Question

Prove that, A relation “is a subset of” in a set of sets is not an equivalence relation.

Answer 1

Hint:
A relation is an equivalence relation if and only if it is reflexive, symmetric and transitive. We’ll show that given relation is not any one of these relations.

Complete step by step solution:
Let A be a set.
In mathematics, a binary relation R over a set X is reflexive if it relates every element of X to itself. Formally, this may be written \[\forall x\; \in \;X\;:\;x{\text{ }}R{\text{ }}x\], or as \[I \subseteq \;R\;\] where I is the identity relation on X.
As, Every set is a subset of itself, i.e., \[A \subset A\].
 So, R is reflexive on A.
A symmetric relation is a type of binary relation. An example is the relation "is equal to", because if \[a\; = \;b\;\] is true then \[b\; = \;a\;\] is also true. Formally, a binary relation R over a set X is symmetric if:
\[\forall a,b \in X(aRb \Leftrightarrow bRa).\]
If \[\;{R^T}\;\] represents the converse of R, then R is symmetric if and only if \[R\; = \;{R^T}\].

For any two sets A, B. If \[A \subset B\]. It is not necessary that \[B \subset A\]. Hence not symmetric and therefore not an equivalence relation.

Note:
An equivalence class is a name that we give to the subset of S which includes all elements that are equivalent to each other. “Equivalent” is dependent on a specified relationship, called an equivalence relation. If there’s an equivalence relation between any two elements, they’re called equivalent.
A simple equivalence class might be defined with an equals sign. We could say
‘The equivalence class of a consists of the set of all x, such that \[x{\text{ }} = {\text{ }}{a^{'}}\]