Point Set Topology MCQ Quiz in मल्याळम - Objective Question with Answer for Point Set Topology - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation :

For Option (1) - 

\(\left\{\left(0, \frac{1}{2}-\frac{1}{n+1}\right) \cup\left(\frac{1}{n}, 1\right): n \in \mathbb{N}\right\} \)

For every n ≥ 7 they cover (0, 1) e.g. For n = 7

⇒ \( \left(0, \frac{1}{2}-\frac{1}{8}\right) \cup\left(\frac{1}{7}, 1\right)\)= \(\left(0, \frac{3}{2}\right) \cup\left(\frac{1}{7}, 1\right)\)

which cover whole (0, 1).

⇒ We get a cover for n = 7

Similarly for n ≥  8

Hence option (1) is true.

(2) \(A_n=\left\{\left(\frac{1}{n}, 1-\frac{1}{n+1}\right): n \in \mathbb{N}\right\}\)

\(\{a_n\}=\left\{\frac{1}{n}\right\} \) is stictly decreasing sequence and \(\left\{b_n\right\}=\left\{1-\frac{1}{n+1}\right\}\) is stictly increasing sequence.
It is clearly observed that 

A₁ ⊆ A₂ ⊆ A₃ ⊆ A₄ ⊆ ...           ...(2) 

As n → ∞ A_n covers (0, 1) (because of 2) But if we fix any n, it is not possible to find any finite subcover.

⇒ No finite subcover exist

⇒ Hence option (2) is not true.

(3) Given \(\left\{\left(\sin ^2\left(\frac{n \pi}{100}\right), \cos ^2\left(\frac{n \pi}{100}\right)\right): n \in \mathbb{N}\right\}\)

For n = 100 we have

\(\left(\sin ^2\left(\frac{100 \pi}{100}\right), \cos ^2\left(\frac{100 \pi}{100}\right)\right)\)

= (sin2 π, cos2 π) 

= ((sin π)2,(cos π)2) 

=(0, 1) which covers (0, 1)

Hence option (3) is true.

(4) A_n = \(\left\{\left(\frac{1}{2} e^{-n}, 1-\frac{1}{(n+1)^2}\right): n \in \mathbb{N}\right\}\)

\(\{a_n\}=\left\{\frac{1}{2} e^{-n}\right\}=\left\{\frac{1}{2 e^n}\right\}\) ; a_n is strictly decreasing and

{b_n} = \(\left\{1-\frac{1}{(n+1)^2}\right\} ; b_n\) is strictly increasing.

∴ By similar argument as above in option (2) no finite sub cover exist.

Hence option (4) is not true.

Explanation:

p : ℝ2 → ℝ be the function defined by p(x, y) = x

(1): A1 = {(x, y) ∈ ℝ2 | x2 + y2 < 1} so it is an open unit disk.

So, p(A1) = (-1, 1)

Then by the definition of limit point, for each γ ∈ p(A1), there exists a positive real number ε such that (γ - ε, γ + ε) ⊆ p(A1). 

(1) is correct 

(2): A2 = {(x, y) ∈ ℝ2 | x2 + y2 ≤ 1} is closed unit disk.

So, for the endpoints -1 and 1, for each γ ∈ p(A3), there does not exist a positive real number ε such that (γ - ε, γ + ε) ⊆ p(A3).

(2) is false

(3):  A3 ={(x, y) ∈ ℝ2 | xy = 0}.

So,  A3 = {(x, 0) | x ∈ ℝ} ∪ {(0, y) | y ∈ ℝ}

p(A3) = (-∞, ∞)

So, every point is an interior point

Hence for each γ ∈ p(A3), there exists a positive real number ε such that (γ - ε, γ + ε) ⊆ p(A3).

(3) is correct

(4): A4 = {(x, y) ∈ ℝ2 | xy = 1}.

p(A4) = (-∞, 0) ∪ (0, ∞)

Hence for each γ ∈ p(A4), there exists a positive real number ε such that (γ - ε, γ + ε) ⊆ p(A4).

(4) is correct

Explanation:

From the properties of the interior point, derived set and close set we know that 

\( (A \cap B)^{\circ}=A^{\circ} \cap B^{\circ} \)

\( (A \cap B) \subseteq A^{\prime} \cap B^{\prime}\)

\(\overline{(A \cap B)} \subseteq \bar{A} \cap \bar{B}\)

and \((A \cup B)^{\circ} \supseteq A^{\circ} \cup B^{\circ}\)

(1), (2), (3), (4) are correct

Explanation:

(1): pn{0,1} = {a₀ + a₁x + a₂x² + ... + a_nxⁿ : a_i ∈ {0, 1}}

Each a_i (for i = 1, 2, ..., n) has 2 choices so card (pn{0,1}) = 2n

(1) is correct

(2): Poly{0,1} = P₀ ∪ P1 ∪ P2 ∪ ... 

So card (Poly{0,1}) = N0

(3) is correct, (2) is false

(4): pn{ℤ} = {a0 + a1x + a2x2 + ... + anxn : ai ∈ ℤ}

So card(pn{ℤ}) = N0

(4) is correct

Explanation:

Let f(x) = -2 then X = {ϕ} which is closed set

(1) is true.

(2) If X is empty set the X is open

If X is non-empty then since f is continuous so X has infinite number of elements.

The set  X is an open set if for any element x in the set there exists an open interval centered around the element  x which is entirely in the set X.

Consider any arbitrary point  p ∈ X. Since  p ∈ X, f(p) > 0.
Since the function f is continuous, for any  ϵ > 0, there exists a δ > 0, such that  |f(x)−f(p)| < ϵ whenever |x−p| < δ.
We will choose ϵ < f(p)/2. Then  f(p)/2 < f(x) < 3f(p)/2 whenever |x−p| < δ.
Consequently, f(x) > 0 whenever |x−p| < δ.
Thus it can be seen that there exists an open interval  (p−δ, p+δ) centered around the element p which is entirely in the set  X.
Since  p is an arbitrary element of the set  X, this is applicable for all elements of the set  X. It is thus clear that for any element  x in the set  X, there exists an open interval centered around the element x which is entirely in the set X.
Hence, X is an open set.

 (2) is true.

Let f(x) = x²

Then X = (- ∞, 0) ∪ (0, ∞) so X is disconnected.

(3) is false, (4) is true 

Explanation:

We know that Derived set of any set S is always closed set.

(3) is correct

Concept:

A subset A of a topological space X is called dense if \(\bar A=X\).

Explanation:

\(\overline{\mathbb N}=\mathbb N\) and \(\overline{\mathbb Z}=\mathbb Z\)

So, ℕ and ℤ  are not dense in ℝ.

\(\overline{ℝ| ℤ}= \overline{ℤ^c}={ℤ^0}^c=\phi^c=\mathbb R\)

Hence ℝ \ ℤ is dense in ℝ.

(3) is 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.

Concept use:

A Neighborhood of a point is a set of points containing that point where one can move some amount in any direction away from that point without leaving the set.

Explanation:

for a Set to be a neighborhood of Q' if it contains each element of the set in it only.

Q' is not a neighborhood of Q because it contains rational also which doesn't belongs to Q.

Q is also not neighbourhood of Q' because it contains rational also in it with irrational.

Hence, The Correct Answer is option 1.

Explanation:

A = [0, 1) ∪ (1, 2] ∪ (Q ∩ [3, 4]) ∪ {5}

Here A, CI(A) and Int(A) are pair wise distinct

(1) is correct