Problem 10 - A Primer of Real Analysis

Question

Suppose \(A_{i} \subset \mathbb{R}, i \in \mathbb{Z}^{+},\) and let $$ B=\bigcup_{i=1}^{\infty} A_{i} $$ Show that $$ \bigcup_{i=1}^{\infty} \overline{A_{i}} \subset \bar{B} $$ Find an example for which $$ \bar{B} \neq \bigcup_{i=1}^{\infty} \overline{A_{i}} $$

Step-by-Step Solution

Answer

\(\bigcup_{i=1}^{\infty} \overline{A_{i}} \subset \bar{B}\) holds. Example: \(A_i = (\frac{1}{i}, 1-\frac{1}{i})\).

Step 1: Understand the Definitions

To solve this problem, we need to understand what \(B\) and \(\overline{A_i}\) mean. \(B\) is the union of all the sets \(A_i\). The notation \(\overline{A_i}\) represents the closure of \(A_i\), which is the set of all limit points of \(A_i\) together with \(A_i\) itself.

Step 2: Aim of the First Part

Our goal is to show that the closure of the union \(B\), denoted \(\bar{B}\), contains the union of the closures of the individual sets \(A_i\). Formally, show that \( \bigcup_{i=1}^{\infty} \overline{A_i} \subset \overline{B} \).

Step 3: Prove \(\bigcup_{i=1}^{\infty} \overline{A_i} \subset \overline{B}\)

Let's take any point \(x\) in \(\bigcup_{i=1}^{\infty} \overline{A_i}\). This means \(x\) belongs to \(\overline{A_j}\) for some index \(j\). Since \(x\) is in the closure of \(A_j\), every neighborhood of \(x\) will contain a point from \(A_j\). Consequently, every neighborhood of \(x\) will also contain points from \(B\) (since \(A_j \subset B\)), implying \(x\) is in \(\overline{B}\). Therefore, \(x \in \overline{B}\), verifying our assertion.

Step 4: Construct an Example

For the second part, we need an example where \(\overline{B} eq \bigcup_{i=1}^{\infty} \overline{A_i}\). Consider \(A_i = \left(\frac{1}{i}, 1 - \frac{1}{i}\right)\) for each positive integer \(i\). Then, \(B = \bigcup_{i=1}^{\infty} A_i = (0, 1)\) and \(\overline{B} = [0, 1]\). Now, \(\overline{A_i} = \left[\frac{1}{i}, 1 - \frac{1}{i}\right]\), so \(\bigcup_{i=1}^{\infty} \overline{A_i} = (0, 1)\), which is not \(\overline{B}\).

Key Concepts

Real Analysis

Real Analysis is a fascinating branch of mathematics. It deals with real numbers and real-valued sequences and functions. In this subject, we explore concepts like limits, continuity, and convergence. Set theory is foundational in Real Analysis as it helps us understand the properties of sets within the real numbers. Here, we examine unions, closures, and limit points to understand how different sets interact with each other in the real number space. It's a discipline that combines rigorous logic with insightful techniques.

Set Union

The union of sets is a crucial concept in set theory. When we say the union of multiple sets, we refer to a set containing all elements from all sets. In mathematical terms, if we have sets \(A_1, A_2, \ldots\), their union, \(\bigcup_{i=1}^{\infty} A_i\), includes every element that belongs to at least one of the sets.
Here are some key points to remember about the union of sets:

• The union does not allow duplicate elements; each element appears only once in the union.
• The union is commutative; \(A \cup B = B \cup A\).
• The union is associative; \((A \cup B) \cup C = A \cup (B \cup C)\).
• The union is flexible, allowing us to combine finite or infinite collections of sets.

Closure of a Set

In mathematical analysis, the closure of a set is a fundamental concept. The closure \(\overline{A}\) of a set \(A\) includes all limit points of \(A\) as well as the original set itself. This set captures all the points where you can 'approach' \(A\).
Here are critical insights about set closures:

• The closure of a set is always closed; it contains all its limit points.
• For any set \(A\), \(A \subseteq \overline{A}\).
• The closure is the smallest closed set that contains \(A\).
• If \(A\) is already closed, \(\overline{A} = A\).
• The closure operation is idempotent, meaning \(\overline{\overline{A}} = \overline{A}\).

Limit Points

Limit points, also known as accumulation points, provide insights into the nature of a set's boundaries. A point \(x\) is a limit point of a set \(A\) if any neighborhood of \(x\) contains at least one point of \(A\) distinct from \(x\) itself.
Understanding limit points is crucial because:

• They help in identifying the closure of a set; \(\overline{A}\) includes all limit points of \(A\).
• They can exist outside the original set \(A\).
• A set does not need to contain any of its limit points to be valid. For example, open sets can have their limit points lying outside the set.
• Recognizing limit points assists in formulating the continuous nature of functions and sequences.