Problem 1 - Calculus One Again

Question

Let \(\boldsymbol{A}\) and \(\boldsymbol{B}\) be non-empty sets of real numbers. Put $$ -A=\\{-x: x \in A\\}, \quad A-B=\\{a-b:(a, b) \in A \times B\\} $$ Prowe that 1\. If \(A\) is bounded above, then \(-A\) is bounded below and \(\sup A=\) \(-\inf (-A)\) 2\. If \(\boldsymbol{A}\) and \(\boldsymbol{B}\) are bounded above then \(\boldsymbol{A} \cup \boldsymbol{B}\) is also bounded above and \(\sup (A \cup B)=\max (\sup A, \sup B)\). 3\. If \(\boldsymbol{A}\) is bounded above and \(\boldsymbol{B}\) is bounded below, then \(\boldsymbol{A}-\boldsymbol{B}\) is bounded above and \(\sup (A-B)=\sup A-\inf B\).

Step-by-Step Solution

Answer

1. \(-A\) is bounded below; \(\sup A = -\inf(-A)\). 2. \(A \cup B\) is bounded above; \(\sup(A \cup B) = \max(\sup A, \sup B)\). 3. \(A - B\) is bounded above; \(\sup(A - B) = \sup A - \inf B\).

Step 1: Understanding the Sets

We are given two sets of real numbers, \(\boldsymbol{A}\) and \(\boldsymbol{B}\). We need to manipulate these sets using the operations of negation and subtraction of elements.

Step 2: Prove Part 1 - Boundedness of Negative Set

Assume \(\boldsymbol{A}\) is bounded above. Let \(M\) be an upper bound of \(A\). This implies \(x \leq M\) for all \(x \in A\). Then \(-x \geq -M\) for all \(-x \in -A\). Thus, \(-M\) is the lower bound of \(-A\).

Step 3: Prove Part 1 - Supremum and Infimum Relation

By definition of supremum, \(\sup A\) is the least upper bound of \(A\). Since \(-A\) consists of all \(-x\) such that \(x \in A\), the greatest lower bound of \(-A\) is \(-\sup A\). This implies \(-\inf (-A) = \sup A\).

Step 4: Prove Part 2 - Boundedness of Union

Assume \(\boldsymbol{A}\) and \(\boldsymbol{B}\) are both bounded above. Let \(M\) and \(N\) be their respective upper bounds. For any element \(x \in A \cup B\), \(x \leq \max(M, N)\). Thus, \(A \cup B\) is bounded above by \(\max(M, N)\).

Step 5: Prove Part 2 - Supremum of Union

The supremum of \(A \cup B\) is the least upper bound. Since any upper bound of \(A\) and \(B\) is an upper bound for \(A \cup B\), the smallest such bound is \(\max(\sup A, \sup B)\).

Step 6: Prove Part 3 - Boundedness of Difference

Assume \(\boldsymbol{A}\) is bounded above by \(M\) and \(\boldsymbol{B}\) is bounded below by \(m\). For any element \(a - b \in A - B\), \(a \leq M\) and \(b \geq m\) imply \(a - b \leq M - m\). Thus, \(A - B\) is bounded above by \(M - m\).

Step 7: Prove Part 3 - Supremum of Difference

The supremum of \(A - B\) is the least upper bound. Consider any sequence \(a_n - b_n\) with \(a_n \to \sup A\) and \(b_n \to \inf B\). Then \(a_n - b_n \to \sup A - \inf B\), implying this is the least upper bound. Therefore, \(\sup(A - B) = \sup A - \inf B\).

Key Concepts

Supremum and Infimum

In real analysis, when studying sets of numbers, the concepts of supremum and infimum are fundamental.
The supremum (or least upper bound) of a set, denoted as \( \sup \), is the smallest number that is greater than or equal to every element in the set. On the other hand, the infimum (or greatest lower bound), denoted as \( \inf \), represents the largest number that is less than or equal to every element in the set.
To grasp these ideas, suppose we have a set \( A \) that is bounded above, meaning there's a number \( M \) such that no element in \( A \) is greater than \( M \). Here, \( \sup A \) is the smallest among all possible upper bounds of \( A \).

• If the highest number \( A \) can reach is not actually in the set, \( \sup A \) will be that number.
• If this number is within \( A \), then \( \sup A \) is simply the maximum element of \( A \).

If you have a set \( -A \) derived from \( A \) by multiplying each element by \(-1\), the least upper bound of \( A \) becomes the greatest lower bound of \( -A \). In simple terms, this relationship can be described as: \( \sup A = -\inf(-A) \).
Similarly, for a set bounded below, these concepts can be thought of in reverse. Comprehending supremum and infimum is crucial for manipulating and understanding bounds in sets.

Bounded Sets

A set of numbers, or a "bounded set," is described as either bounded above or bounded below, according to the context of the problem.
When a set is bounded above, there is a number greater than or equal to every element in the set. If bounded below, then there's a number less than or equal to every element in the set.
Bounded sets play a vital role in real analysis because they help us define limits and understand continuity.

• If both \( A \) and \( B \) are bounded above, the set operation \( A \cup B \) (the union of two sets) is also bounded above.
• In the case of subtraction, when \( A \) is bounded above and \( B \) is bounded below, the difference set \( A - B \) remains bounded. It will have an upper bound that is determined by the subtraction of their respective bounds.

Understanding bounded sets is foundational, as these rules help maintain logical consistency when performing set operations, like unions and differences.

Set Operations

In real analysis, set operations such as union and difference are essential for understanding how sets interact.
Consider the union of two sets, \( A \cup B \). This operation gathers all members from both sets. If \( A \) and \( B \) are both bounded above, then \( A \cup B \) is also bounded above, with its supremum being the largest supremum among the individual sets. Mathematically, \( \sup (A \cup B) = \max(\sup A, \sup B) \).
For the difference between two sets, \( A - B \), elements are formed by subtracting each element in \( B \) from each element in \( A \). If \( A \) is bounded above and \( B \) is bounded below, the resulting set \( A - B \) will be bounded above by \( \sup A - \inf B \).

• This shows how operations can fundamentally change what bounds a set.
• Understanding these interactions is critical for solving complex real analysis problems.

Overall, set operations allow the combination and manipulation of sets under specific rules, paving the road to advanced mathematical analysis.