Problem 6 - A Primer of Real Analysis

Question

Suppose, for \(n=1,2,3, \ldots,\) the intervals \(I_{n}=\left[a_{n}, b_{n}\right]\) are such that \(I_{n+1} \subset I_{n} .\) If \(a=\sup \left\\{a_{n}: n \in \mathbb{Z}^{+}\right\\}\) and \(b=\inf \left\\{b_{n}: n \in \mathbb{Z}^{+}\right\\},\) show that $$ \bigcap_{n=1}^{\infty} I_{n}=[a, b] $$

Step-by-Step Solution

Answer

The intersection \(\bigcap_{n=1}^{\infty} I_n = [a, b]\).

Step 1: Understanding Nested Intervals

The problem involves nested intervals: each interval \(I_{n+1}\) is contained within \(I_{n}\). This implies \(a_n \leq a_{n+1}\) and \(b_{n+1} \leq b_n\). These are non-decreasing and non-increasing sequences respectively.

Step 2: Defining Supremum and Infimum

By definition, \(a = \sup \{a_n : n \in \mathbb{Z}^{+}\}\) indicates that \(a_n \leq a\) for all \(n\) and for any \(\epsilon > 0\), there exists some \(N\) such that \(a - \epsilon < a_N\). Similarly, \(b = \inf \{b_n : n \in \mathbb{Z}^{+}\}\) means \(b_n \geq b\) for all \(n\) and for any \(\epsilon > 0\), there exists some \(M\) such that \(b + \epsilon > b_M\).

Step 3: Showing the Intersection Contains \([a, b]\)

For any point \(x \in \bigcap_{n=1}^{\infty} I_{n}\), \(x\) must be within every interval \([a_n, b_n]\). Hence, \(a_n \leq x \leq b_n\) for all \(n\), implying \(a \leq x \leq b\). Thus, \(x \in [a, b]\). This shows \([a, b]\) is contained in the intersection.

Step 4: Showing \([a, b]\) Contains the Intersection

Conversely, consider any \(x\) in the interval \([a, b]\). From the properties of \(a\) and \(b\), there exists some \(N\) such that \(a_n \leq x \leq b_n\) for all \(n \geq N\). Hence, \(x\) will be in \(I_n = [a_n, b_n]\) for sufficiently large \(n\). Therefore, every point in \([a, b]\) is also in the intersection.

Step 5: Concluding the Solution

Steps 3 and 4 establish that \(\bigcap_{n=1}^{\infty} I_n = [a, b]\). Hence, \([a, b]\) includes all points in the intersection and vice versa.

Key Concepts

Supremum

The concept of supremum, often called the least upper bound, is important when dealing with sequences like the one in the problem. For the sequence \(a_n\), where \(n\) is a positive integer, the supremum, denoted by \(a = \sup \{a_n : n \in \mathbb{Z}^{+}\}\), is the smallest number that is greater than or equal to every number in that sequence. In simpler terms, no number bigger than \(a\) can serve as an upper bound for all \(a_n\).

• \(a\) captures the highest point reached by the values \(a_n\) without exceeding the true upper limits set by any term in the sequence.
• The property means for any small number \(\epsilon > 0\), we can find some \(a_N\) close to \(a\) such that \(a_N > a - \epsilon\).

The supremum is essential to understanding how the sequence progresses and finds its natural limit without overshooting, which is crucial in defining the interval where the sequence converges in the context of nested intervals.

Infimum

The infimum, or greatest lower bound, is the reverse concept of the supremum. For the sequence \(b_n\), the infimum, denoted by \(b = \inf \{b_n : n \in \mathbb{Z}^{+}\}\), is the greatest number that is less than or equal to every term in the sequence. You could think of it as the floor of the sequence.

• \(b\) acts as the lowest point approached by each \(b_n\) in the series.
• This property ensures there exists some \(b_M\) such that \(b_M < b + \epsilon\) for any \(\epsilon > 0\).

This concept is crucial in ensuring that the intervals 'shrink' down to this lower bound as we examine more intervals. Just as the supremum helps tower over its sequence, the infimum ensures we can articulate where these sequences inherently find their lower limit as addressed in the nested interval theorem.

Intersection of Intervals

The term intersection of intervals refers to the common 'overlapped' region when multiple intervals are layered over each other. In the context of our problem, we have a nested sequence of intervals \(I_n = [a_n, b_n]\) where each subsequent interval is contained within the previous one, represented by \(I_{n+1} \subset I_n\). As we take the intersection over all these \(n\) intervals, denoted by \(\bigcap_{n=1}^{\infty} I_n\), we essentially find the region where all intervals overlap.

• In this setup, the intersection will shrink to precisely match \([a, b]\).
• It maintains that any point within the intersection belongs to each interval, assuredly within the bounds of \(a \) and \(b\).

This property of nested intervals shows that despite the shrinking behavior, the intersection cannot be 'emptied out' and instead solidifies what remains as the sequence approaches infinity. So, \[\bigcap_{n=1}^{\infty} I_n = [a, b]\].