Let A = (0, 1) be the open interval in ℝ with the standard topology. Which statement about its boundary ∂A is correct?
A∂A = ∅, because A is open and open sets have no boundary
B∂A = {0, 1}, because these points are in cl(A) but not in int(A)
C∂A = (0, 1), because every point of A is a limit point and thus a boundary point
D∂A = [0, 1], because the closure of A is the full closed interval
The interior of (0,1) is (0,1) itself (every point has an open neighborhood entirely inside A). The closure of (0,1) is [0,1] (the endpoints are limit points not in A). The boundary is cl(A) \ int(A) = [0,1] \ (0,1) = {0, 1}. The endpoints are exactly the boundary points: every neighborhood of 0 or 1 reaches into both (0,1) and its complement. Option A is the key misconception — open sets *contain none* of their boundary points, but that doesn't mean the boundary is empty.
Question 2 Multiple Choice
A set U in a topological space is open, meaning ∂U ∩ U = ∅. What does this tell us about where the boundary points of U are located?
AU has no boundary points at all
BThe boundary points of U lie entirely in the complement of U
CU is also closed, since open sets in Hausdorff spaces are closed
DU contains all of its limit points and is therefore complete
∂U ∩ U = ∅ means the boundary and U are disjoint — boundary points exist but are not in U. Since every point is either in U or not in U, and ∂U misses U entirely, the boundary points must lie in the complement X \ U. This makes sense geometrically: a boundary point of U has neighborhoods that reach into both U and its complement, so it can't be an interior point of U and therefore can't be in an open U.
Question 3 True / False
The open disk {(x,y) : x²+y² < 1} and the closed disk {(x,y) : x²+y² ≤ 1} have different boundaries.
TTrue
FFalse
Answer: False
Both have the same boundary: the unit circle {(x,y) : x²+y² = 1}. For the open disk B: int(B) = B, cl(B) = closed disk, so ∂B = cl(B) \ int(B) = unit circle. For the closed disk A: int(A) = open disk, cl(A) = A itself, so ∂A = A \ open disk = unit circle. The boundary is the same object — it separates both from their complement. What differs is whether the set *contains* its boundary: A does (it's closed), B does not (it's open).
Question 4 True / False
A point p is a boundary point of A if and only if every open neighborhood of p intersects both A and its complement.
TTrue
FFalse
Answer: True
This is the equivalent characterization ∂A = cl(A) ∩ cl(X \ A) in neighborhood language. A point is in cl(A) iff every neighborhood hits A; it is in cl(X \ A) iff every neighborhood hits the complement. A boundary point must satisfy both — it is on the 'edge' in the strongest sense, with no neighborhood small enough to land entirely on one side. This characterization is often more intuitive than the formula ∂A = cl(A) \ int(A) and is equivalent to it.
Question 5 Short Answer
Explain why the open disk and the closed disk have the same boundary. What does this reveal about the relationship between a set and its boundary?
Think about your answer, then reveal below.
Model answer: The boundary ∂A = cl(A) ∩ cl(X \ A) depends on where A meets its complement, not on whether A includes those meeting points. For both disks, the closure includes the unit circle and the closure of the complement also includes the unit circle — so the intersection is the unit circle in both cases. The difference is that the closed disk *contains* its boundary (so it is closed) while the open disk *excludes* its boundary (so it is open). The boundary itself is a neutral separator that belongs intrinsically to neither set.
The deeper insight is that a set's boundary is determined by the topology of the surrounding space and the structure of the set, not by whether the set 'claims' the boundary points. The same circle separates inside from outside regardless of which side of the line you draw. A closed set is precisely one that absorbs its boundary; an open set is precisely one that rejects it. This is why clopen sets (both open and closed) have empty boundaries — there are no transitional points at all.