Questions: Separation Axioms: T₀, T₁, and T₂ (Hausdorff)

Question 1

The cofinite topology on ℝ (open sets are those with finite complement, plus ∅) — which separation axioms does it satisfy?

• A. T₀ only — only one point can be separated from the other
• B. T₁ but not T₂ — every point has a neighborhood excluding any other, but no two disjoint open sets exist
• C. T₂ (Hausdorff) — cofinite sets are large enough to separate any two points
• D. None of T₀, T₁, or T₂ — the cofinite topology is too coarse

Answer: (B)

For any two points x ≠ y in ℝ, the set ℝ \ {y} is open in the cofinite topology (its complement {y} is finite) and contains x but not y; similarly ℝ \ {x} contains y but not x. This satisfies T₁. However, T₂ requires two disjoint open sets U ∋ x and V ∋ y. In the cofinite topology on ℝ, any two nonempty open sets have cofinite complements, so their union of complements is finite — meaning their intersection is cofinite (in particular nonempty). No two nonempty open sets are ever disjoint, so T₂ fails.

Question 2

In a topological space, requiring that every singleton {x} is a closed set is equivalent to which separation axiom?

• A. T₀ — distinct points can be topologically distinguished
• B. T₁ — for any two distinct points, each has an open neighborhood not containing the other
• C. T₂ — any two distinct points have disjoint open neighborhoods
• D. Neither — closedness of singletons is unrelated to separation axioms

Answer: (B)

{x} is closed iff its complement is open. In T₁, for every y ≠ x there exists an open set Uᵧ containing y but not x. The union of all such Uᵧ over y ≠ x equals ℝ \ {x} and is open (unions of open sets are open), so {x} is closed. Conversely, if every singleton is closed, then ℝ \ {x} is open and serves as the T₁ neighborhood of every y ≠ x. The equivalence fails for T₀ (which only requires asymmetric separation) and T₂ (which requires the stronger disjoint-neighborhood condition).

Question 3

Most T₁ space is Hausdorff (T₂).

• T. True
• F. False

Answer: False

This is the central subtlety of the separation hierarchy: T₁ does not imply T₂. The cofinite topology on any infinite set provides a canonical counterexample. It satisfies T₁ (each point has an open neighborhood excluding every other point) but fails T₂ because no two nonempty open sets are disjoint — the 'separating neighborhoods' for two points always overlap. T₁ requires separating each point from the other one at a time; T₂ requires achieving separation simultaneously with disjoint sets, a strictly stronger demand.

Question 4

In any Hausdorff (T₂) space, a sequence can converge to at most one limit.

• T. True
• F. False

Answer: True

Uniqueness of limits is exactly what T₂ guarantees and what weaker axioms cannot. Suppose xₙ → x and xₙ → y with x ≠ y. By Hausdorff, there exist disjoint open sets U ∋ x and V ∋ y. Since xₙ → x, eventually all xₙ ∈ U; since xₙ → y, eventually all xₙ ∈ V. But U ∩ V = ∅, a contradiction. In non-Hausdorff spaces (like the cofinite topology on ℝ), every sequence can converge to every point simultaneously.

Question 5

Why does satisfying T₁ (each point has an open neighborhood excluding every other point) fail to guarantee that any two points can be simultaneously separated by disjoint open sets?

Model answer: T₁ only requires that for each ordered pair (x, y), there exists an open set containing x but not y. These separating sets for (x, y) and (y, x) are found independently and need not be disjoint. The cofinite topology on an infinite set illustrates this: for any x ≠ y, the set ℝ \ {y} separates x from y, and ℝ \ {x} separates y from x — T₁ holds. But any two nonempty open sets in this topology must intersect (their complements are finite, so they share cofinitely many points). Producing disjoint open neighborhoods requires a global constraint — that the two sets can be chosen simultaneously so neither contains any point of the other — which T₁ does not impose.