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