Equivalence Relations as Identification Rules

Identification Rules and Equivalence Relations

An identification rule says which points of a set are to count as the same point for a later construction. The rule must be consistent: every point counts as the same as itself, the rule must work in either direction, and successive identifications must not contradict one another. Equivalence relations express precisely these requirements. Here we focus on the underlying sets of points and their classes, not on any topology that might later be assigned to a set of classes.

For the examples, \(\mathbb{R}\) has its standard topology, \(I=[0,1]\) has the subspace topology inherited from \(\mathbb{R}\), and \(S^n=\{x\in\mathbb{R}^{n+1}:\|x\|=1\}\) has the subspace topology inherited from Euclidean space. The equivalence relations below are defined by the points themselves; their definitions do not depend on these topologies.

The class \([x]\) consists of exactly the points the rule identifies with \(x\). Reflexivity ensures that \(x\) itself belongs to its class. Symmetry means that “identified with” does not depend on which point is named first. Transitivity ensures that if \(x\) and \(y\) are identified, and \(y\) and \(z\) are identified, then \(x\) and \(z\) are identified as well.

Why Equivalence Classes Form a Partition

Proof. Assume first that \(X\ne\varnothing\). For any \(x\in X\), reflexivity gives \(x\sim x\), so \(x\in[x]\); hence \([x]\) is nonempty and every point belongs to a class. Now let \(x,y\in X\), and suppose \([x]\cap[y]\ne\varnothing\). Choose \(z\in[x]\cap[y]\). Then \(z\sim x\) and \(z\sim y\). Symmetry gives \(x\sim z\), and transitivity with \(z\sim y\) gives \(x\sim y\). We show \([x]=[y]\): if \(w\in[x]\), then \(w\sim x\), and \(x\sim y\), so \(w\sim y\) by transitivity; hence \(w\in[y]\). Conversely, if \(w\in[y]\), then \(w\sim y\), and \(y\sim x\) by symmetry, so \(w\sim x\); hence \(w\in[x]\). Therefore the classes are equal. If \(X=\varnothing\), there are no points and thus no classes, so the empty collection covers \(X\) and has no intersecting distinct members. \(\square\)

This result gives a useful way to check a proposed identification: its classes must divide the original set into nonoverlapping groups. Two classes cannot overlap in just one point and still be different. If they share even one point, they are the same class.

Generating an Equivalence Relation from Specified Pairs

Often the initial instruction only lists pairs that should be identified. Those pairs may not yet describe a complete equivalence relation. For example, if \(a\) is identified with \(b\), and \(b\) with \(c\), then transitivity also requires identifying \(a\) with \(c\). The systematic completion is obtained by allowing finite chains of specified identifications, in either direction.

Proof. A chain of length \(m=0\) starts and ends at the same point, so \(\sim_P\) is reflexive. If a chain goes from \(x\) to \(y\), reversing its order gives a chain from \(y\) to \(x\), because each step is allowed in either direction. Thus \(\sim_P\) is symmetric. If one chain goes from \(x\) to \(y\) and another from \(y\) to \(z\), concatenate the chains at \(y\); the resulting finite chain goes from \(x\) to \(z\), so \(\sim_P\) is transitive. Each pair \((x,y)\in P\) gives a chain of length one, so \(x\sim_P y\). Finally, let \(\approx\) be any equivalence relation containing every pair in \(P\). Symmetry of \(\approx\) means that each allowed step in a chain relates its two points under \(\approx\); repeated transitivity then gives \(x\approx y\) for every chain from \(x\) to \(y\), including a chain of length zero by reflexivity. Hence \(\sim_P\) is contained in every equivalence relation containing \(P\), proving it is the smallest one. \(\square\)

The finite-chain rule prevents us from adding arbitrary identifications: it adds exactly what reflexivity, symmetry, and transitivity force. It also gives a direct method for computing classes: start at a point and follow all allowed steps, including steps in reverse.

A Common Mistake: Symmetry Does Not Guarantee Transitivity

A rule can look like a reasonable identification instruction and still fail to be an equivalence relation. For example, on \(\mathbb{R}\) consider the proposed rule \(x\sim y\) if \(|x-y|\) is either \(0\) or \(1\). It is reflexive because \(|x-x|=0\), and symmetric because \(|x-y|=|y-x|\). But \(0\sim1\) and \(1\sim2\), while \(0\not\sim2\) because \(|0-2|=2\), which is neither \(0\) nor \(1\). The rule is not transitive.

A common incorrect calculation is to list \([0]=\{-1,0,1\}\) and \([1]=\{0,1,2\}\), then treat them as different classes even though they overlap. That contradicts the partition theorem: an equivalence relation cannot have overlapping, unequal classes. The correction is not to accept these lists as classes. Instead, include the transitive identifications they force. The smallest equivalence relation containing the pairs at distance \(1\) relates \(x\) to \(y\) exactly when \(x-y\in\mathbb{Z}\): finite chains of steps of \(1\) or \(-1\) have integer total displacement, and every integer displacement can be made by finitely many such steps. In that corrected relation, \([0]=\mathbb{Z}\) and \([1]=\mathbb{Z}\), as the shared points require.

These examples illustrate three distinct forms of identification: one prescribed pair of endpoints, all integer translates, and each point together with its antipode. In each case, the first task is to check the three defining properties; the next is to compute the full class, including every point forced by the rule. The resulting classes are the basic data used when a later construction forms a space of identified points.