MMT-004 Solved Assignment January 2026 | REAL ANALYSIS | IGNOU MSCMACS

Q1. State whether the following statements are true or false. Justify your answers.

a)) The outer measure m* of the set A = {x ∈ R : x² = 1}[−3,2] is 0. (80 words)
b)) A finite subset of a metric space is totally bounded. (80 words)
c)) A connected subspace in a metric space which is not properly contained in any other connected subspace is always open. (80 words)
d)) The surface given by the equation x+y+z-sin(xyz) = 0 can also be described by an equation of the form z = f (x, y) in a neighbourhood of the point (0,0). (80 words)
e)) A real valued function f on [a,b] is continuous if it is integrable on [a,b]. (80 words)

Key Points:

The outer measure of a finite (or countable) set in R is always zero.
A finite subset of any metric space is always totally bounded by using itself as an ε-net.
Connected components of a metric space are always closed, but not necessarily open.
The Implicit Function Theorem allows expressing one variable as a function of others if a specific partial derivative is non-zero.

Answer: This question assesses fundamental concepts in Real Analysis, covering topics like measure theory, metric spaces, multivariable calculus (Implicit Function Theorem), and properties of functions related to continuity and integrability. Each sub-question requires a precise understanding of definitions and theorems to determine the truth value of the statement and provide a concise justification, often involving counterexamples or direct application of theorems. Careful consideration of the condi...

Q2. Find the interior, closure, the set of limit points and the boundary of the set A = {(x, y) ∈ R² : y = 1} in R² with the standard metric.

Key Points:

Interior of a line in R² with standard metric is always empty.
A point is an interior point if an open ball around it lies entirely within the set.
Every point on the line A is a limit point of A in R².
The set of limit points of A is equal to A itself.

Answer: The set A = {(x, y) ∈ R² : y = 1} represents a straight line in R² parallel to the x-axis. We determine its topological properties using standard metric definitions from MMT-004. First, the **interior** of A, denoted A°, is the set of points for which an open disk centered at that point is entirely contained within A. For any point (x₀, 1) ∈ A, any open disk around it will necessarily contain points where y ≠ 1. Therefore, no such open disk can be fully contained in A, making the interior A° = ...

Q2. Consider f: R³ → R³ given by f(x, y, z) = (2x+3y + z, xy, yz, xz) Find f (2,0,-1).

Key Points:

Interior of a line in R² with standard metric is always empty.
A point is an interior point if an open ball around it lies entirely within the set.
Every point on the line A is a limit point of A in R².
The set of limit points of A is equal to A itself.

Answer: The set A = {(x, y) ∈ R² : y = 1} represents a straight line in R² parallel to the x-axis. We determine its topological properties using standard metric definitions from MMT-004. First, the **interior** of A, denoted A°, is the set of points for which an open disk centered at that point is entirely contained within A. For any point (x₀, 1) ∈ A, any open disk around it will necessarily contain points where y ≠ 1. Therefore, no such open disk can be fully contained in A, making the interior A° = ...

Q2. Does Cantor's intersection theorem hold for the metric space X = (0,1] with the standard metric? Justify your answer.

Key Points:

Interior of a line in R² with standard metric is always empty.
A point is an interior point if an open ball around it lies entirely within the set.
Every point on the line A is a limit point of A in R².
The set of limit points of A is equal to A itself.

Answer: The set A = {(x, y) ∈ R² : y = 1} represents a straight line in R² parallel to the x-axis. We determine its topological properties using standard metric definitions from MMT-004. First, the **interior** of A, denoted A°, is the set of points for which an open disk centered at that point is entirely contained within A. For any point (x₀, 1) ∈ A, any open disk around it will necessarily contain points where y ≠ 1. Therefore, no such open disk can be fully contained in A, making the interior A° = ...

Q3. Obtain the second Taylor's series expansion for the function given by f(x, y) = xy² + 5xeʸ at (1, -1/2).

Key Points:

Taylor series expands function around a point using its derivatives.
Second-order expansion includes terms up to degree two.
Calculate function value f(a,b) at the expansion point.
Evaluate first partial derivatives f_x, f_y at (a,b).

