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Manifolds 2 | Interior, Exterior, Boundary, Closure [dark version] thumbnail

Manifolds 2 | Interior, Exterior, Boundary, Closure [dark version]

5 min read

Based on The Bright Side of Mathematics's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.

TL;DR

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?

p is an interior point of S if there exists an open set U in the topology T such that p ∈ U and U ⊆ S. The key requirement is that the neighborhood U must lie completely inside S, not just touch it.

What distinguishes boundary points from interior and exterior points?

A boundary point p is neither interior nor exterior. Equivalently: for every open set U containing p, both intersections U ∩ S and U ∩ (X \ S) are non-empty. So no matter how small an open neighborhood is, it always hits S and also hits the complement.

Why can an exterior point never be an accumulation point?

An exterior point p has some open neighborhood U with U ⊆ X \ S, meaning U ∩ S = ∅. But accumulation points require that for every open neighborhood U containing p, U ∩ S is non-empty. These conditions contradict each other.

How are set-level notions like interior and boundary built from point-level definitions?

The interior S° is the set of all points that are interior points of S. The boundary ∂S is the set of all boundary points of S. The derived set S′ is the set of all accumulation points of S. The closure is defined as closure(S) = S ∪ ∂S, combining S with all its boundary points.

In the example topology on ℝ where open sets are (a, ∞), why does S = (0,1) have empty interior?

Any nontrivial open set in this topology has the form (a, ∞), which extends indefinitely to the right. No such set can fit entirely inside (0,1), because it would always include points > 1. Therefore there is no open neighborhood U with U ⊆ S, so S° = ∅.

Under the same example topology, why is the boundary of (0,1) equal to (−∞,1]?

Points strictly greater than 1 are exterior because there are open sets (a, ∞) contained in the complement of (0,1). Points less than 1 (including those < 0) cannot be exterior because open neighborhoods in this topology always extend to the right and will intersect (0,1). Also, points at 1 are included in the boundary because every open neighborhood of 1 intersects both (0,1) and its complement. Putting these together yields ∂S = (−∞,1].

Review Questions

  1. Given p ∈ X and S ⊆ X, write the neighborhood condition that characterizes p as a boundary point of S.
  2. Explain the difference between an accumulation point and an interior point of S using the “for every open set U” condition.
  3. 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. 1

    Interior points of S require an open neighborhood U with p ∈ U and U ⊆ S.

  2. 2

    Exterior points of S require an open neighborhood U with p ∈ U and U ⊆ X \ S.

  3. 3

    Boundary points are characterized by every open neighborhood intersecting both S and X \ S.

  4. 4

    Accumulation points require every open neighborhood intersect S, which rules out exterior points from being accumulation points.

  5. 5

    Set-level interior, boundary, and derived set are formed by collecting points with the corresponding point properties.

  6. 6

    Closure is defined as closure(S) = S ∪ ∂S, so it depends on which points count as boundary under the chosen topology.

  7. 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, ∞)).

Highlights

A boundary point p is forced by topology: every open neighborhood of p must hit both S and its complement.
Accumulation points use a “for every open neighborhood” intersection condition with S, making them a limit-style notion.
In the ℝ topology where open sets are (a, ∞), the interval (0,1) has no interior points at all.
Under that same topology, the boundary of (0,1) becomes (−∞,1], showing how topology reshapes geometric intuition.
Closure is computed directly as S ∪ ∂S, so once the boundary is identified, closure follows immediately.

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