Question

Show that $$ \operatorname{int}(\operatorname{int} A)=\operatorname{int} A \quad \text { and } \quad \operatorname{int}(A \cap B)=\operatorname{int} A \cap \operatorname{int} B $$ Give an example to show that \(\operatorname{int}(A \cup B)=\operatorname{int} A \cup \operatorname{int} B\) is not necessarily true.

Short answer

Question: Prove the following properties of the interior of sets and provide an example for the third statement: 1) The interior of the interior of a set A is equal to the interior of that set. 2) The interior of the intersection of two sets A and B is equal to the intersection of their interiors. 3) The union of the interiors of two sets is not necessarily equal to the interior of their union. Solution: 1) To prove that the interior of the interior of a set A is equal to the interior of that set, we show that \(\operatorname{int}(\operatorname{int} A) = \operatorname{int} A\). This has been demonstrated by proving that \(\operatorname{int}(\operatorname{int} A) \subseteq \operatorname{int} A\) and \(\operatorname{int} A \subseteq \operatorname{int}(\operatorname{int} A)\). 2) To prove that the interior of the intersection of two sets A and B is equal to the intersection of their interiors, we show that \(\operatorname{int}(A \cap B)=\operatorname{int} A \cap \operatorname{int} B\). This has been demonstrated by proving that \(\operatorname{int}(A \cap B) \subseteq \operatorname{int} A \cap \operatorname{int} B\) and \(\operatorname{int} A \cap \operatorname{int} B \subseteq \operatorname{int}(A \cap B)\). 3) An example where the union of the interiors of two sets is not necessarily equal to the interior of their union is given by sets \(A = [0,1)\) and \(B = (1,2]\), where \(\operatorname{int}(A \cup B) = (0,2)\) but \(\operatorname{int} A \cup \operatorname{int} B = (0,1) \cup (1,2)\). This shows that \(\operatorname{int}(A \cup B) \neq \operatorname{int} A \cup \operatorname{int} B\).

Step-by-step solution

Proof of \(\operatorname{int}(\operatorname{int} A)=\operatorname{int} A\)

First, we need to show that \(\operatorname{int}(\operatorname{int} A) \subseteq \operatorname{int} A\). Because \(\operatorname{int} A\) is the largest open subset of A, \(\operatorname{int}(\operatorname{int} A)\) must also be a subset of \(\operatorname{int} A\). Now, let's prove that \(\operatorname{int} A \subseteq \operatorname{int}(\operatorname{int} A)\). Given any point \(x \in \operatorname{int} A\), there exists an open set \(U\) such that \(x \in U \subseteq \operatorname{int} A\). Since \(U\) is open and \(U \subseteq \operatorname{int} A\), then \(U \subseteq \operatorname{int}(\operatorname{int} A)\). Therefore, we conclude that \(\operatorname{int}(\operatorname{int} A) = \operatorname{int} A\).

Proof of \(\operatorname{int}(A \cap B)=\operatorname{int} A \cap \operatorname{int} B\)

First, we need to show that \(\operatorname{int}(A \cap B) \subseteq \operatorname{int} A \cap \operatorname{int} B\). If \(x \in \operatorname{int}(A \cap B)\), then there exists an open set \(U\) such that \(x \in U \subseteq A \cap B\). This means that \(x \in U \subseteq A\) and \(x \in U \subseteq B\), so \(x \in \operatorname{int} A\) and \(x \in \operatorname{int} B\). Therefore, \(x \in \operatorname{int} A \cap \operatorname{int} B\) and \(\operatorname{int}(A \cap B) \subseteq \operatorname{int} A \cap \operatorname{int} B\). Now let's show that \(\operatorname{int} A \cap \operatorname{int} B \subseteq \operatorname{int}(A \cap B)\). If \(x \in \operatorname{int} A \cap \operatorname{int} B\), then there exist open sets \(U_1\) and \(U_2\) such that \(x \in U_1 \subseteq A\) and \(x \in U_2 \subseteq B\). Let \(U = U_1 \cap U_2\). Then \(x \in U\), and \(U\) is an open set, since the intersection of open sets is open. Furthermore, \(x \in U \subseteq A \cap B\). Hence, \(x \in \operatorname{int}(A \cap B)\), and we have \(\operatorname{int} A \cap \operatorname{int} B \subseteq \operatorname{int}(A \cap B)\). Thus, \(\operatorname{int}(A \cap B)=\operatorname{int} A \cap \operatorname{int} B\).

