Question

Prove the principle of nested intervals: If \(\left[a_{n}, b_{n}\right]\) are closed intervals in a complete ordered field \(F\), with $$ \left[a_{n}, b_{n}\right] \supseteq\left[a_{n+1}, b_{n+1}\right], \quad n=1,2, \ldots $$ then $$ \bigcap_{n=1}^{\infty}\left[a_{n}, b_{n}\right] \neq \emptyset $$ [Hint: Let $$ A=\left\\{a_{1}, a_{2}, \ldots, a_{n}, \ldots\right\\} $$ Show that \(A\) is bounded above by each \(b_{n}\). Let \(p=\sup A\). (Does it exist?) Show that $$ (\forall n) \quad a_{n} \leq p \leq b_{n} $$ i.e., $$ \left.p \in\left[a_{n}, b_{n}\right] \cdot\right] $$

Short answer

The intersection of nested intervals is non-empty, as the supremum of left endpoints ensures a common element.

Step-by-step solution

Understand the Concept of Nested Intervals

Consider a sequence of closed intervals \([a_n, b_n]\) in a complete ordered field \(F\) such that each interval \([a_n, b_n]\) contains the next interval \([a_{n+1}, b_{n+1}]\). This implies that the endpoints satisfy \(a_n \leq a_{n+1}\) and \(b_{n+1} \leq b_n\). Our aim is to show that the intersection of all these intervals is not empty.

Define the Set A

Let \(A = \{a_1, a_2, a_3, \ldots\}\) be the set of left endpoints of these intervals. This set is clearly non-empty, as it is the collection of all left endpoints from the sequence of intervals \([a_n, b_n]\).

Show A is Bounded Above

Each element \(a_n\) of the set \(A\) is bounded above by every corresponding \(b_n\), because by definition of the interval, \(a_n \leq b_n\) for each \(n\). Hence, \(A\) is bounded above in the field \(F\).

Existence of Supremum

Since \(F\) is a complete ordered field and \(A\) is non-empty and bounded above, the least upper bound, or supremum, \(p = \sup A\), exists in \(F\).

Verify p within all Intervals

For each \(n\), we have \(a_n \leq p\) due to the definition of supremum, where \(p\) is the least upper bound of all \(a_n\). Also, \(p \leq b_n\) because each \(b_n\) is an upper bound of \(A\). Therefore, for each \(n\), \(p\) satisfies \(a_n \leq p \leq b_n\), which implies \(p \in [a_n, b_n]\).

Conclusion

Since \(p\) exists and satisfies the condition \(a_n \leq p \leq b_n\) for all intervals \([a_n, b_n]\), \(p\) is an element common to all intervals in the intersection. Thus, the intersection \(\bigcap_{n=1}^{\infty} [a_n, b_n]\) is not empty.

Key concepts

Complete Ordered Field

A complete ordered field is a mathematical structure with some special properties. It is a field, meaning it has operations like addition and multiplication, and it is ordered, meaning we can compare elements with each other using \(<\), \(>\), and \(=\). Moreover, it is complete, which means that every non-empty set that is bounded above has a least upper bound, also known as a supremum.

Examples of complete ordered fields include the set of real numbers \(\mathbb{R}\). The completeness property is crucial in analysis because it ensures that there are no "gaps" in the field. In the context of nested intervals, the complete ordered field condition ensures that certain limiting processes, such as finding supremums, can be done within the field.

The principle of nested intervals takes advantage of this completeness to guarantee that if intervals continually nest within each other, their intersection is not empty, provided we are in a complete ordered field like \(\mathbb{R}\). This principle would not hold if the field weren't complete, because supremums might not exist.

Supremum

The supremum, or least upper bound, is a fundamental concept in analysis. It refers to the smallest number that is greater than or equal to every number in a set. If a set \(A\) is bounded above, and you are in a complete ordered field, there will always be a supremum of \(A\). This is important because having a supremum allows us to consider the "limit" that a sequence of numbers approaches but may not necessarily reach.

Let's consider a set of endpoints \(A = \{a_1, a_2, a_3, \ldots\}\). If we say \(p = \sup A\), we mean that \(p\) is the least value satisfying \(a_n \leq p\) for all \(a_n\) in \(A\).

Moreover, if every \(b_n\) is an upper bound for \(A\), then \(p\) being the least of such bounds assures that \(p \leq b_n\) for each \(b_n\). Thus, \(p\) not only exists by virtue of completeness but also helps us verify essential properties like containing it in all intervals \([a_n, b_n]\). Without this concept of supremum, proving the non-empty nature of nested intervals would be challenging.

Bounded Above

In mathematics, when we say that a set is bounded above, it means there exists some number that is greater than or equal to every element in the set. This concept is crucial when dealing with sequences and series, especially in the context of complete ordered fields.

Consider the set of left endpoints, denoted by \(A = \{a_1, a_2, a_3, \ldots\}\). This set is bounded above by any \(b_n\) because for every \(n\), \(a_n \leq b_n\). The fact that \(A\) is bounded above guarantees that we can find its least upper bound or supremum in a complete ordered field.

Bounding a set above is essentially about setting a "ceiling" beyond which no element of the set rises. It creates a controlled environment where the principle of nested intervals can play out. This upper bound provides the means to calculate or identify key elements such as the supremum, thereby helping concretize the theorem of nested intervals.

Intersection of Intervals

The intersection of intervals refers to the set of elements that are common to all intervals involved. In the principle of nested intervals, we consider a sequence of intervals \([a_n, b_n]\), each one contained in the previous, such that their overall intersection is not empty.

Here’s how it works: You have intervals like \([a_1, b_1] \supseteq [a_2, b_2] \supseteq [a_3, b_3]\) and so on. Each interval being contained in the one before means that they "nest" within each other.

The goal is to demonstrate that when you take the intersection of all these intervals—as described mathematically with \(\bigcap_{n=1}^{\infty} [a_n, b_n]\)—there is still at least one number (or point) that they all share. The nested nature and the completeness of the field assure us of this fact. In the context of the exercise, the element present in every single interval is shown to be the supremum of the set \(A = \{a_1, a_2, \ldots\}\), ensuring the intersection is not empty.

• Each interval contains the following interval fully.
• The concept relies on the completeness of the field, like \(\mathbb{R}\).
• The intersection explains the presence of a common element in all intervals.

This principle is vital in real analysis, proving fundamental properties about real numbers and completeness.