Point Set Topology MCQ Quiz - Objective Question with Answer for Point Set Topology - Download Free PDF

Explanation:

We know that

A ⊆ A̅ 

⇒ int(A) = int(A̅)

A̅ \ int(A̅) ⊆ A̅ \ int(A)

Now, ∂(A̅) = \(̅{̅ A}\) \ int(A̅) = A̅ \ int(A̅) ⊆ A̅ \ int(A) ⊆ ∂(A)

So, (4) is true.

Also, we know that A ⊆ A̅ always.

So, (2) is true.

By property, int(X \ A) ⊆ X \ A̅

(1) is true.

But (3) is NOT necessarily true.

Ex: Let A = ℚ then A̅ = ℝ 

∂A = A̅ \ \int(A) = ℝ \ ϕ = ℝ

and ∂ (int(A)) = ϕ

So, ∂A ⊆ ∂ (int(A)) not true.

Explanation:

(A) A closed set either contains an interval or else is nowhere dense.

This statement is correct

This is a consequence of the perfect set theorem and the Baire category theorem

 A closed set in a complete metric space (like the real numbers) can be decomposed into a perfect set (which contains an interval) and a nowhere dense set.

(B) The derived set of a set is closed

This statement is correct

The derived set (the set of limit points) of any set is always closed

(C) The union of an arbitrary family of closed sets is closed.

This statement is incorrect

The union of an arbitrary family of closed sets is not necessarily closed

For example, consider the closed intervals  [-1/n, 1/n] for n = 1, 2, 3, 

Each of these sets is closed, but their union is (-1, 1) which is open, not closed.

(D) The set R of real numbers is open as well as closed.

This statement is correct

In any topological space, the entire space and the empty set are always both open and closed

Therefore, the correct statements are (A), (B), and (D).

Hence Option(1) is correct.

Explanation:

A' : The set of all limit points of the set A .

 \(\overline{A}\) : The closure of the set A , which includes all points of A and all its limit

points.

The closure of a set A , denoted \(\overline{A}\) , is the union of the set A and its limit points.

Hence, we can express \(\overline{A} = A \cup A'\)

This means the closure of A includes all elements of A as well as the limit

points of A .

The correct option is option 4.

Explanation:

1. "Every bounded above set exists its supremum."

This is true by the Least Upper Bound (LUB) property. In the real numbers,

every non-empty set that is bounded above has a supremum (least upper

bound).

2. "Every bounded below set exists its infimum."

This is also true by the Greatest Lower Bound (GLB) property. In the real

numbers, every non-empty set that is bounded below has an infimum (greatest

lower bound).

3. "Every bounded set exists its supremum and infimum."

This is true because a bounded set in the real numbers has both a supremum

and an infimum as long as it is non-empty.

Hence option 5 is correct.

Concept -

If the given set is of this type \(A = \{ m^2 + e^{-n} + \frac{1}{n} | m \in \mathbb{R}, n \in \mathbb{N}\}\) then we make three cases

(i) when m is fixed and n →  ∞ 

(ii) when n is fixed and m →  ∞ 

(iii) when m →  ∞  and n →  ∞ 

Explanation -

We have the set \(A = \{ m^2 + e^{-n} + \frac{1}{n} | \ \ m ∈ \mathbb{R}, n ∈ \mathbb{N}\}\)

Now we have to make the cases for find the limit point for the set A.

Case (i) - when m is fixed and n →  ∞ 

then \(m^2 + e^{-n} + \frac{1}{n} \to m^2\) and m ∈ R

So all positive real numbers are the limit point for the set A.

Case (ii) - when n is fixed and m →  ∞ 

then \(m^2 + e^{-n} + \frac{1}{n} \to \infty\) and m ∈ R

So,  limit  points does not exist in this case.

Case (iii) - when m →  ∞  and n →  ∞ 

then \(m^2 + e^{-n} + \frac{1}{n} \to \infty\) and m ∈ R

So limit point  does not exist in this case.

Hence all the limit point is collection of all the limit points of the all the cases.

Hence the limit points of A are Always real number but uncountable.

Concept -

(1) The collection of all the sequences on two symbols or more than two symbols is uncountable.

