Discrete math equivalence classes

Question

Let $A$ be a finite set of size $k$ and $R$ a relation on the power set $P(A)$ defined by $R=\left\{(A,B) : |A|=|B|\right\}$

1. Show that $A$ is an equivalence relation.
2. Let $a \in A$. What is the size of the equivalence class of $\{a\}$?
3. Let $a, b$ be two different elements of $A$. What is the size of the equivalence class of $\{a, b\}$?

I’m having a lot of trouble with this problem. It says the relation is on the power set, but then I’m finding the size of the equivalence class of elements within $A$, and then of $(a,b)$? I’m honestly completely lost and don’t have any base to build off of. I think I’m a bit confused on the concept of the size of equivalence classes in general.

Answer 1

You mean to prove that $R$ is an equivalence relation. There are three criteria that must be followed in order to prove $R$ is an equivalence relation,

1. Reflexive
2. Symmetric
3. Transitive

To get you started, with generic sets $B,C,D \in \mathcal{P}(A)$, see that we always have $$|B| = |B|$$ so $R$ is reflexive. This would mean $(B,B) \in R$ for some $B \subseteq A$

To show it's symmetric, it requires you to show that if $|B| = |C|$, then $|C| = |B|$.

To show it's transitive, it requires you to show if $|B| = |C|$, and$ |C| = |D|$, then |B| = |D|$

Answer 2

Informally, under this equivalence relation two subsets are equivalent when they have the same size.

Thus, the equivalence class of $\{a\}$ consists of all subsets of $A$ with cardinality/size equal to one. Thus the size of this equivalence class is $k=|A|$.

The equivalence class of $\{a, b\}$ consists of all two element subsets of $A$. Thus the size of this equivalence class is $\binom{k}{2}=\frac{k(k-1)}{2}$.