What is the key structural difference between a T₂ (Hausdorff) space and a T₃ (regular) space?
AT₂ requires separating any two disjoint closed sets; T₃ requires separating any two points
BT₂ separates two distinct points by disjoint open sets; T₃ separates a point from a disjoint closed set by disjoint open sets
CT₃ is weaker than T₂ — it imposes fewer separation requirements
DT₂ and T₃ are equivalent; the different labels are historical artifacts
In T₂ (Hausdorff), you separate two points: given x ≠ y, find disjoint open U ∋ x and V ∋ y. In T₃ (regular, plus T₁), you separate a point from a closed set: given a point x and a closed set C not containing x, find disjoint open U ∋ x and V ⊇ C. T₃ strictly implies T₂ — separating a point from a (possibly large) closed set is harder than separating two individual points. Option C has the implication direction backward. Option D is false — there exist T₂ spaces that are not T₃.
Question 2 Multiple Choice
Urysohn's lemma states that in a normal (T₄) space, given disjoint closed sets A and B, there exists a continuous function f: X → [0,1] with f(A) = 0 and f(B) = 1. What is most remarkable about this construction?
AIt uses the metric to measure distances between A and B, making it specific to metric spaces
BIt constructs a continuous function from purely topological data — normality alone, with no metric or explicit formula
CIt proves that A and B are homeomorphic when they are disjoint closed sets
DIt shows that every normal space is compact
Urysohn's lemma is remarkable because it produces a continuous function purely from topological hypotheses — no metric, no formula, no coordinates. The proof iteratively finds open sets U_r for each dyadic rational r ∈ [0,1] using normality at every inductive step. The function f(x) = inf{r : x ∈ U_r} turns out to be continuous as a consequence of the nesting relationships. This shows that T₄ is the exact threshold where separation axioms become strong enough to guarantee continuous real-valued functions — a purely topological analogue of metric-space constructions.
Question 3 True / False
Every metric space is normal (T₄).
TTrue
FFalse
Answer: True
Given disjoint closed sets A and B in a metric space, define U = {x : d(x,A) < d(x,B)} and V = {x : d(x,B) < d(x,A)}. These sets are open, disjoint, and contain A and B respectively. So every metric space is normal. This is why T₄ feels like the natural baseline for classical analysis — all spaces that arise in analysis and geometry are metric spaces and hence automatically normal.
Question 4 True / False
A regular (T₃) space is also normal (T₄), since both axioms concern separation of closed sets by open sets.
TTrue
FFalse
Answer: False
Regularity and normality are genuinely distinct: T₃ does not imply T₄. A classical example is the Sorgenfrey plane (ℝ with the lower limit topology, squared), which is regular but not normal. The difference is significant: T₃ separates a single point from a closed set; T₄ must separate two arbitrary disjoint closed sets. Separating two large closed sets from each other is a strictly harder requirement. The hierarchy T₁ ⊂ T₂ ⊂ T₃ ⊂ T₄ is strict at every step.
Question 5 Short Answer
Why is normality (T₄) the threshold that enables continuous functions to be constructed from purely topological data, as Urysohn's lemma demonstrates?
Think about your answer, then reveal below.
Model answer: Urysohn's proof constructs a function by induction: at each step, given two disjoint closed sets (the 'level sets' of the function-to-be), normality guarantees the existence of an open set separating them. Without normality, this inductive step fails — there is no guarantee that two disjoint closed sets can be separated by open sets at all. The function f(x) = inf{r : x ∈ U_r} is built from a nested family of open sets U_r; continuity follows from nesting relationships that normality makes possible at each step. T₃ is not sufficient because it only separates points from closed sets, not two closed sets from each other.
The hierarchy of separation axioms marks exactly which topological constructions become available at each level. T₄ is the precise threshold because both Urysohn's lemma and Tietze extension use the separation of two disjoint closed sets at the core of their proofs.