Answer: To obtain the second Taylor's series expansion for f(x, y) = xy² + 5xeʸ at (1, -1/2), we use the formula: f(x, y) ≈ f(a, b) + (x-a)fₓ(a, b) + (y-b)fᵧ(a, b) + (1/2!)[(x-a)²fₓₓ(a, b) + 2(x-a)(y-b)fₓᵧ(a, b) + (y-b)²fᵧᵧ(a, b)]. First, we evaluate the function and its partial derivatives at (a,b) = (1, -1/2): f(1, -1/2) = 1/4 + 5/√e fₓ(x,y) = y² + 5eʸ; thus, fₓ(1, -1/2) = 1/4 + 5/√e fᵧ(x,y) = 2xy + 5xeʸ; thus, fᵧ(1, -1/2) = -1 + 5/√e Next, we calculate the second-order partial derivatives: fₓₓ(x,y...

Q3. Find the Lebesgue integral of the function f given by f = χ[0,1] + 2χ[3,5] + χ[6,9].

Key Points:

Taylor series expands function around a point using its derivatives.
Second-order expansion includes terms up to degree two.
Calculate function value f(a,b) at the expansion point.
Evaluate first partial derivatives f_x, f_y at (a,b).

Answer: To obtain the second Taylor's series expansion for f(x, y) = xy² + 5xeʸ at (1, -1/2), we use the formula: f(x, y) ≈ f(a, b) + (x-a)fₓ(a, b) + (y-b)fᵧ(a, b) + (1/2!)[(x-a)²fₓₓ(a, b) + 2(x-a)(y-b)fₓᵧ(a, b) + (y-b)²fᵧᵧ(a, b)]. First, we evaluate the function and its partial derivatives at (a,b) = (1, -1/2): f(1, -1/2) = 1/4 + 5/√e fₓ(x,y) = y² + 5eʸ; thus, fₓ(1, -1/2) = 1/4 + 5/√e fᵧ(x,y) = 2xy + 5xeʸ; thus, fᵧ(1, -1/2) = -1 + 5/√e Next, we calculate the second-order partial derivatives: fₓₓ(x,y...

Q3. Find and classify the extreme values of f(x, y) = xy Subject to the constraint x²/8 + y²/2 - 1 = 0.

Key Points:

Taylor series expands function around a point using its derivatives.
Second-order expansion includes terms up to degree two.
Calculate function value f(a,b) at the expansion point.
Evaluate first partial derivatives f_x, f_y at (a,b).

Answer: To obtain the second Taylor's series expansion for f(x, y) = xy² + 5xeʸ at (1, -1/2), we use the formula: f(x, y) ≈ f(a, b) + (x-a)fₓ(a, b) + (y-b)fᵧ(a, b) + (1/2!)[(x-a)²fₓₓ(a, b) + 2(x-a)(y-b)fₓᵧ(a, b) + (y-b)²fᵧᵧ(a, b)]. First, we evaluate the function and its partial derivatives at (a,b) = (1, -1/2): f(1, -1/2) = 1/4 + 5/√e fₓ(x,y) = y² + 5eʸ; thus, fₓ(1, -1/2) = 1/4 + 5/√e fᵧ(x,y) = 2xy + 5xeʸ; thus, fᵧ(1, -1/2) = -1 + 5/√e Next, we calculate the second-order partial derivatives: fₓₓ(x,y...

Q4. Let f: R³ \{(0,0,0)} → R³ \{(0,0,0)} be given by f(x, y, z) = (x/(x²+y²+z²), y/(x²+y²+z²), z/(x²+y²+z²)). Show that f is locally invertible at all points in R³ \{(0,0,0)}.

Key Points:

Local invertibility is determined by the Inverse Function Theorem.
Inverse Function Theorem requires C¹ function and non-zero Jacobian determinant.
Function f(x) = x/||x||² is continuously differentiable (C¹) on its domain.
Jacobian determinant for f(X) = X/||X||² in R³ is det(Jf(X)) = -1/||X||⁶.

Answer: To show that the function f: R³ \{(0,0,0)} → R³ \{(0,0,0)} defined by f(x, y, z) = (x/(x²+y²+z²), y/(x²+y²+z²), z/(x²+y²+z²)) is locally invertible at all points, we utilize the Inverse Function Theorem. This theorem states that a continuously differentiable (C¹) function is locally invertible at a point if its Jacobian determinant at that point is non-zero. The given function f can be written as f(X) = X/||X||², where X = (x,y,z) and ||X||² = x²+y²+z². Since ||X||² is non-zero on the domain, a...

