Question

For any two subsets \(A\) and \(B\) of an ordered field \(F,\) let \(A+B\) denote the set of all sums \(x+y\) with \(x \in A\) and \(y \in B ;\) i.e., $$ A+B=\\{x+y \mid x \in A, y \in B\\} $$ Prove that if \(\sup A=p\) and \(\sup B=q\) exist in \(F,\) then $$ p+q=\sup (A+B) $$ similarly for infima. [Hint for sup: By Theorem 2 , we must show that (i) \((\forall x \in A)(\forall y \in B) x+y \leq p+q\) (which is easy) and \(\left(\mathrm{ii}^{\prime}\right)(\forall \varepsilon>0)(\exists x \in A)(\exists y \in B) x+y>(p+q)-\varepsilon\) Fix any \(\varepsilon>0 .\) By Theorem 2 , $$ (\exists x \in A)(\exists y \in B) \quad p-\frac{\varepsilon}{2}\left(p-\frac{\varepsilon}{2}\right)+\left(q-\frac{\varepsilon}{2}\right)=(p+q)-\varepsilon $$ as required.]

Short answer

The supremum of \( A + B \) is \( p + q \) because \( p+q \) is an upper bound for sums and can be approached arbitrarily closely.

Step-by-step solution

Understanding the Problem Statement

We have subsets \( A \) and \( B \) of an ordered field \( F \). We need to prove that \( \sup A = p \) and \( \sup B = q \) implies \( \sup (A+B) = p+q \). The task requires validation and demonstration of a condition involving the supremum under addition of elements from both sets.

Verify Upper Bound Condition

We are tasked with showing \((\forall x \in A)(\forall y \in B), x+y \leq p+q\). Since \( x \leq p \) and \( y \leq q \) hold for all \( x \in A \) and \( y \in B \) respectively, adding these inequalities gives \( x+y \leq p+q \) for all \( x \in A \) and \( y \in B \). Thus, \( p+q \) is an upper bound for \( A+B \).

Prove Approaching Supremum

The goal is to prove \( (\forall \varepsilon>0)(\exists x \in A)(\exists y \in B), x+y > (p+q)-\varepsilon \). For a given \( \varepsilon > 0 \), choose \( x \in A \) such that \( p - \frac{\varepsilon}{2} < x \) and \( y \in B \) such that \( q - \frac{\varepsilon}{2} < y \). By construction, this means \( x > p - \frac{\varepsilon}{2} \) and \( y > q - \frac{\varepsilon}{2} \).

Combine and Conclude

The previous inequalities allow us to conclude \( x+y > \left(p - \frac{\varepsilon}{2}\right) + \left(q - \frac{\varepsilon}{2}\right) = (p+q)-\varepsilon \). Thus, for any \( \varepsilon > 0 \), there exist elements of \( A \) and \( B \) whose sum exceeds \((p+q)-\varepsilon\). This implies that \( p+q \) is the least upper bound of \( A+B \) and therefore, \( \sup (A+B) = p+q \).

Key concepts

Supremum

When discussing ordered fields and subsets, the concept of a "supremum" is crucial. The supremum, or least upper bound, of a set is the smallest number that is greater than or equal to every number in the set. In other words, if you have a collection of numbers, the supremum is the smallest value that can "cap" the entire collection from above.

To find the supremum, observe these steps:

• Verify that the candidate number is an upper bound for the set, meaning it's greater than or equal to every element in the set.
• Ensure that there's no smaller number that can also act as an upper bound.

In the context of our original exercise, the supremum of the sum set, \( A + B \), is the sum of the suprema of \( A \) and \( B \). This means if \( \sup A = p \) and \( \sup B = q \), then \( \sup (A+B) = p+q \). The proof involves showing both that \( p+q \) is an upper bound for \( A+B \) and that for every small positive number \( \varepsilon \), you can find elements such that their sum gets arbitrarily close to \( p+q \).

This establishes \( p+q \) as the smallest possible value that cannot be exceeded by the sums of elements from \( A \) and \( B \). Thus, it serves as the supremum of \( A+B \).

Infimum

The infimum is another important concept in the context of ordered fields. While the supremum is the least upper bound, the infimum is the greatest lower bound of a set. It represents the largest number that is less than or equal to every number in the set.

To find the infimum, consider:

• Check if the candidate number is a lower bound for the set, meaning it's less than or equal to every element in the set.
• Ensure there's no greater number that can be a lower bound.

In the particular exercise we are discussing, the process of proving \( \inf (A+B) \) follows a similar logic to proving the supremum. If \( \inf A = p \) and \( \inf B = q \), then \( \inf (A+B) = p+q \). This involves demonstrating that \( p+q \) is a lower bound for \( A+B \) and, for any number slightly smaller than \( p+q \), there are elements in the individual subsets whose sum is greater than this smaller number.

By these steps, we confirm that \( p+q \) is indeed the greatest lower bound for \( A+B \). This makes it the infimum.

Subset

A "subset" is a fundamental, basic idea that often comes up in mathematical discussions, especially in set theory. Simply put, a subset is a set where every element in it is also contained in another set, called the "parent" or "superset."

Think of subsets like folders within folders on a computer:

• Every file in the subset folder is also present in the parent folder.
• It's entirely possible for a subset to be empty, containing no elements but still classified as a subset.
• Conversely, any set is a subset of itself because all of its elements are included within the set itself.

In our problem, the subsets \( A \) and \( B \) exist within the context of an ordered field \( F \). The concept of \( A+B \) broadens the view by demonstrating how combining subsets can form another subset within the same field. This combined subset helps in exploring and establishing properties like supremum and infimum for new, composite sets.

Understanding subsets helps in organizing data and developing insights into the relationships and operations possible within and across different sets.