Question

Let \(S\) and \(T\) be nonempty sets of real numbers and define $$ S+T=\\{s+t \mid s \in S, t \in T\\} $$ (a) Show that $$ \sup (S+T)=\sup S+\sup T $$ if \(S\) and \(T\) are bounded above and $$ \inf (S+T)=\inf S+\inf T $$ if \(S\) and \(T\) are bounded below. (b) Show that if they are properly interpreted in the extended reals, then (A) and (B) hold if \(S\) and \(T\) are arbitrary nonempty sets of real numbers.

Short answer

Question: Prove that for nonempty sets \(S\) and \(T\), the supremum of the set \(S+T\) equals the supremum of \(S\) plus the supremum of \(T\), while the infimum of the set \(S+T\) equals the infimum of \(S\) plus the infimum of \(T\). Answer: To prove that the supremum of \(S+T\) equals the supremum of \(S\) plus the supremum of \(T\), and the infimum of \(S+T\) equals the infimum of \(S\) plus the infimum of \(T\), we first need to show that \(M + N\) is an upper bound for \(S + T\) and that \(A + B\) is a lower bound for \(S + T\) where \(M = \sup S\), \(N = \sup T\), \(A = \inf S\), and \(B = \inf T\). Next, we need to prove that \(M + N\) and \(A + B\) are the least upper bound and greatest lower bound, respectively. This is done by assuming there exists another upper bound (\(L\)) or lower bound (\(G\)) for \(S+T\), and showing that this leads to a contradiction. In the case of the general nonempty set, we can use the extended reals to encompass all possible values, including \(+\infty\) and \(-\infty\), and the relations still hold.

Step-by-step solution

(Step 1: Bounded Above Sets and Supremum Proof)

Given that the sets \(S\) and \(T\) are nonempty and bounded above, let's denote the supremum of \(S\) as \(M\) and the supremum of \(T\) as \(N\). We want to show that \(\sup(S+T) = M +N\). Recall that if a set is bounded above, there exists a number \(\alpha\) that is greater than or equal to every element in the set. Here, \(M\) and \(N\) are upper bounds for \(S\) and \(T\), respectively. First, let's show that \(M + N\) is an upper bound for \(S + T\). For every element \(s \in S\) and \(t \in T\), we have \(s \leq M\) and \(t \leq N\). Therefore, when we add these inequalities, \(s+t \leq M+N\) for every element in \(S+T\). This confirms that \(M+N\) is an upper bound for \(S+T\).

(Step 2: Lowest Upper Bound Property Proof)

We need to show that \(M+N\) is the least upper bound for the set \(S+T\). Suppose there exists another upper bound \(L\) for \(S+T\) such that \(L 0\). Similarly, we can choose an element \(t \in T\) such that \(N-\epsilon < t \leq N\). Combining these inequalities, we get: $$ (M-\epsilon) + (N-\epsilon) < s+t \leq M+N $$ Which simplifies to: $$ M + N - 2\epsilon < s+t \leq M+N $$ This contradicts the assumption that \(L<M+N\) since \(s+t\) is an element of the set \(S+T\). Therefore, there cannot be an upper bound for \(S+T\) that is less than \(M+N\). Hence \(M+N\) is the least upper bound i.e., $$ \sup(S+T) = \sup S + \sup T $$

(Step 3: Bounded Below Sets and Infimum Proof)

Given that the sets \(S\) and \(T\) are nonempty and bounded below, let's denote the infimum of \(S\) as \(A\) and the infimum of \(T\) as \(B\). We want to show that \(\inf(S+T) = A +B\). Similar to step 1, if a set is bounded below, there exists a number \(\alpha\) that is less than or equal to every element in the set. Here, \(A\) and \(B\) are lower bounds for \(S\) and \(T\), respectively. First, let's show that \(A + B\) is a lower bound for \(S + T\). For every element \(s \in S\) and \(t \in T\), we have \(s \geq A\) and \(t \geq B\). Therefore, when we add these inequalities, \(s+t \geq A+B\) for every element in \(S+T\). This confirms that \(A+B\) is a lower bound for \(S+T\).