Q4. For the equation x² + y² + z³ = 0, at which points on its solution set, can we be assured that there is a neighbourhood of the point in which the surface given by the equation can be described by an equation of the form z = f (x, y).

Key Points:

Local invertibility is determined by the Inverse Function Theorem.
Inverse Function Theorem requires C¹ function and non-zero Jacobian determinant.
Function f(x) = x/||x||² is continuously differentiable (C¹) on its domain.
Jacobian determinant for f(X) = X/||X||² in R³ is det(Jf(X)) = -1/||X||⁶.

Answer: To show that the function f: R³ \{(0,0,0)} → R³ \{(0,0,0)} defined by f(x, y, z) = (x/(x²+y²+z²), y/(x²+y²+z²), z/(x²+y²+z²)) is locally invertible at all points, we utilize the Inverse Function Theorem. This theorem states that a continuously differentiable (C¹) function is locally invertible at a point if its Jacobian determinant at that point is non-zero. The given function f can be written as f(X) = X/||X||², where X = (x,y,z) and ||X||² = x²+y²+z². Since ||X||² is non-zero on the domain, a...

Q4. Find the Fourier series of f(t) = t² on [−π,π].

Key Points:

Local invertibility is determined by the Inverse Function Theorem.
Inverse Function Theorem requires C¹ function and non-zero Jacobian determinant.
Function f(x) = x/||x||² is continuously differentiable (C¹) on its domain.
Jacobian determinant for f(X) = X/||X||² in R³ is det(Jf(X)) = -1/||X||⁶.

Answer: To show that the function f: R³ \{(0,0,0)} → R³ \{(0,0,0)} defined by f(x, y, z) = (x/(x²+y²+z²), y/(x²+y²+z²), z/(x²+y²+z²)) is locally invertible at all points, we utilize the Inverse Function Theorem. This theorem states that a continuously differentiable (C¹) function is locally invertible at a point if its Jacobian determinant at that point is non-zero. The given function f can be written as f(X) = X/||X||², where X = (x,y,z) and ||X||² = x²+y²+z². Since ||X||² is non-zero on the domain, a...

Q5. Prove that if an open set U can be written as the union of pairwise disjoint family V of open connected subsets, then these subsets must be the components of U. Use this theorem to find the components of the set D U E where D = {(x, y) ∈ R² : ||(x+1, y)|| = 1} and E = {(x, y) ∈ R² : ||x −1, y|| = 1}.

Key Points:

A connected component is a maximal connected subset of a topological space.
The proof relies on a connected subset intersecting at most one disjoint open component.
The theorem explicitly requires the set U to be open for direct application.
Circles in R² are connected but are closed sets, not open sets.

Answer: A connected component of a topological space U is its maximal connected subset. The theorem states that if an open set U is a union of a pairwise disjoint family V = {Vᵢ} of open connected subsets, then these Vᵢ must be the components of U. To prove this, we show each Vᵢ is maximal connected. Assume C is a connected subset of U with Vᵢ ⊆ C. Since U is the disjoint union of open Vⱼ's, any connected set within U can intersect at most one Vⱼ. If C intersected distinct Vⱼ and Vₖ, Vⱼ and (U \ Vⱼ) wou...

Q5. Which of the following subsets of R are compact w.r.t. the metric given against them. Justify your answer.

i)) A = (0,1) in R with standard metric. (50 words)
ii)) A = [3,4] with discrete metric. (50 words)
iii)) {(x, y) ∈ R² : y > 0} with standard metric. (50 words)

Key Points:

A connected component is a maximal connected subset of a topological space.
The proof relies on a connected subset intersecting at most one disjoint open component.
The theorem explicitly requires the set U to be open for direct application.
Circles in R² are connected but are closed sets, not open sets.

Answer: A connected component of a topological space U is its maximal connected subset. The theorem states that if an open set U is a union of a pairwise disjoint family V = {Vᵢ} of open connected subsets, then these Vᵢ must be the components of U. To prove this, we show each Vᵢ is maximal connected. Assume C is a connected subset of U with Vᵢ ⊆ C. Since U is the disjoint union of open Vⱼ's, any connected set within U can intersect at most one Vⱼ. If C intersected distinct Vⱼ and Vₖ, Vⱼ and (U \ Vⱼ) wou...

Q6. If E is a subset of R with standard metric, then show that (E bar) = (E circle) circle.

Key Points:

Ē is the closure of E (smallest closed set containing E).
E° is the interior of E (largest open set contained in E).
The interior operator is idempotent: (A°)° = A°.
Therefore, the question simplifies to showing Ē = E°.

