Questions: Closure, Interior, and Boundary

Question 1

Let A = {(x, y) : x² + y² ≤ 1} (the closed unit disk) in ℝ². What is the interior of A?

• A. The boundary circle {(x, y) : x² + y² = 1}
• B. The open disk {(x, y) : x² + y² < 1}
• C. A itself, because A is closed and therefore equals its own interior
• D. The empty set, because only open sets have a non-empty interior

Answer: (B)

The interior of A is the largest open set contained in A. The closed disk includes its boundary circle, but boundary points fail the interior test: no open ball around a boundary point stays entirely inside the disk. Strip those boundary points away and you get the open disk {x² + y² < 1}, which is open and fits entirely inside A. The closed set A is not its own interior — closed sets contain their boundary, but interior points require a neighborhood that stays inside the set.

Question 2

A point p has the property that every open neighborhood of p contains points in A and points not in A. What does this tell us about p?

• A. p is in the interior of A, because it is reachable from within A
• B. p is in the exterior of A, because it has access to points outside A
• C. p is on the boundary of A, because every neighborhood straddles A and its complement
• D. p is in the closure of A, but we cannot determine whether it is on the boundary without more information

Answer: (C)

The neighborhood characterization of the boundary is exactly this: a point p is a boundary point of A if and only if every open neighborhood of p intersects both A and the complement of A. Interior points have a neighborhood entirely inside A; exterior points have a neighborhood entirely outside A; boundary points are precisely those where no neighborhood can avoid touching both sides. This neighborhood definition is the one that generalizes cleanly to any topological space.

Question 3

For any set A in a topological space, every point in the interior of A is also in the closure of A.

• T. True
• F. False

Answer: True

The interior of A is a subset of A (it is the largest open set contained in A), and A is a subset of its own closure (since the closure adds limit points to A, never removes them). Therefore, interior(A) ⊆ A ⊆ closure(A). Interior points — those with an open neighborhood contained in A — certainly satisfy the closure condition: their neighborhoods intersect A (trivially, since the neighborhoods lie inside A).

Question 4

If p is in the closure of A, then p should be an element of A.

• T. True
• F. False

Answer: False

The closure of A consists of A together with all its limit points — points every neighborhood of which intersects A. A limit point of A need not be in A itself. For example, take A = (0, 1) in ℝ. The point 0 is in the closure of A (every neighborhood of 0 intersects (0,1)), but 0 ∉ A. This is precisely why 'closure' adds something: it includes the limit points on the 'edge' of A that A itself may not contain.

Question 5

Explain the duality between the interior and closure operations, and why it means you never need to define both independently.

Model answer: The interior of A equals the complement of the closure of the complement of A: int(A) = (cl(Aᶜ))ᶜ. Intuitively, a point is inside A if and only if it is not in the closure of the outside. This algebraic relationship means each operation determines the other: once you know how to compute closures (e.g., as intersections of closed sets), you can compute interiors via complement, and vice versa.

The duality reflects the symmetry between open and closed sets in topology: taking complements swaps open sets for closed sets. Because the interior is the largest open set inside A and the closure is the smallest closed set containing A, and because complements turn open sets into closed sets, the two operations are mirror images of each other. This is why most topology texts define closure axiomatically and derive interior from it.