*arrow_back*

*arrow_downward*

# The fundamental theorem of equivalence relations

**Equivalence relations** are “essentially the same” as **partitions**. More precisely, every equivalence relation on a set \(X\) yields a partition of \(X\), and every partition of \(X\) yields an equivalence relation on \(X\). Our goal here is to prove:

\(\tab\) **Theorem (the fundamental theorem of equivalence relations)**. Every equivalence relation yields a partition. Every partition yields an equivalence relation.

This is very surprising, particularly because the definitions of **equivalence relation** and **partition** are nothing alike.

\(\tab\) **Definition**. An **equivalence relation** on a set \(X\) is a relation on \(X\) that is **reflexive**, **symmetric**, and **transitive**.

\(\tab\) **Definition**. A **partition** of a set \(X\) is a **pairwise-disjoint** **cover** of \(X\).

What all of this means, we explain in the sequel.

# Every equivalence relation yields a partition. Example

Every equivalence relation \((R, X, X)\) on an arbitrary set \(X\) gives rise a partition of \(X\), and that partition is the quotient set \(X/R\) of all equivalence classes of \(R\). This means the **cells** of the partition are the **equivalence classes** of \(R\).

\(\tab\) *Example*. Let \(X\) be the set

\[X \hskip8pt := \hskip8pt \{0, 1, 2, 3, 4\},\]

let \((R, X, X)\) be the **equivalence relation** on \(X\) whose **graph** \(R\) is given by

\[R \hskip8pt := \hskip8pt \{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 3), (3, 4), (4, 3), (4, 4)\}.\]

We’ll show that \(R\) gives rise to a specific partition of \(X\) that is intimately related to \(R\). But first, it’s not clear that \(R\) is an equivalence relation, so let’s prove that \(R\) satisfies the properties of an equivalence relation. For \(R\) to satisfy the properties of an equivalence relation, \(R\) must be: \(1)\) reflexive, \(2)\) symmetric, and \(3)\) transitive.

For \(R\) be to be **reflexive**, the following statement must hold: \(a\) is \(R\)-related to \(a\) (for arbitrary \(a \in X\)).

For \(R\) be to be **symmetric**, the following statement must hold: if \(a\) is \(R\)-related to \(b\), then \(b\) is \(R\)-related to \(a\) (for arbitrary \(a,b \in X\)).

For \(R\) be to be **transitive**, the following statement must hold: if \(a\) is \(R\)-related to \(b\) and \(b\) is \(R\)-related to \(c\), then \(a\) is \(R\)-related to \(c\) (for arbitrary \(a,b,c \in X\)).

The equivalence relation \((R, X, X)\) gives rise to *two* **equivalence classes**, meaning \(X/R\) has *two* elements. And the equivalence classes of \(R\), which are precisely the elements of \(X/R\), are the following: \(\{0, 1, 2\}\) and \(\{3, 4\}\). So, \(X/R\) is the set

\[X/R \hskip8pt := \hskip8pt \{\{0, 1, 2\}, \{3, 4\}\}.\]

Why are the equivalence classes of \(R\) the subsets \(\{0, 1, 2\} \subseteq X\) and \(\{3, 4\} \subseteq X\)? To answer this, we can compute the equivalence class \(a/R\) of *each* element \(a\) of \(X\), also denoted \([a]_R\) or \([a]\), and see that the only results are \(\{0, 1, 2\}\) or \(\{3, 4\}\).

The equivalence class \(0/R\) of \(0\) under \(R\) is the set of all elements of \(X\) that \(0\) is \(R\)-related to. Looking at the definition of \(R\), we see that

\(0\) is \(R\)-related to \(0\) because \((0, 0) \in R\), and

\(0\) is \(R\)-related to \(1\) because \((0, 1) \in R\), and

\(0\) is \(R\)-related to \(2\) because \((0, 2) \in R\), and

\(0\) is *not* \(R\)-related to \(3\) because \((0, 3) \notin R\), and

\(0\) is *not* \(R\)-related to \(4\) because \((0, 4) \notin R\).

So, the equivalence class \(0/R\) of \(0\) under \(R\) is the set \(\{0, 1, 2\}\), meaning \(0/R = \{0, 1, 2\}\).

The equivalence class \(1/R\) of \(1\) under \(R\) is the set of all elements of \(X\) that \(1\) is \(R\)-related to. Looking at the definition of \(R\), we see that

\(1\) is \(R\)-related to \(0\) because \((1, 0) \in R\), and

\(1\) is \(R\)-related to \(1\) because \((1, 1) \in R\), and

\(1\) is \(R\)-related to \(2\) because \((1, 2) \in R\), and

\(1\) is *not* \(R\)-related to \(3\) because \((1, 3) \notin R\), and

\(1\) is *not* \(R\)-related to \(4\) because \((1, 4) \notin R\).

So, the equivalence class \(1/R\) of \(1\) under \(R\) is the set \(\{0, 1, 2\}\), meaning \(1/R = \{0, 1, 2\}\).

