In a topological space X, a sequence (xₙ) converges to both x and y, where x ≠ y. Which conclusion must follow?
AX must be disconnected, since x and y must lie in separate components
BX is not compact, since compact spaces always have unique limits
CX is not a Hausdorff space, since the Hausdorff condition guarantees uniqueness of sequential limits
DX must be finite, since infinite spaces always separate distinct points
Uniqueness of sequential limits is a direct consequence of the Hausdorff condition. If X were Hausdorff and (xₙ) converged to both x and y with x ≠ y, take disjoint neighborhoods U of x and V of y. Convergence to x means xₙ ∈ U eventually; convergence to y means xₙ ∈ V eventually. But U ∩ V = ∅, a contradiction. So any space where a sequence converges to two distinct points cannot be Hausdorff. Compactness and connectedness are independent properties that do not alone guarantee unique limits.
Question 2 Multiple Choice
Let K be a compact subset of a Hausdorff space X, and let y be a point not in K. Which of the following is guaranteed by the Hausdorff condition combined with compactness?
AK and {y} are contained in disjoint open sets — so y has an open neighborhood entirely disjoint from K
By must be isolated, meaning {y} is an open set in X
CThe union K ∪ {y} is also compact in X
DNo sequence in K can have y as a cluster point
For each k ∈ K, the Hausdorff condition gives disjoint open sets Uₖ ∋ y and Vₖ ∋ k. The sets {Vₖ} cover K. By compactness, finitely many suffice: V_{k₁}, …, V_{kₙ} cover K. Then U = U_{k₁} ∩ … ∩ U_{kₙ} is an open neighborhood of y disjoint from V_{k₁} ∪ … ∪ V_{kₙ} ⊇ K. This proves that compact subsets of Hausdorff spaces are closed — every exterior point has a neighborhood missing K. This interplay between Hausdorff and compactness is one of the central theorems of topology.
Question 3 True / False
Every metric space is a Hausdorff space, because distinct points at distance d > 0 can always be separated by open balls of radius d/2, which are disjoint by the triangle inequality.
TTrue
FFalse
Answer: True
Given distinct points x and y with d(x,y) = d > 0, let U = B(x, d/2) and V = B(y, d/2). If any point z were in both, then d(x,y) ≤ d(x,z) + d(z,y) < d/2 + d/2 = d, contradicting d(x,y) = d. So U and V are disjoint open neighborhoods separating x and y, confirming the Hausdorff condition. This means all spaces studied in classical analysis — ℝⁿ, normed spaces, manifolds — are automatically Hausdorff.
Question 4 True / False
In any T₁ topological space (where nearly every singleton {x} is closed), sequences have unique limits.
TTrue
FFalse
Answer: False
T₁ is strictly weaker than T₂ (Hausdorff), and T₁ alone does not guarantee unique limits. Counterexample: ℝ with the cofinite topology (open sets are ∅ and sets with finite complement) is T₁, since the complement of {x} is cofinite, hence open. But every sequence of distinct points converges to every point in ℝ: any open neighborhood of any point excludes only finitely many elements, so the sequence is eventually in every open set. Unique limits require the Hausdorff condition — the ability to separate two distinct limit points by disjoint open sets.
Question 5 Short Answer
Explain, using the definition of a Hausdorff space, why a sequence in a Hausdorff space cannot converge to two distinct limits.
Think about your answer, then reveal below.
Model answer: Suppose (xₙ) in a Hausdorff space X converges to both x and y, with x ≠ y. By the Hausdorff condition, there exist disjoint open sets U and V with x ∈ U and y ∈ V. Since xₙ → x, all but finitely many terms lie in U. Since xₙ → y, all but finitely many terms lie in V. But U ∩ V = ∅, so no term can lie in both — a contradiction. Therefore x = y.
This proof uses the Hausdorff condition precisely: disjoint neighborhoods of x and y let both convergence assumptions fight each other into a contradiction. In a non-Hausdorff space, x and y cannot be separated, so their neighborhoods overlap and the sequence can 'converge' to both simultaneously. Unique limits are what make Hausdorff spaces behave like the real line and other spaces in classical analysis — this is why Hausdorff is the default assumption in most of topology and geometry.