Answer: In the context of Real Analysis (MMT-004) with the standard metric on R, the notation Ē represents the **closure of E**, which is the smallest closed set containing E. The notation E° represents the **interior of E**, defined as the largest open set contained within E. The expression (E circle) circle is interpreted as the interior of the interior of E, denoted as (E°)°. A key property of the interior operator is its idempotency: for any set A, (A°)° = A°. Thus, (E°)° simplifies to E°. Theref...

Q6. Show that a set A in a metric space is closed if and only if every convergent sequence in A converges to a point of A.

Key Points:

Ē is the closure of E (smallest closed set containing E).
E° is the interior of E (largest open set contained in E).
The interior operator is idempotent: (A°)° = A°.
Therefore, the question simplifies to showing Ē = E°.

Answer: In the context of Real Analysis (MMT-004) with the standard metric on R, the notation Ē represents the **closure of E**, which is the smallest closed set containing E. The notation E° represents the **interior of E**, defined as the largest open set contained within E. The expression (E circle) circle is interpreted as the interior of the interior of E, denoted as (E°)°. A key property of the interior operator is its idempotency: for any set A, (A°)° = A°. Thus, (E°)° simplifies to E°. Theref...

Q6. Find the interior and closure of the set Q of rationals in R with standard metric.

Key Points:

Ē is the closure of E (smallest closed set containing E).
E° is the interior of E (largest open set contained in E).
The interior operator is idempotent: (A°)° = A°.
Therefore, the question simplifies to showing Ē = E°.

Answer: In the context of Real Analysis (MMT-004) with the standard metric on R, the notation Ē represents the **closure of E**, which is the smallest closed set containing E. The notation E° represents the **interior of E**, defined as the largest open set contained within E. The expression (E circle) circle is interpreted as the interior of the interior of E, denoted as (E°)°. A key property of the interior operator is its idempotency: for any set A, (A°)° = A°. Thus, (E°)° simplifies to E°. Theref...

Q7. Let F be the function from R² to R² defined by F(x, y) = (x² + y², xy). Show that F is differentiable at (1,2). Find the differential matrix of F.

Key Points:

A function is differentiable if its first-order partial derivatives exist and are continuous.
Component functions of F(x, y) = (x² + y², xy) are F₁(x, y) = x² + y² and F₂(x, y) = xy.
Partial derivatives are ∂F₁/∂x = 2x, ∂F₁/∂y = 2y, ∂F₂/∂x = y, ∂F₂/∂y = x.
All partial derivatives are polynomials, hence continuous everywhere in R².

Answer: The given function F: R² → R² is defined by F(x, y) = (x² + y², xy). This function can be expressed by its component functions F₁(x, y) = x² + y² and F₂(x, y) = xy. To show that F is differentiable at (1,2), we need to demonstrate that all its first-order partial derivatives exist and are continuous in a neighborhood of (1,2), as per the conditions for differentiability in real analysis. First, we compute the first-order partial derivatives of F₁(x, y) and F₂(x, y) with respect to x and y. The ...

Q7. Show that the function f defined by f(x, y) = xy/(x² + y²) if (x, y) ≠ (0,0) and 0 if (x, y) = (0,0) is not differentiable at (0,0). Does the partial derivatives of f exist at (0,0) or at any other point in R²? Justify your answer.

Key Points:

A function is differentiable if its first-order partial derivatives exist and are continuous.
Component functions of F(x, y) = (x² + y², xy) are F₁(x, y) = x² + y² and F₂(x, y) = xy.
Partial derivatives are ∂F₁/∂x = 2x, ∂F₁/∂y = 2y, ∂F₂/∂x = y, ∂F₂/∂y = x.
All partial derivatives are polynomials, hence continuous everywhere in R².

Answer: The given function F: R² → R² is defined by F(x, y) = (x² + y², xy). This function can be expressed by its component functions F₁(x, y) = x² + y² and F₂(x, y) = xy. To show that F is differentiable at (1,2), we need to demonstrate that all its first-order partial derivatives exist and are continuous in a neighborhood of (1,2), as per the conditions for differentiability in real analysis. First, we compute the first-order partial derivatives of F₁(x, y) and F₂(x, y) with respect to x and y. The ...

Q7. Is the continuous image of a Cauchy sequence a Cauchy sequence? Justify.

