502 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

Definition 19.1.4 A topological space is said to be Hausdorff if whenever p and qare distinct points of X, there exist disjoint open sets U,V such that p ∈U, q ∈V . In otherwords points can be separated with open sets.

Hausdorff

pU

qV

Definition 19.1.5 A subset of a topological space is said to be closed if its comple-ment is open. Let p be a point of X and let E ⊆ X. Then p is said to be a limit point of E ifevery open set containing p contains a point of E distinct from p.

Theorem 19.1.6 If (X ,τ) is a Hausdorff space and if p ∈ X, then {p} is a closedset.

Proof: If x ̸= p, there exist open sets U and V such that x ∈U, p ∈ V and U ∩V = /0.Therefore, {p}C is an open set so {p} is closed. ■

It would have been enough to assume that if x ̸= y, then there exists an open set con-taining x which does not contain y.

Proposition 19.1.7 If (X ,τ) is a Hausdorff space then a point p is a limit point of aset E if and only if every open set containing p contains infinitely many points of E eachdifferent than p.

Proof:⇐ is obvious. Consider⇒. If p is a limit point and if U is an open set containingp but there are only finitely many points of E different than p contained in U,{qi}m

i=1 , thenconsider V ≡ U ∩∩m

i=1 {qi}C which is an open set because each {qi}C is open. This isbecause if x ̸= qi there exists open Vqi containing x such that qi /∈ Vqi and so V is a finiteintersection of open sets. Therefore, there is a qm+1 ∈V \{p} , a contradiction. ■

Theorem 19.1.8 A subset E, of X is closed if and only if it contains all its limitpoints. A set is closed if and only if its complement is open and a set is open if and only ifits complement is closed.

Proof: Suppose first that E is closed and let x be a limit point of E. Is x ∈ E? If x /∈ E,then EC is an open set containing x which contains no points of E, a contradiction. Thusx ∈ E.

Now suppose E contains all its limit points. Is EC open? If x ∈ EC, then x is not a limitpoint of E because E has all its limit points and so there exists an open set, U containingx such that U contains no point of E other than x. Since x /∈ E, it follows that x ∈U ⊆ EC

which implies EC is an open set because this shows EC is the union of open sets.By definition, E closed⇒ EC is open. If EC is open, then no point of EC can be a limit

point of E and so E is closed since it contains all its limit points so EC open⇒ E closed.■

Definition 19.1.9 A topological space (X ,τ) is said to be regular if whenever Cis a closed set and p is a point not in C, there exist disjoint open sets U and V such thatp ∈U, C ⊆V .