The Gold Standard of Topology | Cofinite Spaces, Compactness and Non-Unique Limits

In the usual real-number line, different points have disjoint small intervals around them. A convergent sequence has just one limit, and a compact subset is closed. It is easy to believe these are consequences of the single word “topology”. In fact, separation properties and the metric setting are doing important extra work.

To expose the missing assumptions, keep the set of points but change which subsets count as open. The cofinite topology declares a nonempty set open whenever its complement contains only finitely many points. On an infinite underlying set, every nonempty open set therefore contains nearly every point, but it need not look like a short interval.

This is a mathematical counterexample, not a physical measurement device. Its value is that every step can be checked by basic set arithmetic. We will prove a compactness property and a failure of Hausdorffness in the same space, then see why limits may cease to be unique.

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Let X be the set of positive integers {1,2,3,…}. Define a collection τ of subsets of X as follows: a subset U belongs to τ if U is empty or X∖U is finite. Thus X itself is open because its complement is empty. A set such as X∖{2,5,9} is open because it omits exactly three points.

By contrast, the singleton {5} is not open in this infinite space, because its complement contains infinitely many positive integers. Nor is the set {2,4,6,…} open: it omits infinitely many odd integers. Open here means “empty or all but finitely many points”, not “has an open interval on either side”.

The complement of an open set is closed. Accordingly, every finite subset of X is closed, and X itself is closed. In fact the closed sets in this cofinite topology are exactly all finite subsets together with X. This concrete description will drive the proofs that follow.

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The topology axioms require that both ∅ and X are open, that arbitrary unions of open sets remain open, and that finite intersections of open sets remain open. We have already checked ∅ and X. To examine an arbitrary union, suppose at least one member U is nonempty and cofinite. The complement of the whole union lies inside the finite set X∖U, so that complement is finite; the union is therefore open.

For two nonempty cofinite open sets U and V, the complement of U∩V is (X∖U)∪(X∖V). A union of two finite sets is finite. So U∩V is also cofinite and open. If one input is empty, the intersection is empty and still open. Repeating the same argument handles every finite intersection.

Notice the word finite. The intersection of infinitely many cofinite open sets need not be open. For example U_n=X∖{n} is open for every positive n, but the intersection over all n is empty, which happens to be open; other infinite families may produce nonempty nonopen results. We must not silently promote the finite-intersection axiom to an arbitrary-intersection axiom.

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To make the last warning concrete, set V_n=X∖{1,2,…,n} for every n≥1. Each V_n is open because it removes only n points. Their intersection is empty, so that particular infinite intersection does not violate openness. It simply shows that repeated removal can exhaust the space.

Now instead let W_n=X∖{2,4,6,…,2n}. Each W_n omits only n even numbers and is open. But the intersection of all W_n is the set of odd positive integers. Its complement contains infinitely many even numbers, so the set of odds is not open in the cofinite topology.

The calculation matters later: compactness arguments work with open covers and finite subcovers, not with arbitrary products or intersections of open sets. A familiar formula can fail when its proof quietly assumes that an infinite operation preserves the same topological property.

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A topological space is compact if every open cover contains a finite subcover. Start with any collection of cofinite open sets whose union is all of X. Choose one nonempty member U of that cover; such a member must exist because X is nonempty and covered. The complement F=X∖U is finite.

Every point f in F belongs to some member U_f of the original cover. Select one such cover set for each of those finitely many points. Then U together with the finitely many chosen U_f covers X. That produces a finite subcover without requiring the cover to have a particular pattern.

The proof even gives a rough size bound: if the selected U misses k points, we can finish with at most k+1 original cover members, possibly fewer when one covers several missing points. There may be infinitely many cover sets in the starting family, but the finiteness of the complement of one nonempty open set is enough to end the construction.

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A space is T1 if for each pair of different points, each has an open neighbourhood omitting the other; equivalently, every singleton is closed. Our cofinite X is T1 because the complement X∖{x} of any singleton is open. Thus {x} is closed, and the open set X∖{y} contains x whenever x and y are distinct.

Hausdorffness is stronger: distinct points must have disjoint open neighbourhoods. In the infinite cofinite space, any two nonempty open sets intersect. Their complements are finite, and the union of two finite complements cannot exhaust the infinite set X; some point belongs to both open sets.

Therefore X is compact and T1, yet it is not Hausdorff. This single calculation explains why “compact implies closed” needs an appropriate separation assumption for subsets of an ambient space. Read compact sets, closedness and the Hausdorff hypothesis for the related general principle.