Key Points:

A function is differentiable if its first-order partial derivatives exist and are continuous.
Component functions of F(x, y) = (x² + y², xy) are F₁(x, y) = x² + y² and F₂(x, y) = xy.
Partial derivatives are ∂F₁/∂x = 2x, ∂F₁/∂y = 2y, ∂F₂/∂x = y, ∂F₂/∂y = x.
All partial derivatives are polynomials, hence continuous everywhere in R².

Answer: The given function F: R² → R² is defined by F(x, y) = (x² + y², xy). This function can be expressed by its component functions F₁(x, y) = x² + y² and F₂(x, y) = xy. To show that F is differentiable at (1,2), we need to demonstrate that all its first-order partial derivatives exist and are continuous in a neighborhood of (1,2), as per the conditions for differentiability in real analysis. First, we compute the first-order partial derivatives of F₁(x, y) and F₂(x, y) with respect to x and y. The ...

Q8. Find the directional derivative of the function f : R⁴ → R⁴ defined by f(x, y, z, w) = (x²y, xyz, x² + y², zw²) at the point (1,2,-1,-2) in the direction v = (1,0,-2, 2).

Key Points:

Directional derivative of `f: Rⁿ → Rᵐ` is `Df(x₀)v`.
Jacobian matrix `Df(x₀)` contains all first-order partial derivatives.
Calculate `∂fᵢ/∂xⱼ` for each component function `fᵢ`.
Evaluate the Jacobian matrix at the given point `(1,2,-1,-2)`.

Answer: The directional derivative of a function `f: Rⁿ → Rᵐ` at a point `x₀` in the direction `v` is given by the matrix-vector product `Df(x₀)v`, where `Df(x₀)` is the Jacobian matrix of `f` evaluated at `x₀`. The Jacobian matrix consists of the partial derivatives of each component function. First, we find the partial derivatives for each component of `f(x, y, z, w) = (f₁, f₂, f₃, f₄) = (x²y, xyz, x² + y², zw²)`. The Jacobian matrix `Df(x,y,z,w)` is: `Df(x,y,z,w) = ` `| 2xy x² 0 0 |` `| y...

Q8. Suppose that f : R → R² is given by f(t) = (t,t²) and g : R² → R³ is given by g(x, y) = (x², xy, y² – x²). Compute the derivative of g∘f.

Key Points:

Directional derivative of `f: Rⁿ → Rᵐ` is `Df(x₀)v`.
Jacobian matrix `Df(x₀)` contains all first-order partial derivatives.
Calculate `∂fᵢ/∂xⱼ` for each component function `fᵢ`.
Evaluate the Jacobian matrix at the given point `(1,2,-1,-2)`.

Answer: The directional derivative of a function `f: Rⁿ → Rᵐ` at a point `x₀` in the direction `v` is given by the matrix-vector product `Df(x₀)v`, where `Df(x₀)` is the Jacobian matrix of `f` evaluated at `x₀`. The Jacobian matrix consists of the partial derivatives of each component function. First, we find the partial derivatives for each component of `f(x, y, z, w) = (f₁, f₂, f₃, f₄) = (x²y, xyz, x² + y², zw²)`. The Jacobian matrix `Df(x,y,z,w)` is: `Df(x,y,z,w) = ` `| 2xy x² 0 0 |` `| y...

Q8. Find B[2, 1/2] in R where d is the metric given by d(x, y) = min{1,|x − y |}.

Key Points:

Directional derivative of `f: Rⁿ → Rᵐ` is `Df(x₀)v`.
Jacobian matrix `Df(x₀)` contains all first-order partial derivatives.
Calculate `∂fᵢ/∂xⱼ` for each component function `fᵢ`.
Evaluate the Jacobian matrix at the given point `(1,2,-1,-2)`.

Answer: The directional derivative of a function `f: Rⁿ → Rᵐ` at a point `x₀` in the direction `v` is given by the matrix-vector product `Df(x₀)v`, where `Df(x₀)` is the Jacobian matrix of `f` evaluated at `x₀`. The Jacobian matrix consists of the partial derivatives of each component function. First, we find the partial derivatives for each component of `f(x, y, z, w) = (f₁, f₂, f₃, f₄) = (x²y, xyz, x² + y², zw²)`. The Jacobian matrix `Df(x,y,z,w)` is: `Df(x,y,z,w) = ` `| 2xy x² 0 0 |` `| y...

