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Magazine Chapter 12: Problem 5
Let \(\sim_{1}\) and \(\sim_{2}\) be distinct equivalence relations on \(A\). Define \(\sim_{3}\) by: \(a \sim_{3} b\) iff \(a \sim_{1} b\) and \(a \sim_{2} b\). Prove that \(\sim_{3}\) is an equivalence relation on \(A\). If \([x]_{i}\) denotes the equivalence class of \(x\) for \(\sim_{i}(i=1,2,3)\), prove that \([x]_{3}=[x]_{1} \cap[x]_{2} .\)
Short Answer
Expert verified
\(\sim_3\) is an equivalence relation on \(A\), with \([x]_3 = [x]_1 \cap [x]_2\).
Step by step solution
01
Prove Reflexivity of \(\sim_3\)
To prove reflexivity, we need to show that for all elements \(a \in A\), \(a \sim_3 a\). Since \(\sim_1\) and \(\sim_2\) are equivalence relations, they are reflexive. Thus, \(a \sim_1 a\) and \(a \sim_2 a\). Therefore, by definition of \(\sim_3\), since both \(a \sim_1 a\) and \(a \sim_2 a\) hold, \(a \sim_3 a\) holds as well. Hence, \(\sim_3\) is reflexive.
02
Prove Symmetry of \(\sim_3\)
To prove symmetry, assume \( a \sim_3 b \). This means both \( a \sim_1 b \) and \( a \sim_2 b \). Since \(\sim_1\) and \(\sim_2\) are symmetric, \(b \sim_1 a\) and \(b \sim_2 a\) hold. Therefore, \(b \sim_3 a\) by definition of \(\sim_3\). Hence, \(\sim_3\) is symmetric.
03
Prove Transitivity of \(\sim_3\)
To prove transitivity, assume \( a \sim_3 b \) and \( b \sim_3 c \). This implies \( a \sim_1 b \), \( a \sim_2 b \), \( b \sim_1 c \), and \( b \sim_2 c \). Since \(\sim_1\) and \(\sim_2\) are transitive, we have \( a \sim_1 c \) and \( a \sim_2 c \). Therefore, \( a \sim_3 c \) by definition of \(\sim_3\). Hence, \(\sim_3\) is transitive.
04
Prove \([x]_3 = [x]_1 \cap [x]_2\)
To prove the equivalence classes property, let \([x]_3 = \{y \in A : x \sim_3 y\}\). By definition, \(x \sim_3 y\) iff \(x \sim_1 y\) and \(x \sim_2 y\). Thus, \([x]_3 = \{y \in A : x \sim_1 y\ \text{and}\ x \sim_2 y\}\), which is exactly \([x]_1 \cap [x]_2\). Therefore, \([x]_3 = [x]_1 \cap [x]_2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reflexivity
In mathematics, reflexivity is a fundamental property of equivalence relations. For a relation to be reflexive, every element must be related to itself. In simpler terms, if you have any element "a" in a set, it should always satisfy the relation "a is related to a". Reflexivity ensures that each element is consistently part of its own relationship.
Let's consider the set \(A\) and equivalence relations \(\sim_1\) and \(\sim_2\) on it, which by definition are reflexive. This means for every element \(a\) in \(A\), \(a \sim_1 a\) and \(a \sim_2 a\) hold true. When defining a new equivalence relation \(\sim_3\) using these two, we say \(a \sim_3 a\) if both \(a \sim_1 a\) and \(a \sim_2 a\) are true.
Thus, since \(\sim_1\) and \(\sim_2\) are reflexive, \(\sim_3\) must also be reflexive because it inherits the property from them.
Let's consider the set \(A\) and equivalence relations \(\sim_1\) and \(\sim_2\) on it, which by definition are reflexive. This means for every element \(a\) in \(A\), \(a \sim_1 a\) and \(a \sim_2 a\) hold true. When defining a new equivalence relation \(\sim_3\) using these two, we say \(a \sim_3 a\) if both \(a \sim_1 a\) and \(a \sim_2 a\) are true.
Thus, since \(\sim_1\) and \(\sim_2\) are reflexive, \(\sim_3\) must also be reflexive because it inherits the property from them.
Symmetry
Symmetry in equivalence relations refers to the idea that if one element is related to another, then the second element is also related back to the first. In simpler terms, if "a is related to b", then "b is related to a" must also be true.
For the equivalence relations \(\sim_1\) and \(\sim_2\) on a set \(A\), they inherently have this symmetry property. This means if \(a \sim_1 b\), then \(b \sim_1 a\), and similarly, \(a \sim_2 b\) implies \(b \sim_2 a\).
Now, consider \(\sim_3\) defined such that \(a \sim_3 b\) if both \(a \sim_1 b\) and \(a \sim_2 b\). Given the symmetry of \(\sim_1\) and \(\sim_2\), it follows that \(b \sim_1 a\) and \(b \sim_2 a\), thereby satisfying \(b \sim_3 a\). Thus, the symmetry of \(\sim_3\) is preserved through its definition.
For the equivalence relations \(\sim_1\) and \(\sim_2\) on a set \(A\), they inherently have this symmetry property. This means if \(a \sim_1 b\), then \(b \sim_1 a\), and similarly, \(a \sim_2 b\) implies \(b \sim_2 a\).