(Step 4: Greatest Lower Bound Property Proof)

We need to show that \(A+B\) is the greatest lower bound for the set \(S+T\). Suppose there exists another lower bound \(G\) for \(S+T\) such that \(G>A+B\). Since \(A\) is the greatest lower bound for the set \(S\), we can choose an element \(s \in S\) such that \(A \leq s 0\). Similarly, we can choose an element \(t \in T\) such that \(B \leq t A+B\) since \(s+t\) is an element of the set \(S+T\). Therefore, there cannot be a lower bound for \(S+T\) that is greater than \(A+B\). Hence \(A+B\) is the greatest lower bound i.e., $$ \inf(S+T) = \inf S + \inf T $$

(Step 5: Extended Reals for Arbitrary Nonempty Sets)

To generalize part (a) to arbitrary nonempty sets of real numbers, we can make use of the extended reals. The extended reals consist of real numbers plus the two elements \(+\infty\) and \(-\infty\). We can simply interpret the sum of two sets as the sum of their least upper bounds and greatest lower bounds as: $$ \sup(S+T) = \sup S + \sup T $$ $$ \inf(S+T) = \inf S + \inf T $$ By adopting the standard rules for arithmetic in extending the reals: - If either \(\sup S\) or \(\sup T\) is \(+\infty\), then \(\sup(S+T)\) will be \(+\infty\). - If either \(\inf S\) or \(\inf T\) is \(-\infty\), then \(\inf(S+T)\) will be \(-\infty\). So the relations hold for arbitrary nonempty sets \(S\) and \(T\) of real numbers when properly interpreted in the extended reals.

Key concepts

Supremum and Infimum

In real analysis, the concepts of supremum and infimum are vital when dealing with bounded sets. The supremum, or least upper bound, of a set is the smallest real number that is greater than or equal to every element in the set. Conversely, the infimum, or greatest lower bound, is the largest real number less than or equal to every element in the set.

Consider two nonempty sets of real numbers, \(S\) and \(T\), that are both bounded above. Each has a supremum: \(M = \sup S\) and \(N = \sup T\). To find the supremum of the set \(S + T = \{s + t \mid s \in S, t \in T\}\), one shows that \(M + N\) is its upper bound.

• For every \(s \in S\) and \(t \in T\), \(s \leq M\) and \(t \leq N\). Therefore, \(s + t \leq M + N\), proving \(M + N\) is an upper bound for \(S + T\).
• Moreover, \(M + N\) is the least such bound, meaning no number less than \(M + N\) can serve as an upper bound for \(S + T\).

Similarly, if \(S\) and \(T\) are bounded below, we can explore their infimums. Denote \(A = \inf S\) and \(B = \inf T\). It follows that \(s \geq A\) and \(t \geq B\) for each \(s \in S\) and \(t \in T\). Thus, \(s + t \geq A + B\).

This confirms \(A + B\) as a lower bound, and further shows it is the greatest one, establishing that \(\inf(S + T) = \inf S + \inf T\). Understanding these dual concepts aids in managing more complex analysis involving bounded sets.

Bounded Sets

In mathematics and real analysis, a set is bounded if all its elements lie within some finite limits, either above or below. If every element of a set \(S\) of real numbers is less than or equal to some number, then \(S\) is said to be bounded above. Similarly, \(S\) is bounded below if every element is greater than or equal to some number.

The behavior of sums of elements from two bounded sets is intriguing. Let's take two nonempty sets, \(S\) and \(T\), that are both bounded above. When you form the set \(S + T = \{s + t \mid s \in S, t \in T\}\), this new set is also bounded. Its least upper bound or supremum can be calculated to be the sum of the supremums of \(S\) and \(T\).

• This principle applies to bounded below sets as well: the infimum of the sum is the sum of the infimums.
• Understanding how bounds relate across combined sets is crucial for tasks in both theoretical analysis and practical calculations.

The concept of boundedness helps in structuring and understanding sets, especially when considering bounded intervals in real numbers. This allows mathematicians and analysts to deduce properties and perform operations on these sets with precision, maintaining the integrity of numerical estimates.

Extended Real Numbers

The extended real numbers are an expansion of the real number system that includes \(+\infty\) and \(-\infty\). These symbols allow for a convenient way to describe limits and behaviors at extremes in various mathematical contexts. In extended real number arithmetic, certain rules apply, particularly addressing how infinities interact with finite numbers.

In real analysis, extended reals are pivotal when sets of numbers are unbounded. For instance, if a set \(S\) is unbounded above,

• its supremum might be understood as \(+\infty\).
• Similarly, for a set \(T\) unbounded below, its infimum could be \(-\infty\).

When examining arbitrarily large or small sets, the concept extends naturally. Even if \(S\) or \(T\) in our problem setup includes infinite bounds, the formulas for supremum and infimum mentioned previously still hold, with certain considerations for infinity arithmetic:

• The sum of a finite number and \(+\infty\) is \(+\infty\).
• The only time the sum of extremes affects the result is \(+\infty + (-\infty)\), which is undefined.

By thoughtfully placing these infinities into real set analysis, students gain the advantage of managing broader mathematical situations. It prepares them for advanced studies where such extended concepts are not just helpful but necessary to comprehend the complete picture.