The equivalence class \(2/R\) of \(2\) under \(R\) is the set of all elements of \(X\) that \(2\) is \(R\)-related to. Looking at the definition of \(R\), we see that

\(2\) is \(R\)-related to \(0\) because \((2, 0) \in R\), and

\(2\) is \(R\)-related to \(1\) because \((2, 1) \in R\), and

\(2\) is \(R\)-related to \(2\) because \((2, 2) \in R\), and

\(2\) is *not* \(R\)-related to \(3\) because \((2, 3) \notin R\), and

\(2\) is *not* \(R\)-related to \(4\) because \((2, 4) \notin R\).

So, the equivalence class \(2/R\) of \(2\) under \(R\) is the set \(\{0, 1, 2\}\), meaning \(2/R = \{0, 1, 2\}\).

The equivalence class \(3/R\) of \(3\) under \(R\) is the set of all elements of \(X\) that \(3\) is \(R\)-related to. Looking at the definition of \(R\), we see that

\(3\) is *not* \(R\)-related to \(0\) because \((3, 0) \notin R\), and

\(3\) is *not* \(R\)-related to \(1\) because \((3, 1) \notin R\), and

\(3\) is *not* \(R\)-related to \(2\) because \((3, 2) \notin R\), and

\(3\) is \(R\)-related to \(3\) because \((3, 3) \in R\), and

\(3\) is \(R\)-related to \(4\) because \((3, 4) \in R\).

So, the equivalence class \(3/R\) of \(3\) under \(R\) is the set \(\{3, 4\}\), meaning \(3/R = \{3, 4\}\).

The equivalence class \(4/R\) of \(4\) under \(R\) is the set of all elements of \(X\) that \(4\) is \(R\)-related to. Looking at the definition of \(R\), we see that

\(4\) is *not* \(R\)-related to \(0\) because \((3, 0) \notin R\), and

\(4\) is *not* \(R\)-related to \(1\) because \((3, 1) \notin R\), and

\(4\) is *not* \(R\)-related to \(2\) because \((3, 2) \notin R\), and

\(4\) is \(R\)-related to \(3\) because \((3, 3) \in R\), and

\(4\) is \(R\)-related to \(4\) because \((3, 4) \in R\).

So, the equivalence class \(4/R\) of \(4\) under \(R\) is the set \(\{3, 4\}\), meaning \(4/R = \{3, 4\}\).

This sums up the situation:

\(\tab\tab\) \(0/R \hskip8pt = \hskip8pt \{0, 1, 2\}\)

\(\tab\tab\) \(1/R \hskip8pt = \hskip8pt \{0, 1, 2\}\)

\(\tab\tab\) \(2/R \hskip8pt = \hskip8pt \{0, 1, 2\}\)

\(\tab\tab\) \(3/R \hskip8pt = \hskip8pt \{3, 4\}\)

\(\tab\tab\) \(4/R \hskip8pt = \hskip8pt \{3, 4\}\)

\(\tab\tab\) \(X/R \hskip8pt = \hskip8pt \{0, 1, 2, 3, 4\}/R \hskip8pt = \hskip8pt \{\{0, 1, 2\}, \{3, 4\}\}.\)

And here comes the punchline. The set \(\{0, 1, 2\}\) and the set \(\{3, 4\}\), which are the elements of the set \(X/R\), form a **partition** \(P\) of \(X\), so that the set

\[P \hskip8pt := \hskip8pt \{\{0, 1, 2\}, \{3, 4\}\}\]

is a partition of \(X\), because the elements of \(P\) are pairwise disjoint (since \(\{0, 1, 2\} \cap \{3, 4\} = \{\}\)), and \(\cup P\) is \(X\) (since \(\cup P\) is \(\cup\{\{0, 1, 2\}, \{3, 4\}\}\), which is \(\{0, 1, 2, 3, 4\}\), which is \(X\)).

So, the equivalence relation \((R, X, X)\) yields a partition of \(X\), and this partition is \(X/R\). More precisely, the **cells** of the partition \(\{\{0, 1, 2\}, \{3, 4\}\}\) are precisely the **equivalence classes** of \(R\), so that the **partition** \(\{\{0, 1, 2\}, \{3, 4\}\}\) is equal to the **quotient set** \(X/R\), which is the set of all equivalence classes of \(R\), which is also the set \(\{\{0, 1, 2\}, \{3, 4\}\}\) because the equivalence classes of \(R\) are \(\{0, 1, 2\}\) and \(\{3, 4\}\).

\(\tab\) \(\square\)

# Every equivalence relation yields a partition. Proof

\(\tab\) **Theorem 1**. Let \(R\) be an **equivalence relation** on a set \(X\). Now \(R\) yields a **partition** of \(X\). More precisely, the **equivalence classes** of \(R\) form a partition of \(X\). In other words, if \(R\) is an equivalence relation, then the **quotient set** \(X/R\) is a **partition** of \(X\).

