Manifolds 2 | Interior, Exterior, Boundary, Closure [dark version]
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Interior points of S require an open neighborhood U with p ∈ U and U ⊆ S.
Briefing
Topology in this lesson gets practical by naming four kinds of points relative to a chosen subset S inside a fixed topological space (X, T): interior points, exterior points, boundary points, and accumulation points. A point p is an interior point of S if there exists an open set U (from the topology T) with p ∈ U and U ⊆ S. It is an exterior point of S if there exists an open set U with p ∈ U and U ⊆ X \ S. Boundary points are the ones that sit in neither category: every open set U containing p must intersect both S and its complement X \ S. Accumulation points add a limit-style condition: for every open set U containing p, the intersection U ∩ S must be non-empty—so p cannot be isolated from S. Notably, an exterior point can never be an accumulation point because some open neighborhood around it avoids S entirely.
Once these point-level definitions are in place, the lesson translates them into set-level constructions. The interior of S (written as S°) is the collection of all points of X that are interior points of S. The exterior of S is the set of all exterior points of S. The boundary of S (written ∂S) consists of all boundary points of S. The derived set of S (written S′) collects all accumulation points of S, often described informally as a “derivative” of the set.
The final step is a concrete example showing why every one of these notions depends on the topology, not just on the underlying set. The space is the real line X = ℝ, but the topology T is nonstandard: besides ∅ and ℝ, the open sets are exactly half-bounded intervals of the form (a, ∞), where the left endpoint a is not included. With this topology, consider S = (0, 1) (so 0 and 1 are excluded). S has no interior points: there is no non-empty open set (a, ∞) that can fit entirely inside (0, 1), because any such open set extends to the right without bound. The exterior behaves differently: points in (1, ∞) have open neighborhoods contained in the complement of S, so the exterior of S is (1, ∞). Boundary points are those that are neither interior nor exterior; in this topology, that turns out to be (−∞, 1]—including 1 but excluding points like those strictly greater than 1, which are already exterior. Finally, the closure of S (written 5S) is defined as S ∪ ∂S, and in this example it coincides with the boundary-determined result, matching the same set described for the closure.
The takeaway is that interior, exterior, boundary, accumulation, and closure are not universal properties of a subset; they are relative to the chosen topology. Change which sets count as open, and the “geometry” of points and limits changes with it—sometimes in surprising ways, like making everything to the left of 1 become boundary for (0, 1) under this particular topology.
Cornell Notes
Relative to a fixed topological space (X, T) and a subset S ⊆ X, points are classified by how open sets around them interact with S. Interior points have an open neighborhood U with U ⊆ S, while exterior points have an open neighborhood U with U ⊆ X \ S. Boundary points are those where every open neighborhood intersects both S and its complement. Accumulation points are those where every open neighborhood intersects S, so exterior points can’t be accumulation points. These point notions become set notions: S° (interior), exterior(S), ∂S (boundary), S′ (derived set of accumulation points), and closure(S) = S ∪ ∂S. A nonstandard topology on ℝ makes (0,1) have empty interior and boundary (−∞,1], illustrating the dependence on the chosen topology.
How does a point p become an interior point of S in a topological space?
What distinguishes boundary points from interior and exterior points?
Why can an exterior point never be an accumulation point?
How are set-level notions like interior and boundary built from point-level definitions?
In the example topology on ℝ where open sets are (a, ∞), why does S = (0,1) have empty interior?
Under the same example topology, why is the boundary of (0,1) equal to (−∞,1]?
Review Questions
- Given p ∈ X and S ⊆ X, write the neighborhood condition that characterizes p as a boundary point of S.
- Explain the difference between an accumulation point and an interior point of S using the “for every open set U” condition.
- In the topology on ℝ with open sets (a, ∞), determine whether a point x > 1 is interior, exterior, boundary, or accumulation for S = (0,1).
Key Points
- 1
Interior points of S require an open neighborhood U with p ∈ U and U ⊆ S.
- 2
Exterior points of S require an open neighborhood U with p ∈ U and U ⊆ X \ S.
- 3
Boundary points are characterized by every open neighborhood intersecting both S and X \ S.
- 4
Accumulation points require every open neighborhood intersect S, which rules out exterior points from being accumulation points.
- 5
Set-level interior, boundary, and derived set are formed by collecting points with the corresponding point properties.
- 6
Closure is defined as closure(S) = S ∪ ∂S, so it depends on which points count as boundary under the chosen topology.
- 7
Changing the topology changes which neighborhoods are open, which can radically alter interior, boundary, and closure (as in the ℝ example with open sets (a, ∞)).