The Gold Standard of Topology | Closed Equivalence Relations and Hausdorff Quotients

Let X be a topological space, let ∼ be an equivalence relation and let q: X → Y = X/∼ be the canonical quotient map. The quotient topology declares U open in Y precisely when q⁻¹(U) is open in X.

Represent the relation by R = {(x,z) ∈ X×X : x ∼ z}. Calling R closed means closed in the product topology on X×X. This statement concerns pairs moving together, not only one class at a time.

For the main theorem, assume X is compact Hausdorff. Under these hypotheses, Y is Hausdorff if and only if R is closed. We will identify where compactness enters rather than treating it as an optional adjective.

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Suppose Y is Hausdorff. Its diagonal ΔY = {(y,y) : y ∈ Y} is closed in Y×Y. To see this, two unequal coordinates have disjoint open neighbourhoods, whose product avoids the diagonal.

The map q×q: X×X → Y×Y is continuous. Moreover, R = (q×q)⁻¹(ΔY), because two source points are related exactly when they have the same quotient image.

The inverse image of a closed set under a continuous map is closed. Hence R is closed. This direction does not require compactness of X. Writing the diagonal equation makes both the conclusion and the actual assumptions visible.

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Fix x ∈ X. If R is closed, the map z ↦ (x,z) is continuous, so its inverse image [x] is closed in X. When X is compact, this class is also compact.

Conversely, knowing that every class is closed gives a T₁ quotient: a singleton {[x]} in Y is closed exactly when its preimage [x] is closed. The equivalence follows by taking complements in the quotient-topology definition.

T₁ means individual quotient points are closed. Hausdorff separation asks for disjoint open neighbourhoods of two distinct points. The second requirement is stronger. The counterexample later will satisfy the first and fail the second.

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Assume now that X is compact Hausdorff and R is closed. For any closed F ⊆ X, its saturation Sat(F) = q⁻¹(q(F)) consists of points related to some point of F.

It is the first-coordinate projection of R ∩ (X×F). The product X×F is compact, and intersecting it with the closed relation gives a compact set. Its continuous projection is compact in X, hence closed because X is Hausdorff.

Thus Sat(F) is closed for every closed F. This is where compactness supplies a useful global control: projecting a closed set need not preserve closedness without additional conditions. The proof does not assume that projection is always a closed operation.

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The quotient definition also tests closed sets: C ⊆ Y is closed if and only if q⁻¹(C) is closed in X. Apply this to C = q(F). Its preimage is Sat(F), which Chapter 4 proved closed.

Therefore q sends closed subsets of X to closed subsets of Y. In other words, q is a closed map. This is a conclusion under the stated assumptions, not part of the definition of every quotient map.

We can now shrink a source neighbourhood while keeping whole classes together. If a class A lies inside an open set O, define V = Y ∖ q(X ∖ O). The set V is open because q is closed, contains the quotient point represented by A, and has q⁻¹(V) ⊆ O.

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Take distinct classes A and B. They are disjoint compact subsets of the Hausdorff space X. Such compact sets have disjoint open neighbourhoods Oₐ and Oᵦ. One proof chooses pointwise Hausdorff separations, then uses finite subcovers first over B and then over A.

Use the construction in Chapter 5: Vₐ = Y ∖ q(X ∖ Oₐ) and Vᵦ = Y ∖ q(X ∖ Oᵦ). These are open neighbourhoods of the two quotient points. Their preimages lie in the disjoint sets Oₐ and Oᵦ.

If Vₐ and Vᵦ had a common point, surjectivity of q would provide a representative in both preimages, a contradiction. Therefore the neighbourhoods are disjoint and Y is Hausdorff. This completes the sufficient direction.

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Let X = [0,2] with its usual topology. For each integer n ≥ 2, identify 1/n with 1 + 1/n. Leave every other point alone. The paired sets are disjoint, so this specifies an equivalence relation with classes of size one or two.

Every class is finite and therefore closed in X. However, the related pairs (1/n, 1 + 1/n) converge to (0,1) in X×X. The points 0 and 1 were left in different singleton classes, so (0,1) does not belong to R.

Thus R is not closed even though each individual class is closed. The failure is visible only when both coordinates vary. Checking one fibre at a time misses this limiting pair.

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Let U and V be any quotient-open neighbourhoods of [0] and [1], respectively. Their source preimages are open in [0,2], contain 0 and 1, and are saturated.

For sufficiently large n, the point 1/n lies in q⁻¹(U), while 1 + 1/n lies in q⁻¹(V). Since those points represent the same quotient point, that point belongs to both U and V.

No such neighbourhoods can be disjoint. Hence the quotient is not Hausdorff. Nevertheless, all its singletons are closed by Chapter 3, so it is T₁. This direct argument establishes the distinction without relying solely on the theorem’s contrapositive.

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Modify the example by also identifying 0 with 1, leaving the other paired classes unchanged. The new relation is the diagonal together with all forward pairs, all reverse pairs and the two new limit pairs (0,1) and (1,0).

Each sequence of off-diagonal pairs now includes its only accumulation point. The forward sequence plus its limit is compact, as is the reverse sequence plus its limit. The diagonal is closed, so their finite union is closed in X×X.

The compact-Hausdorff criterion now applies: the repaired quotient is Hausdorff. Adding one class identification can restore separation because it removes a limiting inconsistency in the relation. This conclusion follows from the rechecked relation; it is not a claim that arbitrary extra identifications improve separation.

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On X = [0,1], identify only 0 and 1 and leave other points alone. Write the relation as the diagonal ΔX together with the two pairs (0,1) and (1,0).

The diagonal is closed because X is Hausdorff. Each added pair is a closed singleton in X×X, and a finite union of closed sets is closed. The source is compact Hausdorff, so the quotient is Hausdorff.

This proves separation, not the full claim that the quotient is a circle. To establish that stronger description, construct the induced map [t] ↦ (cos 2πt, sin 2πt), check continuity and bijectivity, then use the compact-to-Hausdorff homeomorphism theorem. Each conclusion has its own proof obligation.

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Is a quotient of a Hausdorff space always Hausdorff? No. Both the dense-class example and the finite-class example show how separation can fail after identification.

Does a closed relation always suffice without compactness? The theorem proved here assumes a compact Hausdorff source. Its proof uses compactness to obtain closed saturations. Do not discard that hypothesis without replacing the argument with an appropriate result.

Do sequences always detect closedness? They do in metric spaces such as the square used in the counterexample. General topological spaces may require a broader argument. The main proof uses closed sets and compactness rather than assuming a sequential characterisation everywhere.

Is a closed quotient map the same as an open quotient map? No. These describe images of different kinds of sets. The proof here obtains closedness and uses it to create saturated neighbourhoods.

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The linked topology references supply the compactness, separation and quotient framework. The proof above explains the projection and neighbourhood steps explicitly. The paired-sequence counterexample and its repair show why closed fibres alone are insufficient.

Compare dense equivalence classes for a quotient with extremely few open sets, quotient seams for saturation near a join, and compactness through the finite intersection property for another use of finite-subcover reasoning.

For transfer, identify which condition fails before choosing a theorem. In the unrepaired example the source is compact Hausdorff and every class is closed, but the full relation is not closed. Locating that exact failure is the useful learning outcome. Return to Topology for the wider subject directory.

References checked 22 October 2026

Harvard Math 131: compact Hausdorff quotient criterion, Question 1

Compactness and separation: algebraic topology lecture notes

Continue through Topology

Dense equivalence classes and quotient separation

Quotient seams and saturated sets

Compactness and the finite intersection property

Topology root and full subject directory · The Gold Standard of Education · How X Works

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