Sequence and series bsc (MDU) previous year question unit 1

 Unit 1 (2024) 

Q. 1  Let A and B be bounded subsets of R. Prove that the set A+B=(x+y:x∈ A,y ∈ B} is also bounded and Supremum (A+B) = Sup (A) + Sup (B).

We are given that A and B are bounded subsets of ℝ, and we define:

A + B = {x + y : x ∈ A, y ∈ B}

We want to prove two things:

1. A + B is bounded

Since A and B are bounded subsets of ℝ, there exist real numbers such that:

There exist m₁, M₁ ∈ ℝ with:

∀x ∈ A,  m₁ ≤ x ≤ M₁

There exist m₂, M₂ ∈ ℝ with:

∀y ∈ B,  m₂ ≤ y ≤ M₂

Now consider any element z ∈ A + B, so z = x + y for some x ∈ A and y ∈ B.

Then:

  x + y ≤ M₁ + M₂

  x + y ≥ m₁ + m₂

So:

  ∀z ∈ A + B, we have m₁ + m₂ ≤ z ≤ M₁ + M₂

Hence, A + B is bounded.

2. sup(A + B) = sup(A) + sup(B)

Let:

s = sup(A)

t = sup(B)

We want to show: sup(A + B) = s + t

(i) Show sup(A + B) ≤ s + t:

Take any x ∈ A, then x ≤ s

Take any y ∈ B, then y ≤ t

So x + y ≤ s + t

⇒ Every element of A + B is ≤ s + t

Thus, s + t is an upper bound of A + B,

⇒ sup(A + B) ≤ s + t(ii) Show sup(A + B) ≥ s + t:

We’ll show that for any ε > 0, there exists x ∈ A, y ∈ B such that:

  x + y > s + t − ε

Since s = sup(A), there exists x ∈ A such that:

  x > s − ε/2

Since t = sup(B), there exists y ∈ B such that:

  y > t − ε/2

Then:

  x + y > (s − ε/2) + (t − ε/2) = s + t − ε

So, for any ε > 0, we can find an element in A + B that exceeds s + t − ε,

⇒ sup(A + B) ≥ s + t

Conclusion:

From both directions:

  sup(A + B) = s + t = sup(A) + sup(B)

And since A + B is between finite lower and upper bounds, it's bounded.

Q. 2 Define interior point of a set. Prove that for any set A. its interior Aº is open.

Definition: Interior Point

A point x ∈ ℝ is called an interior point of a set A ⊆ ℝ if there exists an open interval (x − ε, x + ε) (for some ε > 0) such that:

  (x − ε, x + ε) ⊆ A

Interior of a Set (A°)

The interior of A, denoted by A°, is the set of all interior points of A:

  A° = {x ∈ A : x is an interior point of A}

Theorem: A° is open

Proof:

We want to prove that A° is open, i.e., every point in A° is an interior point of A°.

Let x ∈ A°.

Then by definition of interior, there exists ε > 0 such that:

  (x − ε, x + ε) ⊆ A

But since x ∈ A° and the interval is entirely within A, it means:

∀y ∈ (x − ε, x + ε), the point y also satisfies the same condition —

  i.e., there is some open interval around y that also lies in A

⇒ So (x − ε, x + ε) ⊆ A°

Hence, x is an interior point of A°, and this is true for all x ∈ A°

⇒ A° is open, by definition of open sets.

Conclusion:

The interior of any set A, denoted A°, is an open set.

Q.E.D.

Q. 3 Prove that every infinite bounded subset of real numbers has a limit point. 

This is a classic result in real analysis and is a consequence of the Bolzano–Weierstrass Theorem.

Theorem:

Every infinite bounded subset of real numbers has at least one limit point.

Definitions:

A limit point (also called an accumulation point) of a set A ⊆ ℝ is a point x ∈ ℝ such that every open interval around x contains at least one point of A different from x.

A set is bounded if it lies within some finite interval [m, M].

Proof:

Let A ⊆ ℝ be infinite and bounded.

Since A is bounded, it lies within some closed interval [m, M].

The interval [m, M] is a closed and bounded interval in ℝ, hence compact.

Now, consider the set A as a subset of the compact interval [m, M].

By the Bolzano–Weierstrass Theorem:

Every infinite bounded subset of ℝ has a convergent subsequence (or has at least one limit point in ℝ).

In our context, this means that:

There exists a real number x ∈ ℝ such that some sequence of distinct points from A converges to x.

Therefore, x is a limit point of A.

Conclusion:

Every infinite bounded subset of ℝ has at least one limit point in ℝ.

Q. 4 Define a compact set. Show that the set of integers is not a compact set.

Definition: Compact Set

A set K ⊆ ℝ is called compact if every open cover of K has a finite subcover.

In ℝ, thanks to the Heine–Borel Theorem, this is equivalent to:

A set K ⊆ ℝ is compact if and only if it is closed and bounded.

Claim: The set of integers ℤ is not compact.

Proof:

We will show that ℤ is not bounded, so it cannot be compact.

The set of integers is:

  ℤ = {..., −3, −2, −1, 0, 1, 2, 3, ...}

Clearly, ℤ is not bounded, because there is no real number M such that |n| ≤ M for all n ∈ ℤ.

Since compact subsets of ℝ must be bounded, and ℤ is unbounded, ℤ is not compact.

Alternate argument using open covers:

Define the open intervals:

  Uₙ = (n − 0.5, n + 0.5) for each n ∈ ℤ

Then:

Each Uₙ is an open interval containing the integer n

The collection {Uₙ : n ∈ ℤ} is an open cover of ℤ

But no finite subcollection of these intervals can cover all of ℤ (since ℤ is infinite)

So, this open cover has no finite subcover.

Conclusion:

The set of integers ℤ is not compact because it is unbounded (and also because some open covers of ℤ do not admit finite subcovers).