The Gold Standard of Topology | Closure of Intersections and Dense Sets

Consider the open real intervals A=(0,1) and B=(1,2). They are disjoint because neither includes the common boundary point 1. Thus A∩B is empty and its closure is also empty.

However closure(A)=[0,1] and closure(B)=[1,2]. Their closures meet at the point 1. Consequently closure(A)∩closure(B)={1}, which is not the same set as closure(A∩B)=∅.

This simple counterexample should interrupt a common algebraic habit: an operation that distributes across unions or finite sums need not distribute across intersections. We must ask for a proof of every interchange rather than relying on visual closeness.

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In a topological space X, a point x belongs to the closure of A if every open neighbourhood of x meets A. We can also characterise closure(A) as the smallest closed subset of X containing A.

Every point of A belongs to its closure, but closure can add boundary or accumulation points that are not members of A. In the usual real line, closure((0,1))=[0,1], because every neighbourhood of 0 or 1 meets the interval.

The notation closure(A) will always mean the closure relative to the stated surrounding space. If X changes from the real line to a subspace, the collection of open neighbourhoods changes and so may the closure.

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For any two subsets A and B of a topological space, A∩B⊆A and A∩B⊆B. Closure is monotone: if S⊆T, then closure(S)⊆closure(T).

Apply monotonicity twice. We obtain closure(A∩B)⊆closure(A) and closure(A∩B)⊆closure(B). Therefore closure(A∩B)⊆closure(A)∩closure(B). No metric, distance or special real-line structure is needed.

There is a second proof: closure(A)∩closure(B) is closed and contains A∩B. Since closure(A∩B) is the smallest closed set containing A∩B, it must be contained in that closed intersection.

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Compare the operations exactly. Intersecting first asks whether a point is in both original sets; taking the closure afterward adds only points approached by the shared part. Closing first separately permits the two sets to approach the same point from different sides.

For A=(0,1) and B=(1,2), the common boundary point 1 is approached by A from the left and by B from the right. There is no point belonging to A∩B anywhere, so the shared part has no points from which to approach 1.

The universal relation is inclusion, not equality. When you write a set identity in a solution, the equality symbol is a claim requiring stronger evidence than either one of the two subset directions.

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Suppose x belongs to closure(A)∩closure(B). Every neighbourhood U of x meets A and meets B. But the point of U in A may differ from the point of U in B; those two witnesses need not be one common point.

To conclude x∈closure(A∩B), we need every neighbourhood U of x to contain a point in A∩B. The two separate statements “U∩A is nonempty” and “U∩B is nonempty” cannot simply be multiplied into one joint statement.

The interval example exposes the failure at x=1. No matter how tightly we focus around 1, there are A-points and B-points nearby, but not a single point belonging to both open intervals.

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Let A=Q, the rational numbers, and let B=R∖Q, the irrational numbers. Work in the ordinary real line. Every nonempty open interval contains a rational number and an irrational number.

Therefore closure(A)=R and closure(B)=R: both A and B are dense in R. Yet A∩B=∅ because no real number can be simultaneously rational and irrational.

It follows that closure(A∩B)=∅, while closure(A)∩closure(B)=R. Both separate closures are as large as possible, but their original sets remain disjoint. Density by itself does not make intersections nonempty.

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The first counterexample uses bounded intervals and a single point of disagreement. The second uses two disjoint dense subsets and disagrees at every real point. Both are strict-inclusion examples, but the mechanisms are easy to diagnose separately.

For the dense-set example, choose any real x and any radius r>0. Inside (x−r,x+r) there is some rational q and some irrational t. Existence of q and t proves x lies in both closures, not that q=t.

In each example the step that fails is replacing two independent witnesses with one shared witness. Naming that failed inference is more useful than memorising only the final formula.

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Let U=R∖{0} and V=R∖{1}. Both are open and dense in the usual real line. Their intersection is R∖{0,1}, which is also dense. Consequently closure(U∩V)=R=closure(U)∩closure(V).

Why does this work? Whenever O is nonempty and open, O∩U remains a nonempty open set because U is open and dense. Since V is dense, that set meets V. Thus O meets U∩V, establishing density.

The proof generalises: a finite intersection of open dense subsets is dense in any topological space. Here the extra openness permits a fresh nonempty open neighbourhood to be passed to the second density condition.

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The finite argument repeatedly intersects an open dense set with another open dense set. Extending it infinitely requires a separate theorem or additional hypotheses. In complete metric spaces and more generally Baire spaces, countable intersections of open dense sets are dense.

Without a Baire-space assumption the statement can fail. In the subspace Q of the real line, enumerate its points q₁,q₂,… . Each U_n=Q∖{q_n} is open and dense in Q, because this space has no isolated points. Yet the intersection of all U_n is empty.

This is not a counterexample to the finite-intersection proof, nor to Baire’s theorem: Q with its usual metric is not complete and is not a Baire space. It teaches how easy it is to smuggle an unstated limit into a perfectly valid finite argument.

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For finite unions, closure(A∪B)=closure(A)∪closure(B). The right side is a finite union of closed sets and contains the union, while monotonicity gives the reverse inclusion. The reason the rule works differs from the failed intersection identity.

For an infinite union even the analogous union equality can fail. Let A_n={1/n} in R. Every A_n is closed, so the union of the closures is the set {1,1/2,1/3,…}; nevertheless the closure of that union also contains 0.

And even for an infinite intersection, we always have closure(∩A_i)⊆∩closure(A_i). What fails is assuming equality without further conditions. The set algebra and the topology must be checked together.

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Practice: let A=(0,1) and B=(1/2,2). Their intersection is (1/2,1). Its closure is [1/2,1]. The two separate closures are [0,1] and [1/2,2], whose intersection is also [1/2,1]. This time equality holds; verify both set operations directly.

FAQ: Does equality require disjoint sets? No; disjoint sets often provide strict counterexamples. Are rational and irrational numbers both dense? Yes, in the ordinary real topology. Does “both dense” imply “their intersection dense”? No, unless the extra hypotheses required by a suitable theorem hold.

Three frequent errors are replacing neighbourhood witnesses with one imagined shared witness, confusing closure with membership, and applying finite-intersection statements to countably many sets without checking whether the ambient space is Baire.

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A strong proof states the ambient space and topology, identifies the direction that always holds, and supplies a counterexample before adding assumptions that restore equality. In an application, distinguish separate approximation evidence from observations shared by all filters or constraints.

Read The Gold Standard of Topology, compactness and the finite intersection property, continuous bijections and their inverse, and coordinatewise convergence. Each explores a different hidden hypothesis.

See the University of Toronto notes on closure and density, the counterexample proof, and Stony Brook topology notes for dense-open intersections and Baire arguments. The transferable discipline is to validate the quantifiers before performing algebra on sets.

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