A topologist wants to construct a continuous function f: X → [0,1] that equals 0 on one closed set F₁ and 1 on a disjoint closed set F₂. What is the minimum topological property X must have?
AHausdorff (T₂): any two distinct points can be separated by open sets
BRegularity (T₃): any point and disjoint closed set can be separated by open sets
CNormality (T₄): any two disjoint closed sets can be separated by open sets
DCompactness: every open cover has a finite subcover
Urysohn's lemma states precisely this: in a normal space, disjoint closed sets can be separated by a continuous function. Regularity only separates a point from a closed set, which is insufficient. Normality is both necessary and sufficient for Urysohn's lemma.
Question 2 Multiple Choice
In the metric-space proof that all metric spaces are normal, what are the separating open sets constructed for disjoint closed sets F₁ and F₂?
AOpen balls of radius 1 centered at each point of Fᵢ
BUᵢ = {x : d(x, Fᵢ) < d(x, Fⱼ)} for i ≠ j
CThe complements X \ Fᵢ, which are open since Fᵢ are closed
DSets constructed inductively using Zorn's lemma
The sets Uᵢ = {x : d(x, Fᵢ) < d(x, Fⱼ)} work: each contains its Fᵢ (since d(x, Fᵢ) = 0 < d(x, Fⱼ) for x ∈ Fᵢ when the sets are disjoint closed), they are open (the distance function is continuous), and they are disjoint (a point can't be closer to each set than to the other simultaneously).
Question 3 True / False
In a normal space, any continuous function defined on a closed subspace can be extended to a continuous function on the entire space.
TTrue
FFalse
Answer: True
This is the Tietze extension theorem, which follows directly from normality via Urysohn's lemma. The proof iteratively constructs extensions with geometrically shrinking error. This is one of the primary reasons normality matters in analysis.
Question 4 True / False
Most regular (T₃) space is also normal (T₄), because if you can separate a point from a closed set, you can separate two closed sets from each other.
TTrue
FFalse
Answer: False
Regularity does not imply normality. Separating a point from a closed set is easier than separating two arbitrary closed sets. There exist regular spaces that are not normal (e.g., the Sorgenfrey plane). The implication goes the other way: compact Hausdorff spaces are normal, but compact Hausdorff implies more than T₃.
Question 5 Short Answer
Why is normality the key hypothesis in Urysohn's lemma, and what does the lemma produce that makes normality so valuable for analysis?
Think about your answer, then reveal below.
Model answer: Normality guarantees that two disjoint closed sets can be surrounded by disjoint open sets, which lets you build a nested family of open sets U_r indexed by rationals in [0,1] with F₁ ⊂ U_r and cl(U_r) ⊂ U_s for r < s. Defining f(x) = inf{r : x ∈ U_r} gives a continuous function separating F₁ and F₂. The lemma produces a continuous real-valued function with prescribed values on closed sets — bridging purely topological separation to the analytic tools of continuous functions.
Without normality you cannot build the required nesting of open sets, so the construction breaks down. The value of the lemma is that it converts a topological condition (open-set separation) into an analytic object (a continuous function), which then enables the Tietze extension theorem and makes normal spaces behave well in functional analysis.