\(\tab\) *Proof*. “If \(R\) is an equivalence relation, then the quotient set \(X/R\) is a partition of \(X\)”. This is a bold claim, so we better make sure we understand it. If \(X\) is a set and \(R\) is an equivalence relation, what does \(X/R\) look like? Well, \(X/R\) is the set of all **equivalence classes** of \(R\). So, every **element** of the set \(X/R\) is a **subset** of the set \(X\). So, \(X/R\) is a *set of subsets* of \(X\), and so \(X/R\) is a *subset of the power set* \({ \mathcal P } X\). This means the set \(X/R\) is a candidate to be a partition of \(X\), because a partition of \(X\) must be made up of subsets of \(X\). But there’s three more requirements:

\(\tab\tab1)\) the [**(unary) union**](set_theory00.html) \(\bigcup X/R\) must be *all* of \(X\),

\(\tab\tab2)\) the elements of \(X/R\) must *all* be disjoint with each other (ie. pairwise disjoint), and

\(\tab\tab3)\) every element of \(X/R\) must be nonempty.

First, we’ll prove two facts:

\(\tab\tab1)\) Every element of \(X\) is in *at least one* equivalence class of \(R\).

\(\tab\tab2)\) Every element of \(X\) is in *at most one* equivalence class of \(R\).

Together, \(1)\) and \(2)\) imply:

\(\tab\tab3)\) Every element of \(X\) is in *exactly one* equivalence class of \(R\).

And \(3)\) implies:

\(\tab\tab4)\) The equivalence classes of \(R\) form a partition of \(X\).

And \(4)\) is the desired result.

To prove \(1)\), let \(a \in X\) be an (arbitrary, but fixed) element of \(X\). Since \(R\) is an equivalence relation, then in particular \(R\) is **reflexive**. So \(a\) is \(R\)-related to \(a\) (by the definition of *reflexive*), or equivalently \((a, a) \in R\), or equivalently \(aRa\).

Recall that \([a]\) is the set of all elements of \(X\) that are \(R\)-related to \(a\). Since \(a\) is an element of \(X\) that is \(R\)-related to \(a\), then \(a\) is in \([a]\), by the definition of \([a]\).

So the element \(a \in X\) is at least in the equivalence class \([a]\). In particular, the element \(a \in X\) is in *at least one* equivalence class. This proves \(1)\).

To prove \(2)\), let \(a \in X\) be an (arbitrary, but fixed) element of \(X\). From \(1)\) we know that \(a\) is in the equivalence class \([a]\).

Now comes the magic: suppose \(a\) is in some other equivalence class, in addition to the equivalence class \([a]\). This equivalence class would be the set of all elements that are \(R\)-related to *some* (mysterious) element; we don’t know what this (mysterious) element is, or even if it exists, but let’s call it \(b\). To be specific, suppose \(a\) is in the equivalence class \([b]\), which is the set of all elements of \(X\) that are \(R\)-related to \(b \in X\). (Notice that the element \(b \in X\) is not arbitrary, but it’s a very specific element that depends on what \(a\) we’ve chosen.) By the definition of \([b]\), we know that \(a\) is \(R\)-related to \(b\), because \(a\) is in \([b]\). But \(R\) is **symmetric** (because \(R\) is an equivalence relation), and this implies that \(b\) is also \(R\)-related to \(a\).

Well, this *contradicts* the assumption that \(a\) is in some other equivalence class, which proves that \(a\) can’t be in some other equivalence class. More specifically, it proves that, if we dare to dream that \(a\) is in some equivalence class \([b]\), then \(b\) must actually be \(a\) (meaning \(b = a\)), and so \([b]\) must actually be \([a]\) (meaning \([b] = [a]\)).

# Every partition yields an equivalence relation. Example

\(\tab\) *Example*. Let \(X\) be the set \(\{0, 1, 2, 3, 4\}\), let \(P\) be the **partition** of \(X\) given by \(P \hskip8pt := \hskip8pt \{\{0, 1, 2\}, \{3, 4\}\}\).

This was an example of a general theorem that applies to all sets and all partitions of sets: every partition \(P\) of an arbitrary set \(X\) gives rise to an equivalence relation on \(X\), and the equivalence classes of that equivalence relation are the cells of the partition \(P\).

# Every partition yields an equivalence relation. Proof

\(\tab\) **Theorem 2**. Let \(P\) be a partition of a set \(X\). Now the partition \(P\) induces an equivalence relation on \(X\).

\(\tab\) *Proof*. TODO

# The fundamental theorem of equivalence relations

We can now prove:

\(\tab\) **Theorem (the fundamental theorem of equivalence relations)**. Every equivalence relation yields a partition. Every partition yields an equivalence relation.

\(\tab\) *Proof*. Theorem \(1\) together with theorem \(2\) prove this result.

\(\tab\) \(\square\)