Topology · Lesson 05

Regular spaces

The ladder continues by replacing one of the two points with a closed set. A space is regular when a point outside a closed set can be separated from the whole of it by disjoint open sets. The useful form of the condition is a shrinking statement: inside any open set around a point there is a smaller open set whose closure still fits.

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x \in U

F \subseteq V

U \cap V = \emptyset

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Intuition

The ladder continues by replacing one of the two points with a closed set. A space is regular when a point outside a closed set can be separated from the whole of it by disjoint open sets. The useful form of the condition is a shrinking statement: inside any open set around a point there is a smaller open set whose closure still fits.

It is one thing to be given a seat away from one other person, and another to be given one away from an entire party. The second demand is what regularity makes.

sets diagram: a-point-away-from-a-closed-set

Regularity: a point and a closed set not containing it are put inside open sets that share no point. The closed set is drawn solid because it holds its own edge, and the open set round it is drawn with a dashed rim because it does not.

A point against a closed set

A space is regular when for every closed $C$ and every point there are disjoint open sets and . Regular together with is called ; the is added because regularity alone says nothing when there are few closed sets to speak of.

x \in / F

C

x\notin C

U\ni x

V\supseteq C

T_1

T_3

T_1

x\notin C\ \text{closed}\implies\exists U\ni x,\ V\supseteq C:\ U\cap V=\varnothing

What regularity amounts to

Equivalently: for every open $U$ around $x$ there is an open $V$ with $x\in V\subseteq\overline{V}\subseteq U$ . That is the shrinking form, proved below, and it is how the condition is used.
Every metric space is regular. Take $r=\tfrac13 d(x,C)$ , which is positive because the closed set misses the point, and use the ball of that radius against the union of the balls of that radius round the points of the set.
Every compact Hausdorff space is regular: a closed subset of a compact space is compact, and the previous lesson separates a point from a compact set.
An indiscrete space with two points is regular and not $T_1$ : the only closed sets are the empty set and everything, so the demand is never made. This is why the rung is usually stated with $T_1$ attached.
Hausdorff does not imply regular. Counterexamples exist on the line with a cleverly chosen topology; they are named in later courses and are not needed here.

Regularity is a shrinking condition

x\in U\ \text{open}\implies\exists V:\ x\in V\subseteq\overline{V}\subseteq U

Take an open set around the point. Its complement is closed and does not contain the point, so regularity gives two disjoint open sets: one around the point, one containing the complement. The first of them is the shrunk set. Its closure stays inside the complement of the second set, because that complement is closed and already contains the first set. And the complement of the second set is inside the original open set, since the second contains everything the original leaves out. Chaining the three containments gives the statement.

Proof steps

C=X\setminus U\ \text{closed},\quad x\notin C

The complement of the open set is closed and misses the point.

V\ni x,\ W\supseteq C,\ V\cap W=\varnothing

Regularity separates the point from that closed set.

V\subseteq X\setminus W

The two are disjoint, so the first sits inside the complement of the second.

\overline{V}\subseteq X\setminus W

That complement is closed and contains the first set, so it contains its closure.

\overline{V}\subseteq X\setminus C=U

The complement of the second set avoids the closed set, which puts the closure inside the original open set.

Applications

Practice

A Point and a Closed Set

Regularity separates a point from a closed set missing it, with disjoint open sets.

U\ni x,\ V\supseteq C,\quad U\cap V=\varnothing

Try it

What does regularity ask for?