Q9. Give an example of a family f_λ of subsets of a set X which has finite intersection property. Justify your choice of example.

Key Points:

FIP requires any finite subcollection's intersection to be non-empty.
Let X = ℝ and the family be f_n = [0, 1/n] for n ∈ ℕ, n ≥ 1.
For any finite subcollection {f_{n₁}, ..., f_{n_k}}, identify M = max(n₁, ..., n_k).
The intersection of this finite subcollection is precisely [0, 1/M].

Answer: In Real Analysis, a family of subsets f_λ = {A_λ : λ ∈ Λ} of a set X is said to have the Finite Intersection Property (FIP) if for every finite subcollection {A_λ₁, A_λ₂, ..., A_λₙ} from the family, their intersection is non-empty. That is, A_λ₁ ∩ A_λ₂ ∩ ... ∩ A_λₙ ≠ ∅. This property is fundamental in concepts like compactness, as explored in MMT-004. Let's consider an example of a family of subsets with the FIP. Let our universal set X be the set of real numbers, ℝ. Define the family of subset...

Q9. Verify the hypothesis and conclusions of the Fatou's lemma for the sequence {f_n} given by f_n(x) = 2n for x ∈ (1/(2n), 1/n) and 0 for x ∈ [0, 1/(2n)] U (1/n, 1].

Key Points:

FIP requires any finite subcollection's intersection to be non-empty.
Let X = ℝ and the family be f_n = [0, 1/n] for n ∈ ℕ, n ≥ 1.
For any finite subcollection {f_{n₁}, ..., f_{n_k}}, identify M = max(n₁, ..., n_k).
The intersection of this finite subcollection is precisely [0, 1/M].

Answer: In Real Analysis, a family of subsets f_λ = {A_λ : λ ∈ Λ} of a set X is said to have the Finite Intersection Property (FIP) if for every finite subcollection {A_λ₁, A_λ₂, ..., A_λₙ} from the family, their intersection is non-empty. That is, A_λ₁ ∩ A_λ₂ ∩ ... ∩ A_λₙ ≠ ∅. This property is fundamental in concepts like compactness, as explored in MMT-004. Let's consider an example of a family of subsets with the FIP. Let our universal set X be the set of real numbers, ℝ. Define the family of subset...

Q10. Let (X,d) be a metric space and A be a subset of X. Show that bdy(A) = ∅ if and only if A is both open and closed.

Key Points:

Boundary (bdy(A)) is defined as cl(A) \ int(A).
A set A is closed if and only if A = cl(A).
A set A is open if and only if A = int(A).
If A is open and closed, then cl(A) = A and int(A) = A.

Answer: Let (X,d) be a metric space and A be a subset of X. We need to show that the boundary of A, denoted bdy(A), is empty if and only if A is both open and closed. First, assume A is both open and closed. By definition, an open set A satisfies A = int(A) (A equals its interior), and a closed set A satisfies A = cl(A) (A equals its closure). The boundary of A is defined as bdy(A) = cl(A) \ int(A). Substituting A for both cl(A) and int(A) in the boundary definition, we get bdy(A) = A \ A = ∅. Thus, i...

Q10. Give an example of an algebra which is not a σ– algebra. Justify your choice of examples.

Key Points:

Boundary (bdy(A)) is defined as cl(A) \ int(A).
A set A is closed if and only if A = cl(A).
A set A is open if and only if A = int(A).
If A is open and closed, then cl(A) = A and int(A) = A.

Answer: Let (X,d) be a metric space and A be a subset of X. We need to show that the boundary of A, denoted bdy(A), is empty if and only if A is both open and closed. First, assume A is both open and closed. By definition, an open set A satisfies A = int(A) (A equals its interior), and a closed set A satisfies A = cl(A) (A equals its closure). The boundary of A is defined as bdy(A) = cl(A) \ int(A). Substituting A for both cl(A) and int(A) in the boundary definition, we get bdy(A) = A \ A = ∅. Thus, i...

Q10. If E is a measurable set and f is a simple function such that a ≤ f(x) ≤ b ∀x∈ E, show that a m(E) ≤ ∫_E f dm ≤ m(E)b.

Key Points:

Boundary (bdy(A)) is defined as cl(A) \ int(A).
A set A is closed if and only if A = cl(A).
A set A is open if and only if A = int(A).
If A is open and closed, then cl(A) = A and int(A) = A.

