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

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.

Concept:

Let (X, τ) be a topological space. X is said to be disconnected if and only if there exist G, H ∈ τ such that

(i) G ≠ ϕ, H ≠ ϕ, G ∩ H = ϕ 

(ii) X = G ∪ H

X is said to be connected if and only if it is not connected.

Explanation:

τ = {ϕ, X, {a, c}, {b}}

{a, c} ∩ {b} = ϕ 

{a, c} ∪ {b} = {a, b, c} = X

Hence (X, τ) is disconnected.

(2) is correct

Concept:

A connected set in a topological space is a set that cannot be split into two non-empty disjoint open subsets within that space.

Explanation:

To analyze the two statements, let's examine the given topology on \(\mathbb{Z}\) (the set of integers) with basis

elements S(a, b) = \(\{ an + b : n \in \mathbb{Z} \}\) , where a, \(b \in \mathbb{Z}\) and a \(\neq\) 0 .

Statement I: S(a, b) is both open and closed for each a,\( b \in \mathbb{Z}\) with a \(\neq\) 0 .

 In this topology, each basis element S(a, b) is a set of integers that form an arithmetic sequence based on a and b .

Since S(a, b) is a basis element, it is open by definition.

Additionally, because S(a, b) is also the complement of a union of other basis sets, it is closed as well.

Therefore, Statement I is TRUE.

Statement II: The only connected set containing x \(\in \mathbb{Z}\) is {x} .

In this topology, each point x \(\in \mathbb{Z}\) is isolated due to the discrete-like structure of the sets S(a, b) . 

Since any larger subset of \(\mathbb{Z}\) can be split into disjoint open subsets (as integers are

isolated here), there are no non-trivial connected subsets in this topology.

Thus, the only connected set containing x is {x} , a single point.

Therefore, Statement II is TRUE.

Hence option 1) is correct.