29: Find the limit points of the following :

(i) N, (ii) (a, b], (iii) R ~ Q, (iv) finite set.

(i) N has no limit point.

(ii) {(a, b]}’ = [a,b]. Hence each point of [a, b] is a limit point of (a, b].

(iii) R ~ Q, consists of all irrational numbers. Every real number p ∈ R is a limit point of R ~ Q, since for each ε > 0, (p - ε, p + ε) contains an irrational number other than p.

(iv) A finite set has no limit point .

Example : State whether the following subsets of R are open, closed :

(i) the set Z of all integers.

Ans. Z is closed, (∵ Z = ϕ ⊂ Z.)

However, Z is not open as Z is not a nbd. of any of its points.
(ii)Ans. S is neither closed nor open.

We have S’ = {0} ⊄ S. Also S is not a nbd. of any of its points.

(iii) The segment (a, b).

Ans. It is an open set, since it is an open interval.

However, it is not a closed set, since { (a, b) }' = [a, b] ⊄ (a, b).

(iv) The set N of natural numbers.

Ans. N is closed. (v N’ = ϕ ⊂ N)

However, N is not open as N is not a nbd. of any of its points.

(v) S ={1, 2, 3, 4}.

Ans. S is a closed set, since it is a finite set and every finite set is closed. However, S is not open, since S is not a nbd. of any of its points.

(vi) The set R of real numbers.

Ans. R is both open and closed.

(vii) S = (0, 1) u (2, 3).

Ans. S being the union of two open sets is an open set.

Now S’ = { (0, 1) }’ u { (2, 3) }’ = [0, 1] u [2, 3] cz S.

Hence. S is not a closed set.
(viii)Ans. S’ = {0} ⊄ S and so S is not closed.

Since S is not a nbd. of each of its points, S is not open.

(ix) The set Q as rational numbers.

Ans. We know Q’ = R ⊄ Q and so Q is not closed.

Further Q is not a nbd. of any of its points and so Q is not open.

(x) S = [0, 1] ∪ [2, 3].

Ans. S being the union of two closed sets is a closed set (Every closed interval is a closed set). Since S is not a nbd. of the points 0, 1,2, and 3, S is not an open set.

Example 30: It F is a closed bounded set, then every infinite subset S of F has limit point in F.

Since S is a subset of the bounded set F, so S is a bounded and infinite set. By Bolzano Weierstrass Theorem, S has a limit point, say p. Since S ⊂ F, so p is also a limit point of F. Since F is a closed set and p is a limit point of F, so p ⇐ F. Hence S has its limit point in F.

Example 31: If A is open and B is closed, show that A ~ B is open and B ~ A is closed.

Since B is closed, R ∼ B is an open set. Now A ~ B = A ∩ (R ~ B), which being    the intersection of two open sets is an open set. Thus A ~ B is an open set.

Again B ~ A = B ∩ (R~A), which being the intersection of two closed sets is a closed set. Thus B ~ A is a closed set.