The Gold Standard of Topology | Dense Equivalence Classes and Separation in a Quotient Space

Let X be the real line ℝ with its usual topology. Define x ∼ y when x − y belongs to the rational numbers ℚ. Reflexivity follows because zero is rational; symmetry follows by negation; transitivity follows by addition.

The equivalence class of x is x + ℚ = {x + r : r ∈ ℚ}. Write Y = ℝ/ℚ for the set of classes and q(x) = [x] for the canonical surjection. Here ℝ/ℚ denotes this additive quotient, not a quotient field or ordinary division.

Give Y the quotient topology: U is open in Y exactly when q⁻¹(U) is open in ℝ. The underlying set and this topology are both essential parts of the problem.

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The class [0] consists of rational numbers. The class [√2] consists of √2 plus rational numbers. They are distinct, because √2 − 0 is irrational. No rational number can lie in both classes: subtracting the two representations would make √2 rational.

Each class is countable because it is a translate of ℚ. The source ℝ is uncountable, so there cannot be only countably many classes; a countable union of countable sets would be countable.

The quotient therefore has many points. A topology with very few open sets need not have a small underlying set. Keep point equality separate from the ability of open sets to distinguish points.

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A subset D of ℝ is dense when every nonempty open interval meets D. Fix a class x + ℚ and an interval (a,b). Since the rational numbers are dense, there is a rational r in (a − x, b − x). Then x + r belongs to both the class and the interval.

Thus every equivalence class is dense in the real line. The classes remain disjoint: dense sets can be disjoint, as the rational and irrational numbers already show.

Density is an intersection property, not equality with the entire line. A class misses all points in every other class while still meeting every nonempty open interval. This distinction is the first place where a spatial picture can become misleading.

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A set A is saturated when it is a union of complete equivalence classes. Its saturation is q⁻¹(q(A)). For our relation, this is A + ℚ, the union of all rational translates of A.

Take any nonempty open interval I. Every class meets I by density, so q(I) = Y. Equivalently, I + ℚ = ℝ. Even a very short interval meets representatives of every quotient point.

This is much stronger than saying I is large in the source. It need not have large length. The identification distributes each class densely throughout ℝ, so a small source neighbourhood already projects onto the entire quotient.

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Let U be a nonempty open subset of Y. Its preimage A = q⁻¹(U) is a nonempty open saturated subset of ℝ. Because A is open and nonempty, it contains a nonempty interval I.

Saturation means A contains every equivalence class that meets I. Every class meets I, so A = ℝ. Surjectivity of q then gives U = Y. The empty set is also open, leaving exactly ∅ and Y.

This is the indiscrete topology. The argument uses openness to obtain an interval and saturation to include all its classes. Merely saying that A is dense would not suffice: many dense subsets of ℝ are proper subsets.

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A T₀ space lets two distinct points be distinguished by an open set containing one but not the other. Our quotient cannot do that, because its only nonempty open set contains every point. Since Y has distinct points, it fails T₀.

A T₁ space has closed singleton sets. Here the only closed sets are ∅ and Y. No singleton is closed, since Y has more than one point. Hence Y fails T₁ as well.

A Hausdorff space gives any two distinct points disjoint open neighbourhoods. Our points only have Y as an open neighbourhood, so Hausdorff separation also fails. The distinct classes have not merged as set elements; their neighbourhood information cannot separate them.

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The map q is continuous by the quotient-topology definition. It is also open. Every nonempty open subset O of ℝ contains an interval, so q(O) = Y; the empty set maps to the empty set. Both images are open in Y.

This supplies an open, continuous, surjective map from the Hausdorff real line to a quotient that is not even T₀. Hausdorffness is therefore not preserved by every continuous open surjection.

Contrast the earlier endpoint-gluing guide, where the quotient map from [0,1] is not open but the resulting circle is Hausdorff. The adjective open describes how a map sends open sets; it does not alone settle the separation properties of its target.

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Let f: Y → ℝ be continuous, with ℝ carrying its usual topology. If f took different values at y₁ and y₂, an open interval around f(y₁) could exclude f(y₂). Its inverse image would be a nonempty proper open subset of Y, which is impossible.

Therefore every continuous real-valued function on Y is constant. More generally the same conclusion holds for a Hausdorff target, using disjoint neighbourhoods of two hypothetical image values.

Equivalently, a continuous function h: ℝ → ℝ that is constant on every rational-difference class must be constant everywhere. For any x and z, choose rationals rₙ approaching z − x. Then h(x + rₙ) = h(x), while continuity gives h(x + rₙ) → h(z).

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Now define x ≈ y when x − y ∈ ℤ. Its classes are x + ℤ, discrete translates rather than dense sets. A short interval no longer meets every class. For instance qℤ((0,0.1)) misses the class of 1/2.

The map t ↦ (cos 2πt, sin 2πt) has precisely these integer-difference fibres. It induces the usual circle quotient. To justify the topology, the map is continuous and open: small open intervals map to open arcs, and arbitrary open sets are unions of such intervals.

Replacing a dense subgroup by ℤ changes the neighbourhood structure. The formulas look similar, but the relevant topological condition is entirely different. Recompute the inverse images rather than transferring the earlier conclusion.

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Let H = √2ℚ and identify x and y when x − y ∈ H. Is the quotient indiscrete? First check that multiplication by the nonzero constant √2 sends the dense set ℚ to a dense subgroup of ℝ.

Every class x + H is therefore dense. The same open-saturation argument shows that any nonempty quotient-open set is the whole quotient. The quotient is indiscrete.

It still has at least two points: 0 and 1 are inequivalent because 1/√2 is irrational. This final check matters. A one-point indiscrete space is Hausdorff; failure of separation needs distinct points. State both the topology and the size condition before applying the conclusion.

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Does indiscrete mean there is only one point? No. It means there are only two open sets, or one when the entire space is empty. This example has uncountably many points.

Does density make different classes equal? No. Density concerns intersections with open sets. Equality concerns which elements belong to the class.

Can the quotient be metrised by an ordinary metric? No. Every metric space is Hausdorff, while this quotient has distinct points that cannot be separated. A pseudometric that is zero everywhere would encode indiscrete neighbourhoods, but would not distinguish the points metrically.

Do all sequences converge here? Every sequence converges to every point in this indiscrete space: the only open neighbourhood to test is the whole space. This illustrates why uniqueness of limits needs separation assumptions.

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Jens Fjelstad’s topology notes provide the inverse-image definition of quotient topology and the surrounding separation framework. The rational-translation argument above is a full derivation from those definitions, with the subgroup comparison and function tests worked out explicitly.

Continue with quotient seams for saturation in a gluing construction, the continuous-bijection example for the role of inverse continuity, and coordinatewise convergence for a different topology defined by neighbourhood tests.

For a transfer check, ask which step fails for ℤ: it is the claim that every class meets every interval. Locating that exact failed step is more useful than memorising that one quotient resembles a circle and another is indiscrete. Return to Topology for the wider subject map.

References checked 22 October 2026

Jens Fjelstad: Notes on Topology, sections 2.8 and 3.3

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Quotient seams and why saturated sets matter

A continuous bijection with a discontinuous inverse

Coordinatewise convergence in infinite products

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