Answer: Let (X,d) be a metric space and A be a subset of X. We need to show that the boundary of A, denoted bdy(A), is empty if and only if A is both open and closed. First, assume A is both open and closed. By definition, an open set A satisfies A = int(A) (A equals its interior), and a closed set A satisfies A = cl(A) (A equals its closure). The boundary of A is defined as bdy(A) = cl(A) \ int(A). Substituting A for both cl(A) and int(A) in the boundary definition, we get bdy(A) = A \ A = ∅. Thus, i...

MMT-004 Solved Assignment — January 2026 / July 2026

REAL ANALYSIS | IGNOU Master of Science (Mathematics with Applications in Computer Science) (MSCMACS)

MMT-004—January 2026 / July 2026

REAL ANALYSIS

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MMT-004 Assignment Questions — January 2026

Q1. State whether the following statements are true or false. Justify your answers.

Q2. Find the interior, closure, the set of limit points and the boundary of the set A = {(x, y) ∈ R² : y = 1} in R² with the standard metric.

Q2. Consider f: R³ → R³ given by f(x, y, z) = (2x+3y + z, xy, yz, xz) Find f (2,0,-1).

Q2. Does Cantor's intersection theorem hold for the metric space X = (0,1] with the standard metric? Justify your answer.

Q3. Obtain the second Taylor's series expansion for the function given by f(x, y) = xy² + 5xeʸ at (1, -1/2).

Q3. Find the Lebesgue integral of the function f given by f = χ[0,1] + 2χ[3,5] + χ[6,9].

Q3. Find and classify the extreme values of f(x, y) = xy Subject to the constraint x²/8 + y²/2 - 1 = 0.

Q4. Let f: R³ \{(0,0,0)} → R³ \{(0,0,0)} be given by f(x, y, z) = (x/(x²+y²+z²), y/(x²+y²+z²), z/(x²+y²+z²)). Show that f is locally invertible at all points in R³ \{(0,0,0)}.

Q4. For the equation x² + y² + z³ = 0, at which points on its solution set, can we be assured that there is a neighbourhood of the point in which the surface given by the equation can be described by an equation of the form z = f (x, y).

Q4. Find the Fourier series of f(t) = t² on [−π,π].

Q5. Which of the following subsets of R are compact w.r.t. the metric given against them. Justify your answer.

Q6. If E is a subset of R with standard metric, then show that (E bar) = (E circle) circle.

Q6. Show that a set A in a metric space is closed if and only if every convergent sequence in A converges to a point of A.

Q6. Find the interior and closure of the set Q of rationals in R with standard metric.

Q7. Let F be the function from R² to R² defined by F(x, y) = (x² + y², xy). Show that F is differentiable at (1,2). Find the differential matrix of F.

Q7. Show that the function f defined by f(x, y) = xy/(x² + y²) if (x, y) ≠ (0,0) and 0 if (x, y) = (0,0) is not differentiable at (0,0). Does the partial derivatives of f exist at (0,0) or at any other point in R²? Justify your answer.

Q7. Is the continuous image of a Cauchy sequence a Cauchy sequence? Justify.

Q8. Find the directional derivative of the function f : R⁴ → R⁴ defined by f(x, y, z, w) = (x²y, xyz, x² + y², zw²) at the point (1,2,-1,-2) in the direction v = (1,0,-2, 2).

Q8. Suppose that f : R → R² is given by f(t) = (t,t²) and g : R² → R³ is given by g(x, y) = (x², xy, y² – x²). Compute the derivative of g∘f.

Q8. Find B[2, 1/2] in R where d is the metric given by d(x, y) = min{1,|x − y |}.

Q9. Give an example of a family f_λ of subsets of a set X which has finite intersection property. Justify your choice of example.

Q9. Verify the hypothesis and conclusions of the Fatou's lemma for the sequence {f_n} given by f_n(x) = 2n for x ∈ (1/(2n), 1/n) and 0 for x ∈ [0, 1/(2n)] U (1/n, 1].

Q10. Let (X,d) be a metric space and A be a subset of X. Show that bdy(A) = ∅ if and only if A is both open and closed.

Q10. Give an example of an algebra which is not a σ– algebra. Justify your choice of examples.

Q10. If E is a measurable set and f is a simple function such that a ≤ f(x) ≤ b ∀x∈ E, show that a m(E) ≤ ∫_E f dm ≤ m(E)b.