Question

In an ordered field \(F,\) let \(\emptyset \neq A \subset F\). Let \(c \in F\) and let \(c A\) denote the set of all products \(c x(x \in A) ;\) i.e., $$ c A=\\{c x \mid x \in A\\} $$Prove that (i) if \(c \geq 0\), then $$ \sup (c A)=c \cdot \sup A \text { and } \inf (c A)=c \cdot \inf A $$ (ii) if \(c<0\), then $$ \sup (c A)=c \cdot \inf A \text { and } \inf (c A)=c \cdot \sup A $$ In both cases, assume that the right-side sup \(A\) (respectively, inf \(A\) ) exists.

Short answer

For \(c \geq 0\), \(\sup(cA) = c \cdot \sup A\) and \(\inf(cA) = c \cdot \inf A\). For \(c < 0\), \(\sup(cA) = c \cdot \inf A\) and \(\inf(cA) = c \cdot \sup A\).

Step-by-step solution

Understand the Definitions

Before we begin the proof, recall the definitions of supremum (sup) and infimum (inf): \(\sup A\) is the least upper bound of the set \(A\), while \(\inf A\) is the greatest lower bound. If \(cA = \{ cx \mid x \in A \}\), \(cA\) is the set formed by multiplying each element of \(A\) by \(c\).

Case 1 - Prove for \( c \geq 0 \)

Assume \(c \geq 0\). For any \(x \in A\), \(cx \geq 0\). Since \(\sup A = M\) means \(x \leq M\) for all \(x \in A\), hence \(cx \leq cM\). Therefore, \(cM\) is an upper bound for \(cA\). Since \(M\) is the least upper bound, \(cM\) must also be the least upper bound in \(cA\) unless a smaller bound exists, which does not, as \(M\) is minimal for \(A\). Thus, \(\sup(cA) = c \cdot \sup A\).

Infimum for \( c \geq 0 \)

Similarly, if \(\inf A = m\) where \(m\) is the greatest lower bound for \(A\), every \(x \in A\) has \(m \leq x\). Then \(cm \leq cx\), so \(cm\) is a lower bound for \(cA\). Since no larger lower bound exists than \(m\), \(cm\) is the greatest lower bound for \(cA\), giving \(\inf(cA) = c \cdot \inf A\).

Case 2 - Prove for \( c < 0 \)

Assume \(c < 0\). The inequality relationships reverse when multiplying by negative numbers. So, if \(\sup A = M\), \(cx \geq cM\) for \(x \leq M\) implies that \(cM\) becomes a lower bound (not upper) for \(cA\). Similar reasoning shows \(c \cdot \inf A\) (infeasible in original ordering) is an upper bound for \(cA\) because all products \(cx\) with minimal \(x\) make \(cx \geq c \cdot \inf A\). Therefore, \(\inf (cA) = c \cdot \sup A\) and \(\sup (cA) = c \cdot \inf A\).

Verify and Conclude

For \( c \geq 0 \), we demonstrate that multiplying the bounds by \(c\) preserves the ordering, so \( \sup(cA) = c \sup A \) and \( \inf(cA) = c \inf A \). For \( c < 0 \), the multiplication reverses ordering, thus \( \sup(cA) = c \inf A \) and \( \inf(cA) = c \sup A \). The problem conditions regarding existence of \( \sup \) and \( \inf \) validate outcomes in these cases.

Key concepts

Supremum and Infimum

When dealing with a set of numbers, two important concepts often arise: supremum and infimum. These help us understand the upper and lower bounds of a set.

The supremum of a set, denoted as \(\sup A\), is the smallest number that is greater than or equal to every number in the set \(A\). It's like the ceiling in a room that limits how tall an object can be without touching it.
If your set is a list of numbers like \( \{1, 2, 3.5, 4.9\} \), then \(\sup A = 4.9\), the largest number that’s still in \(A\) or next larger number outside \(A\) if the set is incomplete.

The infimum, on the other hand, is often thought of as the opposite. It's the greatest number that is less than or equal to every number in the set \(A\). Think of it as the floor that you can't go below.
In our same set, \(\inf A = 1\) is the smallest number in \(A\), or if necessary a slightly smaller number not in \(A\) that serves as the greatest lower bound.

Understanding supremum and infimum helps us grasp how bounds change, especially in ordered fields where these concepts preserve an order involving rigid rules of comparison between numbers.

Scalar Multiplication

Scalar multiplication involves multiplying every element of a set by a single number, often called a "scalar". This action is fundamental in areas such as linear algebra and is used extensively when dealing with ordered fields.

Let's break this down with an example: suppose we have a set \(A = \{1, 2, 3\}\) and a scalar \(c = 2\). Then, the set after scalar multiplication, denoted \(cA\), becomes \(\{2 \times 1, 2 \times 2, 2 \times 3\}\) or \(\{2, 4, 6\}\).

Key effects of scalar multiplication include:

• If the scalar \(c\) is positive, every element in \(A\) is stretched away from zero.
• If \(c\) is zero, the resulting set \(cA\) contains only zero, regardless of \(A\).
• If \(c\) is negative, it not only stretches the set, but reverses the order - flipping all positive numbers to negative and vice versa.

This operation is crucial in understanding how sets transform when each element is affected by the same factor, leading to shifts in both supremum and infimum.

Inequality Reversal

Inequality reversal is a crucial concept when working with ordered fields, particularly in the context of scalar multiplication. It explains how inequalities change when multiplied by different types of numbers.

Here's the essential rule: when you multiply an inequality by a positive number, the direction of inequality remains the same. For example, if you have \(a < b\) and you multiply both sides by a positive \(c\), you still get \(ca < cb\).

However, if you multiply by a negative number, the inequality sign flips. So, \(a < b\) becomes \(ca > cb\) if \(c < 0\).
This flip is because negative times positive turns to negative, requiring a reversal in comparisons.

• Example: For \(c = -2\), transforming \(5 < 8\) gives \(-10 > -16\), illustrating reversal.
• This behavior is crucial for ensuring logical consistency when determining bounds.

Understanding inequality reversal is key in ensuring proper arithmetic operations, particularly when dealing with proofs involving ordered fields. It preserves the truth of inequalities within transformed sets.