Problem 8 - Applied Discrete Structures

Question

Define \(r\) on the power set of \\{1,2,3\\} by \(A r B \Leftrightarrow|A|=|B| .\) Prove that \(r\) is an equivalence relation. What are the equivalence classes under \(r ?\)

Step-by-Step Solution

Answer

The relation \(r\) is an equivalence relation, and the equivalence classes are \(\{\emptyset\}\), \(\{\{1\}, \{2\}, \{3\}\}\), \(\{\{1,2\}, \{1,3\}, \{2,3\}\}\), and \(\{\{1,2,3\}\}\).

Step 1: Define the Relation

The power set of \(\{1,2,3\}\), denoted as \(\mathcal{P}(\{1,2,3\})\), is the set of all subsets of \(\{1,2,3\}\). The relation \(ArB\) is defined as \(|A|=|B|\), meaning that set \(A\) is related to set \(B\) if and only if they have the same number of elements.

Step 2: Prove Reflexivity

For any subset \(A\) of \(\{1,2,3\}\), we have \(|A|=|A|\). Therefore, for all \(A\) in the power set, \(ArA\) holds true. This satisfies the reflexivity property of equivalence relations.

Step 3: Prove Symmetry

Assume \(ArB\). This implies \(|A|=|B|\). By symmetry of equality, it follows that \(|B|=|A|\). Hence, \(BrA\) also holds, satisfying the symmetry property of equivalence relations.

Step 4: Prove Transitivity

Assume \(ArB\) and \(BrC\). This means \(|A|=|B|\) and \(|B|=|C|\). From the transitivity of equality, we have \(|A|=|C|\). Therefore, \(ArC\) holds, satisfying the transitivity property of equivalence relations.

Step 5: Determine Equivalence Classes

The equivalence classes are formed based on the number of elements in each subset of \(\{1,2,3\}\).* The class with 0 elements: \(\{ \emptyset \}\)* The class with 1 element: \(\{\{1\}, \{2\}, \{3\}\}\)* The class with 2 elements: \(\{\{1,2\}, \{1,3\}, \{2,3\}\}\)* The class with 3 elements: \(\{\{1,2,3\}\}\)Each of these classes groups subsets with the same size, as defined by the relation \(r\).

Key Concepts

Reflexivity

To understand reflexivity in the context of equivalence relations, imagine a mirror. Each set is like an individual looking into this mirror. Reflexivity tells us that a set should be related to itself.

In our exercise, we are dealing with the power set of \( \{1, 2, 3\} \), which consists of all possible subsets. So, for any subset \(A\), it must relate to itself because the number of elements in \(A\) is, naturally, equal to the number of elements in \(A\).

Let's break this down further:

• Reflexivity requires that for every subset \(A\), the relation \(ArA\) holds true.
• Since \(|A| = |A|\) for any subset \(A\), the condition is satisfied for every subset.
• This makes \(r\) reflexive for all subsets in the power set.

Each subset sees itself as equal in size, thus fulfilling the reflexive property.

Symmetry

Symmetry in equivalence relations is a friendly two-way street. If one set is related to another, the opposite must also hold true.

In our context, if one subset \(A\) has the same number of elements as another subset \(B\), then \(B\) must also have the same number of elements as \(A\). Think of symmetry as a mutual agreement: "If I'm equal to you, you're equal to me too."

Consider these points:

• Symmetry means if \(ArB\) is true, \(BrA\) must also be true.
• From our definition, if \(|A| = |B|\), then naturally \(|B| = |A|\) because equality itself is symmetric.

This property ensures that any relation between two sets works both ways, making \(r\) symmetric in this exercise.

Transitivity

Transitivity is like a bridge connecting elements in a sequence. It states that if one subset is related to a second, and the second is related to a third, then the first must be related to the third.

Imagine three subsets, \(A\), \(B\), and \(C\). If \(A\) has the same number of elements as \(B\) and \(B\) has the same number as \(C\), transitivity tells us \(A\) must have the same number as \(C\) as well.

This can be further detailed as:

• If you have \(ArB\) and \(BrC\), this implies \(|A| = |B|\) and \(|B| = |C|\).
• Therefore, by the transitive property of equality, \(|A| = |C|\) follows.

Thus, the relation \(r\) satisfies the transitive property, linking all equivalent sizes through a chain of connections.