Topology MCQ Quiz in বাংলা - Objective Question with Answer for Topology - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation -

For option (1) -

Since \(U= \{ (x,y) \in R^2 \ | \ x^2 + y^2 < 1 \}\) is open in usual topology on R² but there is no open right half plane contained in U. So set of open right half plane is not a basis for usual topology on R².

Hence false option.

For option (2) -

Consider \(S= \{ (x,y) \in R^2 \ | \ 0 < y \le1, \ x = 0 \}\) 

Claerly S is a finite straight line parallel to y - axis but there does not exist a open set U in dictionary order topology such that \((0,1) \in U \subseteq S\)

Hence false option.

For option (3) -

This is true since we can inscribe a suitable open disc inside an open rectangle and vice versa, so set of open rectangles is basis for usual topology.

For option (4) -

This is true, Since any open set in dictionary order topology contains finite line parallel to y -axis and for any given line parallel to y -axis without end points we get open set contained in straight line, so given set from basis for dictionary order topology on R².

Hence option (3) and (4) are correct.

Explanation:

By the invariance of domain theorem, a continuous one-to-one map from a compact space onto a Hausdorff space is an open map. Therefore, the given map is an open map.

Option (1) is correct

Explanation:

Let A = {xπ | x ϵ Q} and B = {x\(√{2}\) | x ϵ Q} and B = {x√2| x ∈ Q}  

then A ∩ B = ϕ and both A and B are countable and dense in R

Option (1) is true.

For each n take finite subset B of S such that |B| = n then for each p ∈ B, A_p is countable and dense subset of R.

option (2) is true

Since set of all primes S (say) similar to N.

For each prime p ∈ S consider A_p = {x√p | x ∈ Q} then each Ap is countable and dense in R and also for p ≠ q, Ap ∩  Aq = ϕ.

option (3) is true.  

For each α ∈ Q^c, consider A_α ={αx | x ∈ Q} then A_α is countable and dense in R and for α, β ∈ Q^c such that, α ≠ β, A_α  ∩ A_β =  ϕ    

Option (4) is true

Concept:

A set is called compact set if every open cover has finite sub-cover.

Explanation:

A = \(\mathbb{N}\) ∪ {∞} and B = \(\mathbb{N}\) ∪ {-∞}.

τ be the topology on X consisting of subsets U of X such that either U ⊂ \(\mathbb{N}\) or X\U is finite.

A = (X - U) ∪ \(\cup_{i=1}^{\infty}\)xi, which can be covered by (X - U) ∪ \(\cup_{i=1}^n\)x_i

we will get finite sub-cover.

So A is compact. Option (1) is correct.

X\A = { - ∞} which can be covered by any finite set.

X\A is compact. Option (2) is correct.

A ∪ B = X = \(\mathbb{N}\) ∪ {∞, - ∞} = (X - U) ∪ \(\cup_{i=1}^{\infty}\)xi, which can be covered by (X - U) ∪ \(\cup_{i=1}^n\)xi

A ∪ B is compact. Option (3) is correct.

A ∩ B = \(\mathbb{N}\).

let open cover of A ∩ B be \(\cup_{i=1}^{\infty}\)xi which does not have finite sub-cover. 

A ∩ B is not compact. Option (4) is false. 

Explanation:

A path connected domain is  a domain where every pair of points in the domain can be connected by a path going through the domain.

A set which contain all its limit point is closed and when closed and bounded it is compact.

Option 1) A = {(x ,y) \(\in\) R*R : (x-1)(x-2)(y-3)(y+4)=0 } 

Zeroes of set is x=1, x=2, y=3, y=-4 these all are straight line if we observed by a graph so given set is a path connected. 

Option 1 is correct. 

Option 2) Given set

A = {(x ,y) \(\in\) R*R : (x-1)(x-2)(y-3)(y+4)=0 }

we get x=1, x=2, y=3, y=-4 

These all are straight lines so we can write as 

A= \(\cup\) { \({x=1,x=2, y=3, y=-4}\) } 

we know straight lines are  closed  and  unbounded 

 union of a finite closed set is closed so A is closed and unbounded 

We know closed and bounded is compact so here given set is not compact 

Option 2 is incorrect.  

Option 3) By the similar argument in option 2.

Given set A is closed So, option 3 is correct.

Option 4 ) Set A is closed so closure of set A \(= A \neq R^2\) so it is not dense 

Option 4 is incorrect. 

The correct options are (1) and (3).

Concept:

Let X be a topological space and E be a subset of X.

Let E̅ is not connected implies that 

there exist U and V be a subset of X . such that 

E̅ \( \subseteq U \)\(\cup V\) and \(U\cap \)\(\bar{V} = \emptyset\)

Explanation:

Let X be a topological space and E be a subset of X.

Let E̅ is not connected implies that 

there exist U and V be a subset of X . such that 

E̅ \( \subseteq U \)\(\cup V\) and \(U\cap \)\(\bar{V} = \emptyset\) 

and 

\(\bar{E} \cap\) V \(\neq \emptyset\) and \(\bar{E}\cap\)V \(\neq \emptyset\)

Hence option (1) is correct

Let 

X = R , Z = usual topology 

E = (0, 1) is connected 

∂E = { 0 ,1 } is disconnected .

