Question

Let \(E\) be a nonempty subset of an ordered set; suppose \(\alpha\) is a lower bound of \(E\) and \(\beta\) is an upper bound of \(E\). Prove that \(\alpha \leq \beta\).

Short answer

Question: Given a nonempty subset E of an ordered set, with α as a lower bound and β as an upper bound, prove that α ≤ β. Answer: By the given conditions, the relationship α ≤ x ≤ β holds for all x in the nonempty subset E, which implies that α is always less than or equal to x and x is always less than or equal to β. Since this relationship is true for all elements in the set E, it follows that α is less than or equal to β, or α ≤ β.

Step-by-step solution

Understand the definitions of lower and upper bounds

A lower bound for a set \(E\), denoted by \(\alpha\), is an element such that \(\alpha \leq x\) for all \(x \in E\). Similarly, an upper bound for a set \(E\), denoted by \(\beta\), is an element such that \(x \leq \beta\) for all \(x \in E\).

Use the given conditions and definitions to derive a relationship

Let \(x\) be any element in \(E\). We are given that \(\alpha\) is a lower bound for \(E\), meaning \(\alpha \leq x\) for all \(x \in E\). We are also given that \(\beta\) is an upper bound for \(E\), which means \(x \leq \beta\) for all \(x \in E\).

Combine the inequalities for the lower and upper bounds

Since \(\alpha \leq x\) and \(x \leq \beta\) for all \(x \in E\), we can combine the inequalities to derive the relationship \(\alpha \leq x \leq \beta\) for all \(x \in E\).

Prove that \(\alpha \leq \beta\)

By the relationship \(\alpha \leq x \leq \beta\) for all \(x \in E\), and considering that \(E\) is a nonempty set, we can conclude that \(\alpha \leq \beta\). This is because, for any element \(x\) in the nonempty set \(E\), \(\alpha\) is always less than or equal to \(x\) and \(x\) is always less than or equal to \(\beta\).

Key concepts

Lower Bound

In the context of ordered sets, a lower bound of a subset is a crucial concept. Imagine you have a set of numbers, and you want to determine the smallest point that is not exceeded by any element in your set. This point is known as the lower bound. In mathematical terms, if you have a subset \(E\) of an ordered set and an element \(\alpha\), \(\alpha\) serves as a lower bound of \(E\) if for every element \(x\) in \(E\), \(\alpha \leq x\).
To put it simply, no element in your set is smaller than \(\alpha\). Lower bounds are valuable because they help establish limits and constraints within mathematical problems, assuring that all elements behave within predictable boundaries.

Upper Bound

An upper bound is, in many ways, the opposite of a lower bound. For any given subset in an ordered set, an upper bound sets the ceiling that no element in the subset surpasses. In more formal terms, if \(E\) is a subset and \(\beta\) is an element, then \(\beta\) is an upper bound of \(E\) if \(x \leq \beta\) for every element \(x\) in \(E\).
This means that \(\beta\) acts as a barrier that elements of the set cannot cross upwards. The concept of upper bounds is helpful in understanding containment and restrictions on data sets and is often used to find the greatest limit for potential values in a set.

Inequalities

Inequalities are essential tools in mathematics for expressing the relative size or order of two values. They are used to show when one value is less than, greater than, or equal to another. In our problem, inequalities help compare the lower bound \(\alpha\) and the upper bound \(\beta\).
For any element \(x\) in the set \(E\), we have two inequalities: \(\alpha \leq x\) and \(x \leq \beta\). Together, these inequalities suggest the sequence \(\alpha \leq x \leq \beta\). Since \(E\) is nonempty and \(x\) is an arbitrary element within it, these inequalities collectively show that \(\alpha\) cannot be greater than \(\beta\), thus proving \(\alpha \leq \beta\). Using inequalities in this manner allows mathematicians to deduce relationships between different boundaries and enhance our understanding of the constraints within sets.