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In Problems \(5-8\), assume that $$ \Omega=\\{1,2,3,4,5,6\\} $$ \(A=\\{1,3,5\\}\), and \(B=\\{1,2,3\\}\). Find \((A \cup B)^{c}\).

Short Answer

Expert verified
The complement of the union, \((A \cup B)^{c} = \{4, 6\}\).

Step by step solution

01

Understand the Union of Sets

First, determine the union of sets \(A\) and \(B\). The union \(A \cup B\) is the set of all distinct elements that belong to either \(A\) or \(B\) or both. So, for sets \(A=\{1,3,5\}\) and \(B=\{1,2,3\}\), the union is:\[A \cup B = \{1, 3, 5\} \cup \{1, 2, 3\} = \{1, 2, 3, 5\}\]
02

Determine the Complement of the Union

Next, find the complement of the union set \((A \cup B)\). The complement of a set with respect to \(\Omega\) includes all elements in \(\Omega\) that are not in the set. Thus, \[(A \cup B)^c = \Omega - (A \cup B) = \{1, 2, 3, 4, 5, 6\} - \{1, 2, 3, 5\} = \{4, 6\}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Union of Sets
In set theory, the union of sets is a fundamental operation that combines all elements from two or more sets. When we refer to the union of sets, we're talking about bringing together unique elements from each set involved. For example, consider two sets: Set \(A = \{1, 3, 5\}\) and Set \(B = \{1, 2, 3\}\).

When we find the union of \(A\) and \(B\), denoted as \(A \cup B\), we gather all unique elements from both sets, without any overlapping elements counted twice. In mathematical terms, this can be seen as \(A \cup B = \{1, 3, 5\} \cup \{1, 2, 3\}\), resulting in the set \( \{1, 2, 3, 5\} \).

This operation is particularly useful when we want to know the total collection of elements across several sets. Remember, elements are not repeated in the union, ensuring each is counted only once in the final set.
Complement of a Set
The complement of a set is another important concept in set theory. It involves identifying all elements not present in a given set but within a particular universal set, often denoted as \(\Omega\). The complement is symbolized by a superscript \(^c\) or sometimes as \(A'\) for set \(A\).

For instance, if we have a universal set \(\Omega = \{1, 2, 3, 4, 5, 6\}\) and another set \(S = \{1, 2, 3, 5\}\), the complement of set \(S\) with respect to \(\Omega\) would be \(S^c = \Omega - S\). This means we include all elements from \(\Omega\) that are not in \(S\).

In practice, calculating \((A \cup B)^{c}\) involves determining which elements of the universal set \(\Omega\) are absent in the union set \(A \cup B\). Here, \((A \cup B) = \{1, 2, 3, 5\}\). Therefore, the complement \((A \cup B)^c = \{4, 6\}\), which clearly shows \{4, 6\} are the elements exclusively found in \(\Omega\) but not in \(A \cup B\).

This is particularly significant when delineating elements outside of a specified subset, helping to frame our analysis of set-related problems.
Universal Set
The universal set, often denoted by \(\Omega\), is a comprehensive set containing all possible elements within a particular context or study. It's essentially the 'universe of discourse' for any specific problem. 

Within this universal set, we define other sets whose elements are subsets of \(\Omega\). Consider \(\Omega = \{1, 2, 3, 4, 5, 6\}\) in our example. All sets, such as \(A\), \(B\), or any other derivations like \(A \cup B\), are based on elements from within \(\Omega\).

The purpose of the universal set is to provide a reference for operations like union and complement. Set operations make more sense when considered relative to \(\Omega\), ensuring each element in a discussion is accounted for.

It's crucial for distinguishing what is included and what's excluded in results from a set operation. By defining \(\Omega\), we give context and boundaries to our set theory problems, making universal sets an essential piece of broader problem-solving in mathematics.

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