Explanation -

The set T= {(x1, x2,..., xn,...): x1, x2,..., xn, ... ∈ {1, 3, 5, 7, 9}} 

Now we have 5 symbols and T represents the collection of all the sequences.

Hence the set T is Uncountable.

Concept use:

Bounded set : A set S is bounded if it has both upper and lower bounds. 

Closed set: If a set contain each of its limit point in the set 

Calculations:

S = {x5 - x4<=100 where x ∈ R} is unbounded and Closed 

T = { x2 - 2x <67 where x ∈ (0, ∞)} is Open and bounded

Hence the Intersection of the Closed set and Open Set need not be closed set, but it is bounded also.

So, The Correct option is 2.

Concept -

(i) Image of an interval under continuity is an interval.

(ii)  Image of compact set under continuity is compact.

(iii) Every interval in connected set.

Explanation -

We have B is closed interval in (o, ∞) and f(t) = sin(t) 

So clearly f(t) is a continuous function.

And B is a closed interval in (o, ∞) implies B is compact set because of boundedness.

We know that image of compact set under continuity is compact.

Hence A is compact set ⇒ closed set.

So option(1) and (4) are false.

we know that image of an interval under continuity is interval. Hence A is connected.

So option (3) is false.

Now the closure of \(\bar{A} = A \neq R\)  

Hence A is not dence in R.

Hence option(2) is true.

Concept:

(ii) A function f : X → Y is called surjective if if for every element in y there exist a pre-image in X.

(iii) A function f : X → Y is called bijective if it is both injective and surjective. 

Explanation:

X is a non-empty finite set and 

Y = {f-1(0) : f is a real-valued function on X}.  

i.e., f: X → Y is real valued function.

So, |Y| = 2^|X| 

Therefore Y has 2|x| elements. 

(2) is true.

Since X is non-empty finite so Y is finite.

(1) is false.

We know that if a function f: A → B is surjective then |A| > |B|.

Here |X| \(\ngeq\) |Y| so f can't be surjective.

(4) is false

Since f is not surjective so not bijective.

(3) is false.

Concept -

If the given set is of this type \(A = \{ m^2 + e^{-n} + \frac{1}{n} | m \in \mathbb{R}, n \in \mathbb{N}\}\) then we make three cases

(i) when m is fixed and n →  ∞ 

(ii) when n is fixed and m →  ∞ 

(iii) when m →  ∞  and n →  ∞ 

Explanation -

We have the set \(A = \{ m^2 + e^{-n} + \frac{1}{n} | \ \ m ∈ \mathbb{R}, n ∈ \mathbb{N}\}\)

Now we have to make the cases for find the limit point for the set A.

Case (i) - when m is fixed and n →  ∞ 

then \(m^2 + e^{-n} + \frac{1}{n} \to m^2\) and m ∈ R

So all positive real numbers are the limit point for the set A.

Case (ii) - when n is fixed and m →  ∞ 

then \(m^2 + e^{-n} + \frac{1}{n} \to \infty\) and m ∈ R

So,  limit  points does not exist in this case.

Case (iii) - when m →  ∞  and n →  ∞ 

then \(m^2 + e^{-n} + \frac{1}{n} \to \infty\) and m ∈ R

So limit point  does not exist in this case.

Hence all the limit point is collection of all the limit points of the all the cases.

Hence the limit points of A are Always real number but uncountable.

Concept -

(i) Theorem  - Let A be the subset of R then\((\bar{A^c})^c = A^0\)

(ii) Theorem - Let G is open in X iff \(\overline{(G \cap \bar{A}) } = \overline{(G \cap A)} \) for all A ⊂ X

Explanation -

For option (i) - Let A be the subset of R then\((\bar{A^c})^c = A^0\)

Theorem  -

Let A be the subset of R then\((\bar{A^c})^c = A^0\)

Lets understand the proof -

We know that A⁰ = int(A) is an open set and closer of any set is closed set so \(\bar{A^c}\) is closed set.

And the complement of closed always gives you open set. 

So LHS and RHS are open set in general.

