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Prove that if \(A\) consists of positive elements only, then \(q=\sup A\) iff (i) \((\forall x \in A) x \leq q\) and (ii) \((\forall d>1)(\exists x \in A) q / d

Short Answer

Expert verified
Conditions (i) and (ii) ensure \( q = \sup A \).

Step by step solution

01

Understand the Problem

We need to prove if a number \( q \) is the supremum of a set \( A \) consisting entirely of positive elements, then it satisfies two conditions: every element \( x \) in \( A \) is less than or equal to \( q \), and for every \( d > 1 \), there is an \( x \) in \( A \) such that \( q/d < x \).
02

Definitions and Properties

The supremum \( q \) of a set \( A \) is the least upper bound. This means \( q \) is the smallest number such that all elements of \( A \) are less than or equal to \( q \), and there is no smaller number satisfying this. Also, since \( q \) is the supremum, the values from \( A \) can get arbitrarily close to \( q \).
03

Prove \( q \) is an Upper Bound

To show \( q = \sup A \), we must confirm \( q \) is an upper bound. This requires verifying condition (i): \((\forall x \in A) x \leq q\). Since \( q \) is the supremum, by definition, every \( x \in A \) meets \( x \leq q\). Hence, the first condition is satisfied.
04

Prove \( q \) is the Least Upper Bound

To demonstrate \( q \) is the least such upper bound, verify condition (ii): \((\forall d > 1)(\exists x \in A) q/d < x\). Assume that there is no \( x \in A \) such that \( q/d < x \). Then \( q/d \) would become a smaller upper bound than \( q \), which contradicts \( q \) being the least upper bound. Thus, this condition ensures the elements in \( A \) can come arbitrarily close to \( q \).
05

Conclusion

The conditions (i) and (ii) together ensure that \( q \) is both the upper bound for all \( x \in A \) and the least such bound allowing elements in \( A \) to come arbitrarily close to it. Hence, they are both necessary and sufficient for \( q = \sup A \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Set Theory
In mathematics, set theory is a fundamental area that focuses on the study of sets, which are collections of objects. Sets might contain numbers, symbols, or even other sets. In the context of proving properties of numbers and functions, such as finding a supremum, set theory provides the language and tools needed to articulate these concepts precisely.

When dealing with a set like \( A \), which contains positive elements, understanding the nature of sets helps us precisely define what elements belong to \( A \) and how these elements interact with each other and with other numbers, such as an upper bound or supremum. In set theory, operations and characteristics of sets, such as unions, intersections, and subsets, provide the foundational blocks to build more complex mathematical ideas in other fields, such as real analysis.
Upper Bound
An upper bound is an essential concept in mathematics, particularly in the study of sequences and functions. An upper bound of a set \( A \) of numbers is a value \( b \) such that for every element \( x \) in the set, \( x \leq b \). In other words, no element in the set exceeds this upper bound.

Upper bounds are crucial for dealing with supremum problems because the supremum of a set is a specific type of upper bound called the least upper bound. When we say \( q \) is an upper bound of \( A \), it means \( q \) is a limit that the elements in \( A \) do not surpass. However, there may be multiple upper bounds for any set, but among them, the supremum is the smallest. Thus, understanding upper bounds can help you comprehend the constraints and limitations within which the elements of a set exist.
Least Upper Bound
The least upper bound, also known as the supremum, is the smallest number that serves as an upper bound for a set. Unlike a general upper bound, the supremum is the smallest possible value that can act as an upper bound, without being exceeded by any element of the set.

In proving that \( q \) is the supremum of a set \( A \), we must show that \( q \) is an upper bound that satisfies special conditions. The first condition is straightforward: each element in \( A \) should be less than or equal to \( q \). The second, more sophisticated condition states that for any number greater than 1, there exists an element in \( A \) so close to \( q \) that dividing \( q \) by this number gives a result smaller than that element.

These conditions ensure that no smaller number can be the supremum of the set, as it couldn't accommodate such close elements from \( A \) without being surpassed. Mastering the concept of least upper bound is key in solving problems related to limits, continuity, and optimizations in calculus and analysis.
Real Analysis
Real analysis is a branch of mathematics that deals with real numbers and real-valued functions. Central to this field are limits, continuity, and differentiability, which depend heavily on the properties of sets and bounds within the real number system.

Understanding the supremum involves concepts from real analysis, including the completeness property of real numbers, which ensures that every non-empty set of real numbers bounded above has a least upper bound. It highlights the rigor and precision required in real analysis.

Applying real analysis to set-related exercises fosters a deeper understanding of how mathematical concepts interact. It enables us to not only find solutions but verify them rigorously by analyzing function behavior, their limits, and boundaries, as seen while proving that a number is the supremum of a set. This integration of real analysis is crucial in understanding and proving essential properties within the realm of calculus and beyond.

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Most popular questions from this chapter

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.

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