Regularity, normality and shrinking neighborhoods

Try

Two disjoint closed subsets of a metric space can have distance zero without touching: in R², take the closed ray y=0 with x≥0 and the graph y=1/x with x>0. Both are closed and their vertical separation tends to zero as x grows. Uniform positive distance between entire sets is therefore too strong a requirement for separating them topologically.

Normality asks instead for disjoint open neighborhoods containing the sets. Regularity asks to separate one point from a closed set not containing it. The neighborhoods may have widths that vary from point to point.

Predict

Predict whether Hausdorff separation of individual pairs automatically handles a point versus an infinite closed set. Pairwise choices may not combine into one neighborhood that avoids the entire set.

Predict why compactness could help combine such choices. A finite subcover would reduce infinitely many neighborhood requirements to a finite intersection. That mechanism will later prove compact Hausdorff spaces normal.

Experiment

Fix the convention used in this unit.

Property	Separation requirement, in addition to T₁
Regular	A point and a disjoint closed set have disjoint open neighborhoods
Normal	Two disjoint closed sets have disjoint open neighborhoods
Regular shrinking	x∈V⊆closure V⊆U for a suitable open V
Normal shrinking	A⊆V⊆closure V⊆U for closed A inside open U

Some texts use regular or normal without including T₁ and reserve numbered axioms for the combination. Here T₁ is always included.

Understand

The regular shrinking characterization follows by separating x from the closed complement of U. Let V contain x and W contain that complement, with V∩W empty. Since W is open and misses V, it also misses closure V. Thus closure V⊆X without W⊆U. Conversely, shrinking inside X without a closed set F gives disjoint neighborhoods V and X without closure V of x and F.

For normality, replace the point by a closed A in exactly the same argument. A closed set inside an open set can therefore be enclosed by an open set whose closure still stays inside. This closure control will be central in constructing continuous separating functions.

Every metric space is normal. For disjoint nonempty closed A and B, the distance functions d(x,A) and d(x,B) are continuous. At a point of A, the first is zero and the second is positive, because the point is outside the closed B. Define U by d(x,A)<d(x,B), and V by the reverse strict inequality. They are disjoint open sets containing A and B. No positive lower bound on the distance between the whole sets is required. Empty-set cases are immediate, and metric spaces are T₁.

A closed subspace of a normal space is normal because its relatively closed sets are ambient-closed and can be separated upstairs. An arbitrary subspace need not inherit normality. Regularity, in contrast, does pass to arbitrary subspaces by representing a relative closed set as intersection with an ambient closed set and restricting separating neighborhoods.

Apply

Construct a metric separating function f(x)=d(x,A)/(d(x,A)+d(x,B)). Its denominator is positive at each point because disjoint closed sets cannot both contain that point. Verify f=0 on A and f=1 on B. This anticipates Urysohn’s lemma for normal spaces without a metric.

Repair “Disjoint closed sets have positive distance.” The metric proof uses pointwise distances outside a closed set, not a uniform gap between two possibly unbounded sets. Explain exactly where continuity supplies open inequality sets.

Continue

Countability conditions next provide another way to combine local information. Later, regularity plus a countable basis will lead to metrization. Optional depth: prove normality implies regularity under our T₁ convention by treating the singleton point as one of the two disjoint closed sets.