Example where \(\operatorname{int}(A \cup B) \neq \operatorname{int} A \cup \operatorname{int} B\)

Let \(A = [0,1)\) and \(B = (1,2]\). Then we have \(\operatorname{int} A = (0,1)\) and \(\operatorname{int} B = (1,2)\). Now consider the union \(A \cup B = [0,2]\), and note that \(\operatorname{int}(A \cup B) = (0,2)\). However, \(\operatorname{int} A \cup \operatorname{int} B = (0,1) \cup (1,2) = (0,1) \cup (1,2)\) which is not equal to \((0,2)\). Thus, we have found an example where \(\operatorname{int}(A \cup B) \neq \operatorname{int} A \cup \operatorname{int} B\).

Key concepts

Abstract Analysis

The field of abstract analysis is a sub-discipline of mathematics that deals with the study of mathematical structures using a general and abstract approach. This field encompasses a variety of topics such as real and complex analysis, functional analysis, metric spaces, and topology. One of the fundamental concepts in abstract analysis is the notion of an interior point. The interior of a set consists of all points that have a neighborhood completely contained within the set. One interesting property is that the interior of the interior of a set is the same as the interior of the set; this is portrayed in the exercise solution where it is shown that \(\operatorname{int}(\operatorname{int} A) = \operatorname{int} A\). Here, the set \(A\) does not change when taking the interior twice, leading to the conclusion that interior operations are idempotent in this context.

Understanding idempotence is critical: it is a property of certain operations that, when applied multiple times to any value, yields the same result as if they were applied just once. This is analogous to multiplying a number by one: doing it multiple times doesn't change the value. Viewing the concept in the framework of abstract analysis aids in relating and applying these principles to a broader array of mathematical structures and operations.

Open Sets

In mathematical analysis, particularly in topology, an open set is a fundamental concept that helps define and understand continuity, convergence, and the structure of spaces. An open set is defined with respect to a given topology, and it is a set where, for every point within the set, you can find a neighborhood around that point that is completely contained within the set itself. The idea is illustrated in the solution to the exercise, demonstrating that the interior of any set \(A\) is an open set. Therefore, any point \(x\) inside the interior has an open 'bubble' around it entirely within \(A\).

Open sets are particularly useful because they let us discuss 'local' behavior within a space without being concerned about the 'edge' effects, since by definition, points are comfortably away from the boundaries. When the exercise states that \(\operatorname{int} A\) is an open set, it emphasizes this concept of 'openness' by ensuring that any point within has room to move within the set without stepping out of bounds. For students grappling with the exercise, it's essential to visualize open sets as forming a kind of 'bubble' or 'cushion' around each point.

Set Operations

The exercise delves into set operations, which are procedures that combine, relate, or modify sets in various ways. Mathematical sets are collections of distinct objects, and understanding operations like union, intersection, and the concept of interiors is crucial for solving problems in set theory and related fields. The step by step solution for the exercise explains the relationship between the interior of intersections and unions of sets. Notably, the intersection of the interiors of two sets is equal to the interior of the intersection of the same two sets (\(\operatorname{int}(A \cap B) = \operatorname{int} A \cap \operatorname{int} B\)).

The proof technique used to show this involves taking an element from each set and reasoning about their belonging to open sets that define their interiors. However, the union operation is not as straightforward; the example provided in the exercise demonstrate that the interior of the union of two sets is not always the same as the union of the interiors (\(\operatorname{int}(A \cup B) eq \operatorname{int} A \cup \operatorname{int} B\)). Understanding this nuance is critical, as it cautions against assuming that properties of one set operation (like intersection) automatically transfer to another (like union). This reinforces the importance of carefully considering each set operation in its own right.