Hence option (2) is not correct

Let 

E = { (x , sin\(\frac{1}{x}\)) , x>0 }

\(\bar{E}\) = E \(\cup\) { -1 \(\leq\) y \(\leq\) 1 } 

is not path connected

Hence option (3) is not correct

Let 

X= N 

E = {1} 

\(\bar{E}\) = N 

here E is compact.

There does not any finite subcover for this family of open sets.

Hence option (4) is not correct

Explanation:

Y is a bounded open set implies that the closure of Y is compact.​​

\(\cup_{j \geq 1}^N\) U_j is an open cover of Y̅.​

\(\cup_{i=1}^N U_j\_i\) will be a finite subcover of Y̅. 

Hence option (1) is true

Finite subcover

\(\ U_j1\) , \(U_j2\) , . . . . . . \(U_jN\)

take N large enough. such that 

Y ⊆\(\cup_{j=1}^N\)Uj 

Hence option (2) is true

Taking

(0, 5) ⊆\(\cup_{j=1}^N\)U_J 

{ U_j } = { 2, 3, 4 , 5 ,  . . . }

Hence option (3), (4) are false

Concept:

Particular Point Topology (with point 0):

• A set \( U \subseteq \mathbb{R} \) is open if and only if \( U = \emptyset \) or \( 0 \in U \).
• This makes \( 0 \) a special point; all open sets must contain it (except the empty set).
• Dense Set: A set is dense if it intersects every non-empty open set (i.e., contains 0).
• Compactness: A set is compact if every open cover has a finite subcover — needs careful checking here.
• Hausdorff: A space is Hausdorff if distinct points can be separated by disjoint open neighborhoods.

Calculation:

Given, \( \tau = \{ U \subseteq \mathbb{R} : U = \emptyset \text{ or } 0 \in U \} \)

⇒ Analyze each statement:

Statement 1: The set of all irrational numbers is dense.

⇒ Irrationals do not necessarily include 0.

⇒ Since all non-empty open sets must contain 0, the set must include 0 to be dense.

⇒ False

Statement 2:

For each prime \( p \), the set \( \{0, \sqrt{p}\} \) is dense.

⇒ Contains 0

⇒ intersects all open sets

⇒ True

Statement 3: \( [0,1] \) is compact.

⇒ Take open cover: \( \{ \mathbb{R} \setminus \{x\} \} \) for all \( x \in [0,1] \setminus \{0\} \) and \( \mathbb{R} \)

⇒ No finite subcover can cover all of [0,1]

⇒ False

Statement 4: \( (\mathbb{R}, \tau) \) is Hausdorff.

⇒ All non-empty open sets contain 0

⇒ cannot separate any two non-zero points.

⇒ Cannot find disjoint open sets for any pair

⇒ False

∴ Only correct statement is Option 2

Concept:

Key Topological Concepts:

• Dense Subset: A subset A of a topological space X is dense if every open set in X intersects A.
• Path-connectedness: A space is path-connected if any two points can be joined by a continuous path within the space.
• Connected Space: A topological space is connected if it cannot be expressed as the union of two disjoint non-empty open sets.
• Compact Subset: A subset of a topological space is compact if every open cover has a finite subcover.
• Closed Set: A subset is closed if its complement is open.

Important Notes:

• Option 1: The given set forms a lattice in ℂ but is not dense.
• Option 2: In ℝ³, open connected sets need not be path-connected (true in higher dimensions, especially with "holes").
• Option 3: This is a known result from general topology (fiberwise connectedness implies connectedness for open surjective maps).
• Option 4: Compact subsets in general need not be closed in infinite topological spaces (e.g., cofinite topology).

Calculation:

Given,

Let p: X → ℝ be a continuous surjective open map

Each fiber p⁻¹({a}) is connected ∀ a ∈ ℝ

⇒ We use the theorem:

If p is continuous, surjective, open and all fibers p⁻¹({a}) are connected, then X is connected

⇒ This is a result from general topology

∴ Option 3 is the correct statement:

If p⁻¹({a}) is connected ∀ a ∈ ℝ, then X must be connected.

Explanation:

Let U ⊆ ℝ × ℝ be an open set. 

The projection of this set onto the first coordinate is a subset of ℝ, and it will generally be an open set because projection preserves openness in ℝ.

Hence π is open.

(1) is TRUE, (2) is NOT TRUE.

A function is continuous if the preimage of every open set is open.

The preimage of an open set 𝑉 ⊆ ℝ under π is π^-1(V) = V × ℝ, which is an open set in ℝ × ℝ.

Hence π is continuous.

(3) is TRUE.

The codomain of π is ℝ, and for each x ∈ ℝ, there exists a point (x,y) ∈ ℝ × ℝ such that π(x, y) = x.

Thus, every element of ℝ has a preimage.

Hence the projection π is surjective.

(4) is TRUE.