In the indiscrete topology on X = {1, 2, 3} — where the only open sets are ∅ and X — the constant sequence (1, 1, 1, ...) converges to:
AOnly the point 1, since all terms equal 1
BAll three points 1, 2, and 3 simultaneously — the only neighborhood of any point is all of X, which every term trivially belongs to
CNo point — the sequence is eventually constant and does not truly approach a limit
DPoints 1 and 2 but not 3, since 3 is furthest from the terms
In the indiscrete topology, the only open set containing any point is X itself. The convergence definition requires that for every open neighborhood U of x, all but finitely many terms lie in U. Since the only neighborhood of every point is X, every term of every sequence trivially lies in it — so every sequence converges to every point simultaneously. This is the canonical example of why limits are not unique in general topological spaces, and why Hausdorff spaces (where any two distinct points have disjoint open neighborhoods) are needed to restore uniqueness.
Question 2 Multiple Choice
Let A = {1/n : n ∈ ℕ} = {1, 1/2, 1/3, 1/4, ...} with the standard topology on ℝ. Which correctly identifies the limit points of A?
AAll elements of A are limit points, since they belong to the set
BOnly 0 is a limit point — every neighborhood of 0 contains infinitely many points of A other than 0, while each 1/n is isolated in A
CThe set has no limit points because A is countable
DEvery real number is a limit point of A since A is an infinite set
Being in a set and being a limit point are different things. Each point 1/n is isolated in A: the open interval (1/n − ε, 1/n + ε) for small enough ε contains no other point of A (since consecutive terms 1/n and 1/(n+1) have positive separation). So no element of A is a limit point of A. The point 0 is not in A but is a limit point: any neighborhood of 0, however small, contains 1/n for all sufficiently large n — infinitely many points of A. The closure of A is A ∪ {0}.
Question 3 True / False
In any topological space, if a sequence converges to x and also to y, then x = y.
TTrue
FFalse
Answer: False
This holds in Hausdorff spaces but fails in general. In a non-Hausdorff space — such as the indiscrete topology on any set with more than one point — limits are severely non-unique: every sequence converges to every point simultaneously. The Hausdorff condition precisely ensures that distinct points have disjoint open neighborhoods, which forces any convergent sequence to eventually stay in one neighborhood and out of the other, preventing it from converging to both points at once. Limit uniqueness is not a logical necessity — it is a topological property that must be imposed.
Question 4 True / False
A set S in a topological space is closed if and only if it contains all of its limit points.
TTrue
FFalse
Answer: True
This is one of the equivalent characterizations of closed sets in a topological space, and it provides a neighborhood-based definition of closedness that works without distances. A set is closed iff its complement is open. If S contains all its limit points, any point outside S has a neighborhood disjoint from S (otherwise it would be a limit point), so the complement is open — S is closed. Conversely, if S is closed and x is a limit point of S, any neighborhood of x meets S, so x cannot be in the open complement — x must be in S. The closure of S is then precisely S together with all its limit points.
Question 5 Short Answer
Why does the definition of a limit point of a set A require that every neighborhood of x contains a point of A *other than x itself*, and what goes wrong if we drop this condition?
Think about your answer, then reveal below.
Model answer: Without the 'other than x' clause, every point of A would automatically qualify as a limit point of A — just because the neighborhood contains x itself. This would collapse the distinction between a point being isolated in A (belonging to A but having a neighborhood that misses all other points of A) and being a genuine accumulation point (having every neighborhood intersect A in other points). With the condition, isolated points — like the element 0 in A = {0} ∪ (1, 2) — are correctly excluded from the limit points, allowing the concept to capture genuine accumulation behavior.
The 'other than x' clause ensures that being a limit point is a property about the *surrounding* elements of A, not just x's own membership. It separates isolated points (which belong to A but are surrounded by a gap) from limit points (which A's elements approach from outside). This distinction is essential for computing closures, characterizing closed sets, and analyzing convergence — for example, in proving that a set is closed iff its complement is open, the argument depends critically on limit points being genuinely accumulating, not just present.