Questions: Closure, Interior, and Boundary of Sets

Question 1

Let A = [0, 1) in ℝ with the standard topology. What is the boundary ∂A?

• A. {0} — only the left endpoint, since 1 is not in A
• B. {0, 1} — both endpoints are in the closure but not the interior
• C. ∅ — A contains all its limit points
• D. (0, 1) — all interior points form the boundary

Answer: (B)

The interior A° = (0,1): points strictly inside have open neighborhoods fitting within A. The closure Ā = [0,1]: both endpoints 0 and 1 are limit points (every open interval around them hits A). The boundary ∂A = Ā \ A° = [0,1] \ (0,1) = {0,1}. Both endpoints satisfy the boundary condition: every open neighborhood of 0 and every open neighborhood of 1 intersects both A and its complement. The fact that 1 is not in A is irrelevant to whether it's a boundary point — boundary is defined by the closure-minus-interior formula.

Question 2

Suppose Ā = A for a set A in some topological space. What can you conclude about A?

• A. A is open, because the closure operation produces open sets
• B. A is closed, because A equals the smallest closed set containing itself
• C. A has empty boundary, because no points need to be added to A to close it
• D. A is both open and closed (clopen), because it satisfies a fixed-point condition

Answer: (B)

The closure Ā is defined as the smallest closed set containing A. If A = Ā, then A is equal to a closed set, so A is closed. This is the definition: a set is closed if and only if it contains all its limit points, which is equivalent to equaling its own closure. Option C is tempting but wrong — ∂A = Ā \ A° = A \ A° could still be non-empty if A is not open. Option D overclaims; Ā = A just says A is closed, not that it is also open.

Question 3

A set A is open if and mainly if A equals its own closure.

• T. True
• F. False

Answer: False

This confuses the characterization of open sets with that of closed sets. A set is closed if and only if it equals its own closure (Ā = A). A set is open if and only if it equals its own interior (A° = A). These are distinct properties: closure is about containing limit points, interior is about being 'surrounded' by the set. An open interval (0,1) satisfies A° = A but Ā = [0,1] ≠ A. A closed interval [0,1] satisfies Ā = A but A° = (0,1) ≠ A.

Question 4

The boundary of any open set is always disjoint from the set itself.

• T. True
• F. False

Answer: True

If A is open, then A = A° (an open set equals its own interior). The boundary is ∂A = Ā \ A°. Since A° = A, we have ∂A = Ā \ A. Because Ā \ A and A are always disjoint by set subtraction, ∂A ∩ A = ∅. Intuitively: boundary points have every neighborhood intersecting both A and its complement, so they cannot be interior points — but all points of an open set are interior points. A boundary point of an open set must lie outside the set.

Question 5

Explain why a set A is closed if and only if A equals its own closure, using the definition of closure as the smallest closed set containing A.

Model answer: By definition, Ā is the smallest closed set containing A, so A ⊆ Ā always. If A is closed, then A itself is a closed set containing A, and since Ā is the smallest such set, Ā ⊆ A. Combined with A ⊆ Ā, we get Ā = A. Conversely, if Ā = A, then A equals the closed set Ā, so A is closed. The equivalence follows directly from the minimality in the definition of closure.

The argument is a clean minimality argument: closure is defined as the smallest closed superset, so any closed superset of A is at least as large as Ā. If A is itself closed, it is the smallest closed set containing A, so it must equal Ā. The converse is immediate because Ā is always closed. This characterization is foundational — it makes 'closed set' and 'equals its own closure' synonymous, which is constantly used in continuity proofs.