When combining sets removes, preserves or creates an exposed edge
A complete route through union and intersection boundaries, with interval, square, dense-set and infinite-family checks.
Choose the question you need to answer
Start from the definition
Prove the finite rules
See shared-edge cancellation
Test separation and limits
1. A shared edge can disappear from the outside boundary
Place two closed intervals A=[0,1] and B=[1,2] in the real line. Their boundaries are {0,1} and {1,2}. If boundary distributed over union, the boundary of A∪B would be {0,1,2}.
But A∪B=[0,2], whose boundary is only {0,2}. The shared point 1 is interior to the union, so it disappears from the outside boundary. The two component boundaries contain information that the combined set no longer exposes.
This is the one-dimensional version of two adjacent rooms sharing a wall. The common wall can be a boundary of each room while lying inside the union of the rooms.
2. Start from closure minus interior
For a subset A of a topological space X, the boundary is ∂A=cl(A)∖int(A). Equivalently, ∂A=cl(A)∩cl(X∖A). A point is on the boundary when every neighbourhood meets both A and its complement.
The ambient space matters, as shown in the relative-boundary lesson. In this article all examples state X explicitly before combining sets.
Finite unions commute with closure: cl(A∪B)=cl(A)∪cl(B). Interiors commute with finite intersections: int(A∩B)=int(A)∩int(B). The crossed operations do not generally distribute in the same way.
3. Prove the inclusion for a union
Take x∈∂(A∪B). Every neighbourhood U of x meets A∪B, so U meets A or B. Every U also meets X∖(A∪B), which is contained in both X∖A and X∖B.
If x were in neither ∂A nor ∂B, then x would be interior to A or its complement and interior to B or its complement. The requirement that every U meets the union and its complement rules out the combinations that avoid both boundaries.
Therefore ∂(A∪B)⊆∂A∪∂B. The inclusion can be strict because some points on the right become interior after the sets are joined.
4. Calculate the adjacent-interval example completely
| Set | Closure in R | Interior in R | Boundary in R |
|---|---|---|---|
| A=[0,1] | [0,1] | (0,1) | {0,1} |
| B=[1,2] | [1,2] | (1,2) | {1,2} |
| A∪B=[0,2] | [0,2] | (0,2) | {0,2} |
| A∩B={1} | {1} | ∅ | {1} |
The table prevents a visual shortcut. Point 1 has different roles in the component sets, their union and their intersection.
At point 1, every small interval meets A and its complement, and also B and its complement. But a sufficiently small interval around 1 lies entirely inside A∪B, so 1 is not on the union boundary.
5. Prove the inclusion for an intersection
Take x∈∂(A∩B). Every neighbourhood meets A∩B, so it meets both A and B. Every neighbourhood also meets X∖(A∩B)=(X∖A)∪(X∖B).
If x were on neither ∂A nor ∂B, the neighbourhood behaviour would eventually place x entirely inside or outside each set. Since every neighbourhood meets A∩B, x cannot be locally outside either one. It would then be interior to both, hence interior to A∩B, contradicting boundary membership.
Thus ∂(A∩B)⊆∂A∪∂B. Again equality need not hold: boundaries belonging to portions discarded by the intersection can vanish.
6. Use overlapping intervals to see discarded boundaries
Let A=[0,2] and B=[1,3]. Their intersection is [1,2], so ∂(A∩B)={1,2}. Yet ∂A∪∂B={0,2}∪{1,3}={0,1,2,3}.
Points 0 and 3 are component boundaries, but they lie outside the overlap. They have no reason to remain on the intersection boundary.
The inclusion keeps exactly the safe part of the intuition: a boundary of the combination must come from a boundary of at least one component. It does not say that every component boundary survives.
7. Move from points to a shared planar edge
Let A=[0,1]×[0,1] and B=[1,2]×[0,1] in R². Each square has four boundary edges. Their common vertical edge E={1}×[0,1] lies in both component boundaries.
The union is the rectangle [0,2]×[0,1]. Except for its endpoints, E lies in the interior of that rectangle and therefore disappears from ∂(A∪B). The outer perimeter remains.
The intersection is precisely E. As a subset of R², E has empty interior and is closed, so its boundary is all of E. The same geometric edge cancels for the union and survives completely for the intersection.
8. Record what survives in the square example
| Object | Interior in R² | Boundary role of shared edge E |
|---|---|---|
| Square A | (0,1)×(0,1) | E is part of ∂A |
| Square B | (1,2)×(0,1) | E is part of ∂B |
| Union rectangle | (0,2)×(0,1) | Interior points of E are not boundary |
| Intersection E | ∅ | All of E is boundary |
This example also separates topological boundary from a line drawn in a construction diagram. The ink does not decide whether a point is inside, outside or boundary; neighbourhoods in the stated ambient space do.
If the ambient space were E itself, the boundary of E as a subset of E would be empty. Always keep the ambient space attached.
9. Learn a strong condition that prevents cancellation
If cl(A) and cl(B) are disjoint, the sets are separated strongly enough that joining them cannot turn a boundary point of one into an interior point supplied by the other. Under this condition, ∂(A∪B)=∂A∪∂B.
Ordinary disjointness is not enough. The rational numbers Q and irrational numbers R∖Q are disjoint and both dense in R. Each has boundary R, while their union is R and has empty boundary.
The failure is maximal: every component boundary point cancels in the union. Separation must concern neighbourhoods or closures, not just membership.
10. Infinite families need a new check
For any family {Ai}, the boundary of the union need not be controlled by simply unioning every boundary in a useful equality. Even finite equality already fails; infinite limiting behaviour adds more ways for new closure points to appear.
For example, let An={1/n} in R. Each boundary is {1/n}. The union has boundary {1/n:n≥1}∪{0}; the limiting point 0 is not in any individual boundary.
The finite inclusion ∂(A∪B)⊆∂A∪∂B therefore must not be copied blindly to an infinite union. The missing limit point is the transfer check.
11. Questions that repair common misconceptions
Does a boundary always remain after taking a union? No. It can become interior, as the shared edge does.
Can an intersection create a boundary point that belongs to neither component boundary? For two sets, no: ∂(A∩B) is contained in ∂A∪∂B. But it can select a shared interface as the whole result.
Does disjoint mean boundaries add exactly? No. Use disjoint closures or another appropriate separation condition.
Can a picture settle the answer? Only after the ambient space, topology and set operation are specified. Compute closure and interior, then use the neighbourhood test at disputed points.
12. Continue through closure, space and combination rules
Begin with the Topology root. Compare relative boundaries, closure of intersections and dense sets, and derived sets and limit points.
Authoritative reading: Brian Conrad’s Stanford notes on interior, closure and boundary prove finite-union behaviour for closure and examine where intersections fail; Allen Hatcher’s Cornell topology notes provide the wider point-set framework.
Return to the Gold Standard of Education and the How X Works hub. The transferable rule is simple: component boundaries are candidates, not guaranteed survivors.