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Take the sequence x_n=n: 1,2,3,4,… in the cofinite topology on X. Fix any proposed limit x∈X and any open neighbourhood U of x. Because U is nonempty and open, the excluded set F=X∖U is finite.

Only finitely many sequence terms x_n=n belong to that finite excluded set F. Choose an index N beyond every excluded term. For all n≥N, x_n lies inside U. This is exactly the definition of convergence of x_n to x. Crucially, the argument works for every proposed point x.

So a sequence of ever-larger positive integers converges to 1, 2, 3, 100 and indeed every point of X under this topology. It does not do so in the usual metric topology on the integers. The different answer reflects different neighbourhood systems, not a breakdown in arithmetic.

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A constant sequence x_n=1 always converges to 1 because every neighbourhood of 1 contains every term. It does not converge to 2: the open neighbourhood X∖{1} contains 2 but excludes all terms. The T1 property prevents this particular constant sequence from acquiring another limit.

Next consider 1,2,1,2,…. This alternating sequence converges to no point. For candidate x=1, the open neighbourhood X∖{2} omits infinitely many terms. For x=2, use X∖{1}. For any other x, use X∖{1,2}. None of those proposed limits satisfies eventual neighbourhood membership.

The distinction can be stated more generally. If every point of X occurs only finitely many times in a sequence, then every finite excluded set is visited only finitely many times. Such a sequence converges to every point. A particular topology can therefore permit both nonunique limits and sequences with no limit at all.

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Let A be the even positive integers. It is infinite and proper: it leaves out all odd numbers. A is not closed because the only closed subsets of our infinite cofinite space are finite sets and the whole space. Its closure must therefore be X: no proper closed set can contain the infinite subset A.

The same reasoning shows every infinite subset of X is dense. Every nonempty cofinite open set meets an infinite subset, since removing finitely many points cannot remove all its elements. Thus the even and odd integers are disjoint subsets, yet both are dense in this unusual topology.

This is not a contradiction with the fact that we could not separate distinct points by disjoint nonempty open sets. Dense sets, closure and separation are all being evaluated against the same coarse family of open sets. Their behaviour changes together when the topology changes.

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In the subspace topology, a subset A⊆X inherits opens of the form A∩U with U open in X. If U is cofinite in X, then its intersection with A omits only finitely many points of A. Therefore A itself carries a cofinite topology when A is infinite, and a discrete topology when A is finite.

The earlier finite-subcover proof works in every infinite cofinite subspace, while finite spaces are compact by a direct finite-choice argument. So every subset A of X is compact in its subspace topology. In particular, the infinite set of even positive integers is compact as a subspace of X, even though it is not closed in X.

This is a constructive reminder that the theorem “compact subsets of a Hausdorff space are closed” cannot drop the Hausdorff hypothesis. Here our ambient X lacks Hausdorffness and has compact but nonclosed proper subsets. The source of the failure is visible in the open-set definition, not hidden behind technical notation.

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Suppose f maps our infinite cofinite X continuously into the ordinary real line with its usual Hausdorff topology. Could f assign distinct real values to two points a and b? If f(a)≠f(b), choose disjoint open real intervals around the two values.

Continuity would make their inverse images two nonempty disjoint open subsets of X, one containing a and the other containing b. But any two nonempty open sets in an infinite cofinite space intersect. This contradiction shows that every continuous real-valued function on this particular X must be constant.

That property is not a universal feature of compact spaces: continuous real functions on the ordinary closed interval [0,1] can vary freely within continuity limits. It arises here because the cofinite topology is so weak at separating points, whereas the real codomain has disjoint neighbourhoods for different values.

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Practice: replace X with the finite set {a,b,c}. Every subset now has finite complement, so the cofinite topology becomes the discrete topology. It is compact because it is finite, but unlike the infinite cofinite case it is Hausdorff. The sequence a,b,a,b,… has no limit, and a constant sequence a has only the limit a. The word “infinite” in the earlier counterexample was indispensable.

FAQ: Is a cofinite space T1? Yes, for any underlying set. Is an infinite cofinite space Hausdorff? No. Can sequences have more than one limit? Yes; x_n=n converges to every positive integer. Is every subset compact? Yes, with the inherited cofinite or finite discrete topology. Does a compact set therefore have to be closed? No, unless relevant separation hypotheses are present.

For foundational definitions and examples see the Analysis and Topology lecture notes, which construct the cofinite topology and prove its failure of Hausdorffness, and the cofinite-topology overview for the compactness and separation properties. Connect this proof to the Topology subject root, compactness through finite intersections and why sequences can miss topological closure. The lasting habit is to identify exactly which open sets are allowed before transferring a metric-space conclusion.

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