Countable and Uncountable Sets MCQ Quiz in मल्याळम - Objective Question with Answer for Countable and Uncountable Sets - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

(i) Limit superior of a sequence is the largest limit point.

(ii) A sequence (xn) is convergent if and only if lim sup xn = lim inf xn = lim xn= finite 

Explanation:

\( X=\{\left(x_n\right)_{n \geq 1}:\displaystyle \limsup _{n \rightarrow \infty} x_n=1\), where xn ∈ {0, 1}}

Since \(\displaystyle \limsup _{n \rightarrow \infty} x_n=1\) for xn ∈ {0, 1} then 

xn = 1 for infinitely many n

⇒ there exist an infinite subset S such that x_n = 1 for all x ∈ S

Hence cardinality of X is the nuber of subsets of natural numbers with inifnite cardinality = \(2^{\aleph_0}\)

Therefore X is uncountable.

\(\displaystyle Y=\{\left(x_n\right)_{n \geq 1}: \lim _{n \rightarrow \infty} x_n\), does not exist, where xn ∈ {0, 1}}

\(\lim _{n \rightarrow \infty} x_n\) does not exist for xn ∈ {0, 1}

Then lim sup xn and lim inf xn are not equal

⇒ lim sup xn = 1 and lim inf xn = 0

So, 0 and 1 both are limit points of xn 

Hence there exist an inifnite subset S of natural number such that xn = 1 for all n ∈ S and xn = 0 for all n ∈ S^c

i.e., Y is the set of all infinite subsets of natural number whose complement is also infinite

So Cardinality of Y = \(2^{\aleph_0}\)

Therfeore Y is uncountable

Given:

Various sets to examine for countable infinity.

Concept used:

A countably infinite set is a set which has same cardinality (size) as the set of natural numbers.

Calculation:

The set of all even numbers are countable 

Option A is correct as the set of all even numbers can be put into one-to-one correspondence with the set of all natural numbers

Given:

The set of all integers.

Concept used:

A set is considered countable if there exists an injective function from the set to the natural numbers, and uncountable otherwise.

Calculation:

The set of all integer numbers can be put into one-to-one correspondence with a subset of the set of natural number, so it is a countable set.

for n is odd then take (n + 1)/ 2 

for n is even then take - n/2

Hence, the answer is (A). Countable

Given:

Our options represent various different sets

Concept used:

A finite set is the one that has a specific number of elements.

An infinite set is one that goes on without end.

Calculation:

Among the options, only Option (D) represents a set with a specific number of elements.

In this case, there are 11 elements: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 other options are all sets that represent infinite sets. 

Hence, Option D is Correct.

Given:

The set of all irrational numbers in the interval (0, 1).

Concept used:

A set is uncountable if no injective function exists from the set to the natural numbers.

Calculation:

The set of all irrational numbers between 0 and 1 is not countable because there is no one-to-one correspondence between this set and the set of natural numbers.

Hence, the answer is (B). Uncountable. 

Explanation:

\(\rm S=\{(x, y) \lvert\, x^2+y^2=\frac{1}{n^2}\), where n ∈ ℕ and either x ∈ ℚ or y ∈ ℚ}

ℚ is the set of rational numbers and ℕ is the set of positive integers. 

if x ∈ ℚ

\(x^2+y^2=\frac{1}{n^2}\) ⇒ \(y^2=\frac{1}{n^2}-x^2\)  ⇒ y = ± \(\sqrt{\frac{1}{n^2}-x^2}\) ∈ ℚ

So, S = (x, y) ∈ ℚ

Now, since ℚ is a countable set so S is countable.

Option (2) is true and option (3) and (1) are false

S is non-empty as if we consider n = 1 then (1, 0) ∈ S.

Option (4) is false

Concept Used:

Countable Set: A countable set is a set whose elements can be put in one-to-one correspondence with the natural numbers (1, 2, 3, ...)

Uncountable Set: A set that is not countable is uncountable.

Some Countable Sets are ℕ, \(\mathbb{Q}\) [Set of Rational Numbers], \(\mathbb{Z}\) [Set of Integers], and any subsets of these sets are also countable.

Explanation:

Set 1: {x ∈ ℝ | log(x) = p/q for some p, q ∈ ℕ}

This set is countable. The logarithm function takes on rational values which is a countable set, so the set of all x for which log(x) is a rational number is countable.

Set 2: {x ∈ ℝ | (cos(x))n + (sin (x))n = 1 for some n ∈ ℕ}

This set is uncountable. For n = 2, this equation represents points on the unit circle, and it includes points with transcendental values such as irrational multiples of π. Therefore, the set is uncountable.

Set 3: {x ∈ ℝ | x = log(p/q) for some p, q ∈ ℕ}

This set is countable. Similar to the first set, it involves the logarithm function with rational arguments, making it countable.

Set 4: {x ∈ ℝ | cos(x) = p/q for some p, q ∈ ℕ}

This set is countable. The cosine function takes on rational values only for specific angles (e.g., multiples of π/3, π/4), and these values are countable.

Thus, only uncountable set is {x ∈ ℝ I (cos(x))n + (sin (x))n = 1 for some n ∈ ℕ}.

Explanation:

We know that (0, 1) and [0, 1] both uncountable and ℤ, ℚ both countable.

So The set of all functions f : (0, 1) → ℤ is uncountable

The set of all function f: [0,1] → ℚ is uncountable.

The set of all functions f : (0, 1) → ℚ is uncountable.

So (2) is true only

Explanation:

g(t) = f(x0 + tu), so

g'(t) = \(\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\)

      = \(\lim_{h\to0}\frac{f(x_0+xu+hu)-f(x_0+hu)}{h}\)

     = Duf(x0 + hu) which exist always

So g is differentiable for all x ∈ \(\mathbb R\) and we know that \(\mathbb R\) is countable.

Option (1) is correct and others are false

Concept:

To find the critical points of a function \(f(x)\) we put \(f'(x)\) = 0.

Explanation:  

\(e^x + x = 1\)

We are looking for real values of  \(x \)  that satisfy this equation. Let’s rewrite this as \(e^x = 1 - x\)

To find the real solutions, we need to investigate the points where these two functions intersect, i.e., where,

\(e^x = 1 - x\)

As  \(x \to -\infty\) , \(e^x \to 0 \), and \(1 - x \to \infty \).

So, there is no solution in this region.

For large positive  \(x\) :

As  \(x \to \infty\) , \( e^x \to \infty\) , and  \(1 - x \to -\infty\) .

Again, there is no solution in this region.

Check around  \( x = 0\) :

At \(x = 0 \), \(e^0 + 0 = 1 \), which satisfies the equation.

So,  \(x = 0 \) is a solution.

We can check the behavior of the function \( f(x) = e^x + x\) .

The derivative of this function is \(f'(x) = e^x + 1 \)

Since \(e^x + 1 > 0 \) for all real \(x\), the function is strictly increasing. Therefore, it can only cross  \( y = 1\) at one point,

which implies that the solution  \(x = 0 \) is the only real solution.

Since there is only one real solution, the cardinality of the set of real solutions is 1.

Thus, the correct answer is option 2).