# Equivalence relations and partitions: some core basic theorems

Suppose R is an equivalence relation on a set S. For each a in S, let [a] denote the set of elements of S to which a is related under R, that is, $[a] = \{ x: (a,x) \in R\}$

We call [a] the equivalence class of a in S under R. The collection of all such equivalence classes is denoted by S/R, that is, $S/R = \{ [a]: a \in S\}$. It is called the quotient set of S by R.

The fundamental property of an equivalence relation and its quotient set is contained in the following theorem:

Theorem I:

Let R be an equivalence relation on a set S. Then, the quotient set S/R is a partition of S. Specifically,

(i) For each $a \in S$, we have $a \in [a]$.

(ii) $[a]=[b]$ if and only if $(a,b) \in R$.

(iii) If $[a] \neq [b]$, then [a] and [b] are disjoint.

Proof of (i):

Since R is reflexive, $(a,a) \in R$ for every $a \in S$ and therefore $a \in [a]$.

Proof of (ii):

Assume: $(a,b) \in R$.

we want to show that $[a] = [b]$. That is, we got to prove, (i) $[b] \subseteq [a]$ and (ii) $[a] \subseteq [b]$.

Let $x \in [b]$; then, $(b,x) \in R$. But, by hypothesis $(a,b) \in R$ and so, by transitivity, $(a,x) \in R$. Accordingly, $x \in [a]$. Thus, $[b] \subseteq [a]$.

To prove that $[a] \subseteq [b]$, we observe that $(a,b) \in R$ implies, by symmetry, that $(b,a) \in R$. Then, by a similar argument, we obtain $[a] \subseteq [b]$. Consequently, $[a]=[b]$.

Now, assume $[a] = [b]$.

Then by part (i) of this proof that for each $a \in S$, we have $a \in [a]$. So, also, here $b \in [b]=[a]$; hence, $(a,b) \in R$.

Proof of (iii):

Here, we prove the equivalent contrapositive of the statement (iii), that is:

If $[a] \bigcap [b] \neq \emptyset$ then $[a] = [b]$.

if $[a] \bigcap [b] \neq \emptyset$ then there exists an element $x \in A$ with $x \in [a] \bigcap [b]$. Hence, $(a,x) \in R$ and $(b,x) \in R$. By symmetry, $(x,b) \in R$, and, by transitivity, $(a,b) \in R$. Consequently, by proof (ii), $[a] = [b]$.

The converse of the above theorem is also true. That is,

Theorem II:

Suppose $P = \{ A_{i}\}$ is a partition of set S. Then, there is an equivalence relation $\sim$ on S such that the set $S/\sim$ of equivalence classes is the same as the partition $P = \{ A_{i}\}$.

Specifically, for $a, b \in S$, the equivalence $\sim$ in Theorem I is defined by $a \sim b$ if a and b belong to the same cell in P.

Thus, we see that there is a one-one correspondence between the equivalence relations on a set S and the partitions of S.

Proof of Theorem II:

Let $a, b \in S$, define $a \sim b$ if a and b belong to the same cell $A_{k}$ in P. We need to show that $\sim$ is reflexive, symmetric and transitive.

(i) Let $a \in S$. Since P is a partition there exists some $A_{k}$ in P such that $a \in A_{k}$. Hence, $a \sim a$. Thus, $\sim$ is reflexive.

(ii) Symmetry follows from the fact that if $a, b \in A_{k}$, then $b, a \in A_{k}$.

(iii) Suppose $a \sim b$ and $b \sim c$. Then, $a, b \in A_{i}$ and $b, c \in A_{j}$. Therefore, $b \in A_{i} \bigcap A_{j}$. Since P is a partition, $A_{i} = A_{j}$. Thus, $a, c \in A_{i}$ and so $a \sim c$. Thus, $\sim$ is transitive.

Accordingly, $\sim$ is an equivalence relation on S.

Furthermore,

$[a] = \{ x: a \sim x\}$.

Thus, the equivalence classes under $\sim$ are the same as the cells in the partition P.

More later,

Nalin Pithwa.

# Some foundation mathematics

Well-Ordering Principle:

Every non-empty set S of non-negative integers contains a least element; that is, there is some integer a in S such that $a \leq b$ for all b’s belonging to S.

Because this principle plays a role in many proofs related to foundations of mathematics, let us use it to show that the set of positive integers has what is known as the Archimedean property.

Archimedean property:

If a and b are any positive integers, then there exists a positive integer n such that $na \geq b$.

Proof:

By contradiction:

Assume that the statement of the theorem is not true so that for some a and b, we have $na for every positive integer n. Then, the set $S = \{ b-na : n \in Z^{+}\}$ consists entirely of positive integers. By the Well-Ordering Principle, S will possess a least element, say, $b-ma$. Notice that $b- (m+1)a$ also lies in S; because S contains all integers of this form. Further, we also have $b-(m+1)a=(b-ma)-a contrary to the choice of $b-ma$ as the smallest integer in S. This contradiction arose out of original assumption that the Archimedean property did not hold; hence, the proof. QED.

First Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) Whenever the integer k is in S, the next integer $k+1$ is also in S.

Then, S is the set of all positive integers.

Second Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) If k is a positive integer such that $1,2,\ldots k$ belong to S, then $(k+1)$ must also be in S.

Then, S is the set of all positive integers.

So, in lighter vein, we assume a set of positive integers is given just as Kronecker had observed: “God created the natural numbers, all the rest is man-made.”

More later,

Nalin Pithwa.