Now, consider \(\sim_3\) defined such that \(a \sim_3 b\) if both \(a \sim_1 b\) and \(a \sim_2 b\). Given the symmetry of \(\sim_1\) and \(\sim_2\), it follows that \(b \sim_1 a\) and \(b \sim_2 a\), thereby satisfying \(b \sim_3 a\). Thus, the symmetry of \(\sim_3\) is preserved through its definition.
Transitivity
Transitivity is another critical component of equivalence relations. It means that if an element is related to a second and the second is related to a third, then the first must also be related to the third. In mathematical terms, for elements \(a\), \(b\), and \(c\), if "a is related to b" and "b is related to c", then "a is related to c" as well.
With equivalence relations \(\sim_1\) and \(\sim_2\), transitivity is a given. So if \(a \sim_1 b\) and \(b \sim_1 c\), then \(a \sim_1 c\). The same logic applies for \(\sim_2\).
Defining \(\sim_3\) as before, such that \(a \sim_3 b\) implies both \(a \sim_1 b\) and \(a \sim_2 b\), we can deduce that if both \(b \sim_1 c\) and \(b \sim_2 c\) hold, then \(a \sim_3 c\) must be true because of transitivity in the underlying relations \(\sim_1\) and \(\sim_2\). Thus, \(\sim_3\) retains transitivity.
With equivalence relations \(\sim_1\) and \(\sim_2\), transitivity is a given. So if \(a \sim_1 b\) and \(b \sim_1 c\), then \(a \sim_1 c\). The same logic applies for \(\sim_2\).
Defining \(\sim_3\) as before, such that \(a \sim_3 b\) implies both \(a \sim_1 b\) and \(a \sim_2 b\), we can deduce that if both \(b \sim_1 c\) and \(b \sim_2 c\) hold, then \(a \sim_3 c\) must be true because of transitivity in the underlying relations \(\sim_1\) and \(\sim_2\). Thus, \(\sim_3\) retains transitivity.
Equivalence Classes
Equivalence classes are a subdivision of a set into disjoint classes or groups, where members of each class are equivalent to each other under a given equivalence relation. They form a partition of the set.
Suppose you have an equivalence relation \(\sim\) on a set \(A\). For any element \(x\) in \(A\), the equivalence class \([x]\) contains all elements \(y\) such that \(x \sim y\).
In our case, the relation \(\sim_3\) is built from both \(\sim_1\) and \(\sim_2\). The equivalence class \([x]_3\) for \(\sim_3\) consists of all elements \(y\) such that \(x \sim_3 y\), which occurs if \(x \sim_1 y\) and \(x \sim_2 y\). Hence, \([x]_3\) is effectively the intersection of the equivalence classes \([x]_1\) and \([x]_2\) from the relations \(\sim_1\) and \(\sim_2\). This intersection property arises naturally from the definition of \(\sim_3\).
Suppose you have an equivalence relation \(\sim\) on a set \(A\). For any element \(x\) in \(A\), the equivalence class \([x]\) contains all elements \(y\) such that \(x \sim y\).
In our case, the relation \(\sim_3\) is built from both \(\sim_1\) and \(\sim_2\). The equivalence class \([x]_3\) for \(\sim_3\) consists of all elements \(y\) such that \(x \sim_3 y\), which occurs if \(x \sim_1 y\) and \(x \sim_2 y\). Hence, \([x]_3\) is effectively the intersection of the equivalence classes \([x]_1\) and \([x]_2\) from the relations \(\sim_1\) and \(\sim_2\). This intersection property arises naturally from the definition of \(\sim_3\).
Intersection of Sets
The intersection of sets is a basic operation in set theory that results in a new set containing all elements that are common to the given sets. If you have two sets, say \(X\) and \(Y\), their intersection, denoted by \(X \cap Y\), includes only the items found in both \(X\) and \(Y\).
In terms of equivalence relations and their classes, the intersection plays a crucial role. Given our relations \(\sim_1\) and \(\sim_2\), each defines equivalence classes \([x]_1\) and \([x]_2\) for an element \(x\). The equivalence class \([x]_3\) for the relation \(\sim_3\) is precisely the intersection \([x]_1 \cap [x]_2\).
This means, to find \([x]_3\), we look for elements common to both \([x]_1\) and \([x]_2\). This effectively shows how \(\sim_3\) narrows down the equivalence by requiring elements meet the criteria of both \(\sim_1\) and \(\sim_2\). Thus, the concept of intersection helps understand the formation of equivalence classes in \(\sim_3\).
In terms of equivalence relations and their classes, the intersection plays a crucial role. Given our relations \(\sim_1\) and \(\sim_2\), each defines equivalence classes \([x]_1\) and \([x]_2\) for an element \(x\). The equivalence class \([x]_3\) for the relation \(\sim_3\) is precisely the intersection \([x]_1 \cap [x]_2\).
This means, to find \([x]_3\), we look for elements common to both \([x]_1\) and \([x]_2\). This effectively shows how \(\sim_3\) narrows down the equivalence by requiring elements meet the criteria of both \(\sim_1\) and \(\sim_2\). Thus, the concept of intersection helps understand the formation of equivalence classes in \(\sim_3\).
