Question

Prove, independently, the principle of nested intervals in \(E^{n},\) i.e., Theorem 5 with $$ F_{m}=\left[\bar{a}_{m}, \bar{b}_{m}\right] \subseteq E^{n}; $$ where $$ \bar{a}_{m}=\left(a_{m 1}, \ldots, a_{m n}\right) \text { and } \bar{b}_{m}=\left(b_{m 1}, \ldots, b_{m n}\right) $$ [Hint: As \(F_{m+1} \subseteq F_{m}, \bar{a}_{m+1}\) and \(\bar{b}_{m+1}\) are in \(F_{m} ;\) hence by properties of closed intervals, \(a_{m k} \leq a_{m+1, k} \leq b_{m+1, k} \leq b_{m k}, \quad k=1,2, \ldots, n\) Fixing \(k\), let \(A_{k}\) be the set of all \(a_{m k}, m=1,2, \ldots\) Show that \(A_{k}\) is bounded above by each \(b_{m k},\) so let \(p_{k}=\sup A_{k}\) in \(E^{1}\). Then $$ (\forall m) \quad a_{m k} \leq p_{k} \leq b_{m k} . \text { (Why?) } $$ Unfixing \(k\), obtain such inequalities for \(k=1,2, \ldots, n\). Let \(\bar{p}=\left(p_{1}, \ldots, p_{k}\right)\). Then $$ (\forall m) \quad \bar{p} \in\left[\bar{a}_{m}, \bar{b}_{m}\right], \text { i.e., } \bar{p} \in \bigcap F_{m}, \text { as required. } $$ Note that the theorem fails for nonclosed intervals, even in \(E^{1} ;\) e.g., take \(F_{m}=\) \((0,1 / m]\) and show that \(\bigcap_{m} F_{m}=\emptyset\).]

Short answer

The principle of nested intervals holds when intervals are closed, but it fails for nonclosed intervals.

Step-by-step solution

Understanding the Given Closed Intervals

We start with the closed intervals \( F_m = [\bar{a}_m, \bar{b}_m] \) in \( E^n \), where \( \bar{a}_m = (a_{m1}, \ldots, a_{mn}) \) and \( \bar{b}_m = (b_{m1}, \ldots, b_{mn}) \). The intervals are nested such that \( F_{m+1} \subseteq F_m \).

Analyzing the Properties of Nested Intervals

Since \( F_{m+1} \subseteq F_m \), any interval \( F_m \) satisfies \( a_{mk} \leq a_{m+1,k} \leq b_{m+1,k} \leq b_{mk} \) for each component \( k \). This means each sequence \( \{a_{mk}\}_{m=1}^{\infty} \) is non-decreasing and bounded above, and each sequence \( \{b_{mk}\}_{m=1}^{\infty} \) is non-increasing and bounded below.

Defining the Supremum for Each Component

Fixing a component \( k \), we define \( A_k = \{ a_{mk} : m = 1, 2, \ldots \} \). Since \( A_k \) is bounded above by each \( b_{mk} \), we let \( p_k = \sup A_k \). As a result, for any \( m \), \( a_{mk} \leq p_k \leq b_{mk} \), ensuring \( p_k \) lies within each interval's bounds for that component.

Constructing the Supremum Vector

By performing the process in Step 3 for each component \( k = 1, 2, \ldots, n \), we construct \( \bar{p} = (p_1, p_2, \ldots, p_n) \). This vector holds \( \bar{p} \in [\bar{a}_m, \bar{b}_m] \) for all \( m \), thereby showing \( \bar{p} \in \bigcap_{m} F_m \).

Highlighting Failure Case for Nonclosed Intervals

The principle fails for nonclosed intervals; for example, using \( F_m = (0, 1/m] \), the intersection \( \bigcap_{m} F_m \) is empty, illustrating the importance of closed intervals in the theorem.

Key concepts

Supremum in Analysis

In analysis, the concept of supremum (or least upper bound) plays a crucial role, especially in the context of nested intervals. For any non-empty set of real numbers that has a finite upper bound, the supremum is the smallest real number that is greater than or equal to each number in the set. In the given exercise, each sequence of interval boundaries, denoted as \( \{ a_{mk} \} \) for a fixed component \( k \), is non-decreasing and bounded above by corresponding elements \( \{ b_{mk} \} \). The importance of the supremum here is in proving the convergence of each sequence of lower bounds to a specific point, which ultimately resides in all intervals. To find this supremum, we use the property that every element \( a_{mk} \) must be less than or equal to any upper bound \( b_{mk} \). Thus, \( p_k = \sup A_k \) where \( A_k \) is the set of lower bounds. The supremum \( p_k \) for each component serves as a crucial part in forming the vector \( \bar{p} \), which confirms its existence within every nested interval.

Closed Intervals in Euclidean Space

Closed intervals have special properties that make them very important in Euclidean spaces. In general, a closed interval contains all its limit points, including both endpoints, which is not the case with open intervals. In the context of the Nested Intervals Theorem, this is crucial. A closed interval in Euclidean space, like \( [\bar{a}_m, \bar{b}_m] \), ensures that the endpoint conditions are fullfilled in the theorem. Each vector formed by the endpoints is part of this closed interval, and it's where the supremum of any sequence of lower bounds and the infimum of any sequence of upper bounds belong. The failure case presented in the exercise, with nonclosed intervals such as \((0, 1/m]\), highlights this importance. Here, as \( m \) increases, no intersection point exists within the interval confines, stressing the necessity of closed intervals to ensure non-empty intersections in nested interval principles.

Convergence of Sequences

In mathematical analysis, convergence of sequences is a central theme, particularly when proving the Nested Intervals Theorem. For a sequence to converge in this setting, the sequence must become arbitrarily close to some limit. During the analysis of nested intervals, we have sequences such as \( \{a_{mk}\}_{m=1}^{\infty} \) and \( \{b_{mk}\}_{m=1}^{\infty} \) for each component \( k \). Each sequence due to its boundedness properties either converges to a limit point or maintains a consistent boundary that the supremum \( p_k \) and infimum form. Overall, the convergence here is tightly linked to the nesting property: every next interval is contained within or equal to the previous. The convergence assures the existence of at least one point (\( \bar{p} \)) common to all intervals, which is critical in concluding the theorem. Convergence, along with boundedness, ensures that despite the arbitrary nesting, the sequences will always lead to a definitive overlap point shared by all intervals.