Topology MCQ Quiz in தமிழ் - Objective Question with Answer for Topology - இலவச PDF ஐப் பதிவிறக்கவும்

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.

Concept use:

Compact Spaces
A topological space is compact if every open cover of the space has a finite subcover. Intuitively, this means that you can cover the space with finitely many open sets.

Hausdorff Spaces
A topological space is Hausdorff if for any two distinct points, there exist disjoint open sets, one containing each point. This essentially means that points can be "separated" by open sets.

Normal Spaces
A topological space is normal if for any two disjoint closed sets, there exist disjoint open sets, one containing each closed set. This is a stronger separation property than Hausdorff.

Calculations:

Why Compact Hausdorff Spaces Are Normal
The key idea is that compact Hausdorff spaces have a strong separation property. Given two disjoint closed sets in a compact Hausdorff space, we can use the Hausdorff property to separate points, and the compactness property to "glue" these separations together to obtain disjoint open sets containing the closed sets.

A more detailed proof involves:

Using the Hausdorff property: For each point in one closed set, find an open neighborhood that doesn't intersect the other closed set.
Using compactness: Cover the first closed set with finitely many of these open neighborhoods.
Constructing disjoint open sets: Use the union of these neighborhoods as one open set, and the intersection of the complements of the neighborhoods as the other open set.

Why Metric Spaces Are Normal:

Metric spaces are a special type of topological space where we have a distance function. This distance function allows us to define open balls around points. Given two disjoint closed sets in a metric space, we can find a positive distance between them. Using this distance, we can construct disjoint open balls around each closed set.

In essence, the metric structure provides a powerful tool for separating sets in metric spaces, making them normal.