Eg. - If we take \(A = \mathbb{Q}\) then 

LHS = \((\bar{A^c})^c = \bar{(\mathbb{Q}^c})^c ={\mathbb{R}}^c = \phi\)

RHS = A0 = \(\mathbb{Q} ^0= \phi\)

Hence LHS and RHS are equal also.

Hence option(i) is true.

For option(ii) -

Given that G is open.

Since \(G ∩ A \subseteq G ∩ \bar{A}\) as \(A \subseteq \bar {A}\)

⇒ \(\overline{G ∩ A } \subseteq \overline {G ∩ \bar{A}}\)

Similarly we can prove that \( \overline {G ∩ \bar{A}} \subseteq \overline {G ∩ \bar{A}}\)

Let \(x ∈ ​​ \overline {G ∩ \bar{A}}\) and S_r(x) be a nbd of x, then \(S_r(x) \cap (G \cap \bar{A}) \neq \phi\) as \(x ∈ ​​ \overline {G ∩ \bar{A}}\)

then there exist \(y \in S_r(x) \cap (G \cap \bar{A}) \) there exist r' such that \(S_{r'}(y) \subset (S_r(x) \cap G)\) as G is open  and \(S_{r'}(y) \cap A \neq \phi\)

⇒ \(S_r(x) \cap (G \cap A) \neq \phi\)

⇒ \(x ∈ ​​ \overline {G ∩ A}\)

Hence  \( \overline {G ∩ \bar{A}} = \overline {G ∩ \bar{A}}\)  for all A ⊂ X

Hence option(ii) is true.

For Option (iii) & (iv) -

Given that \(\overline{(G \cap \bar{A}) } = \overline{(G \cap A)} \)  for all A ⊂ X

Substitute A = G^c then 

\(\overline{G \cap \bar{G^c}} =\overline{G \cap {G^c}} = \phi\)

⇒ \({G \cap \bar{G^c}} = \phi \Rightarrow \bar{G^c} = \overline{X\backslash G} \subset X \backslash G = G^c\)

Therefore G^c is closed and hence G is open.

Hence option(iv) is correct and option(iii) is false.

Concept -

(1)  If f: A → B is a one-to-one  then |A| ≤ |B| 

Explanation -

We have If f: A → B is a one-to-one 

then |A| ≤ |B| 

Given that A is a countable set then B is either countable or uncountable. 

Hence option (1) and (2) are false.

But there always exist a subset of B which is countable.

So option(3) is true.

Concept use:

Bounded set : A set is said to be bounded if it has Its Suprimum & Infimum

Limit point: A point is called a limit point of a set if there is no minimum distance from that point to members of the set.

Uncountable set : we do not map the element of the Set to N.

Explanation:

Option 1: take {1, 2} have no limit point. Hence, Incorrect

Option 2: take N have no Limit point. Hence, Incorrect.

Option 3: take 1/n No interior point. Hence, Incorrect.

Option 4:

Case 1:

let a ∈ R And S = Q is countable set 

now (a - ∈ , a + ∈) ∩ S - {a} = infinite means it is a limit point 

Case 2:

and if you take a ∈ S and set S is uncountable 

now (a - ∈ , a + ∈) ∩ S - {a} = infinite 

Hence, it means it has a limit point 

Hence, option 4 is Correct 

Concept -

(1) A set of open disjoint intervals of any uncountable set is countable.

(2) 2ℵ = C

Explanation -

We have A be an infinite set of disjoint open subintervals of (0, 1).

Hence the set A is similar to the set of natural number.

So |A| = ℵ ( Aleph  Naught) 

Now given that B is the power set of A.

So |B| = 2^|A| = 2^ℵ = C ( Continum )

⇒ B is an uncountable set. and it is also similar to (0,1).

The cardinality of B is greater than the cardinality of A.

(4) is true.

Concept -

(1) Every non empty interval is uncountable.

Explanation -

We know that [√2, √3] is an uncountable set because of a non-empty interval.

Now the set of all irrational numbers in [√2, √3] = [√2, √3] ∩ Q^c 

We get infinitely many points in the intersection or uncountably many irrational numbers in this intersection.

Hence The set of all irrational numbers in [√2, √3] is uncountable.