Problem 6 - Analysis with an introduction to proof

Question

Use the axiom for regularity to show that for any set \(x, x \cup\\{x\\} \neq x\).

Step-by-Step Solution

Answer

Using the axiom of regularity, we first define the union between set \(x\) and its own element: \(x \cup \{x\} = \{y | y \in x\text{ or } y = x\}\). Then, we consider two cases: 1. If \(x\) is empty, \(x \cup \{x\} = \{x\} \neq x\). 2. If \(x\) is non-empty, the axiom of regularity gives us an element \(y \in x\) such that \(y\) and \(x\) are disjoint. Since \(y \neq x\), but \(y \in x\) and \(x \in x \cup \{x\}\), we cannot have \(x \cup \{x\} = x\), as it would contradict the disjointness property. Thus, for any set \(x\), \(x \cup \{x\} \neq x\).

Step 1: Define the union between set \(x\) and its own element

Let the set \(x\) be any arbitrary set. We are tasked with proving that \(x \cup \{x\} \neq x\). The union between set \(x\) and its own element can be defined as \(x \cup \{x\} = \{y | y \in x\text{ or } y = x\}\), meaning any element \(y\) within this union is either in \(x\) itself or \(y \equiv x\).

Step 2: Use the axiom of regularity

The axiom of regularity states that for every non-empty set \(A\), there exists an element \(y \in A\) such that \(y\) and \(A\) are disjoint (i.e., they have no elements in common). We apply this axiom to our set \(x\): 1. If the set \(x\) is empty, then \(x \cup \{x\} = \{x\}\) by definition of set union. In this case, \(\{x\} \neq x\), because the set \(\{x\}\) has one element, whereas \(x\) has none. 2. If the set \(x\) is non-empty, we apply the axiom of regularity to it. By the axiom, there exists an element \(y \in x\) such that \(y\) and \(x\) are disjoint. Now let's use this disjointness property to compare the union and the original set.

Step 3: Compare the union and the original set

Since \(y\) and \(x\) are disjoint, it means that \(y \neq x\). However, because \(y \in x\), the element \(y\) is also in the union \(x \cup \{x\}\). Furthermore, recall that by definition, \(x \in x \cup \{x\}\). If \(x \cup \{x\} = x\), then it would imply that \(x\) is an element of \(x\), which contradicts the fact that \(y\) and \(x\) are disjoint, according to the axiom of regularity. Therefore, we must conclude that \(x \cup \{x\} \neq x\), and we have completed the proof.

Key Concepts

Set Theory

Set theory is the mathematical theory that studies sets, which are collections of objects. In set theory, we can define a set by listing its elements or by a property that all its members share. One of the fundamental concepts in set theory is the idea that a set can include other sets as elements.

For example, if we have a set called \( A \) that contains 1, 2, and another set \{3, 4\}, we can write it as \( A = \(1, 2, \{3, 4\}\) \). When we describe sets, we use symbols such as \( \cup \) for the union of sets, and \( \in \) to denote that an element is a member of a set. Understanding how sets operate and relate to one another is crucial for exploring the foundations of mathematics and its applications in many fields.

Proof Strategy

A proof is a logical argument that demonstrates the truth of a statement beyond any doubt. In mathematics, it's essential to choose the right strategy when trying to prove a statement. The axiom of regularity (also known as the axiom of foundation) is one of the methods that can be used in proofs within set theory.

When engaging a proof strategy, it is important to understand what you are assuming (the premises), what you are trying to prove (the conclusion), and the rules or axioms that you can apply. For this particular exercise, we use the axiom of regularity along with the definition of set union to show that a set and the union of itself with its singleton set are not equivalent. This strategy combines direct reasoning with axiom application to reach a logical conclusion.

Union of Sets

The union of sets is a fundamental operation in set theory. It combines all the distinct elements from each of the sets into a new set. If we have two sets \( A \) and \( B \), the union \( A \cup B \) consists of all elements that belong to \( A \) or \( B \) or both. This operation is commutative, meaning \( A \cup B = B \cup A \) and associative, which means you can group the operation in any way without affecting the outcome.

Understanding the operation of the union is key to solving many problems in set theory. The concept that each element in the union belongs to at least one of the contributing sets is crucial when analyzing set relationships or proving properties about sets, like in the given exercise where the uniqueness of the union involving a set and it's singleton is established.