Download SEPARATION AXIOMS 1. The axioms The following categorization

SEPARATION AXIOMS
1. The axioms
The following categorization of topological spaces below is a measure of how “fine”or how “coarse”the
topology the space is. If a topological space X has any of these properties then any space Y homeomorphic to X will also carry the same properties.
(1) A topological space is said to be T0 or Kolmogorov if for any two points x, y ∈ X there
exists an open set U ⊆ X such that either x ∈ U and y ∈
/ U , or y ∈ U and x ∈
/ U.
(2) A topological space is said to be T1 or Fréchet if for any two points x, y ∈ X there exist
open sets Ux , Uy ⊆ X such that x ∈ Ux − Uy and y ∈ Uy − Ux .
(3) A topological space is said to be T2 or Hausdorff if for any two points x, y ∈ X there exist
disjoint open sets Ux , Uy ⊆ X such that x ∈ Ux and y ∈ Uy .
(4) A topological space is said to be T3 if for any closed set A ⊆ X and any point y ∈ X − A
there are disjoint open sets UA , Uy ⊆ X such that A ⊆ UA and y ∈ Uy .
(5) A topological space is said to be T4 if for any two disjoint closed sets A, B ⊆ X there are
disjoint open sets UA , UB ⊆ X with A ⊆ UA and B ⊆ UB .
(6) A topological space is said to be T5 if for any two separated sets A, B ⊆ X (see definition 1.1
below) there are disjoint open sets UA , UB ⊆ X with A ⊆ UA and B ⊆ UB .
(7) A topological space is called regular if it is both T0 and T3 .
(8) A topological space is called normal if it is both T1 and T4 .
Definition 1.1. We say that two subsets A, B ⊆ X of the topological space X are separated if
Ā ∩ B = ∅ = A ∩ B̄.
It should be pointed out that there are examples of topological spaces which satisfy property Ti
but do not have property Ti+1 for i = 0, ..., 4. On the other hand, the various Ti properties are
not completely independent. For example property T1 implies property T0 , property T2 implies T1 ,
T4 together with T2 implies T3 and T5 implies T4 . However T3 does not imply property T2 . These
dependencies are summarized in the table:
T1
T2
T4 + T2
T5
=⇒ T0
=⇒ T1
=⇒ T3
=⇒ T4
For a wealth of examples of topological spaces with their various properties described in detail, see
the excellent book “Counterexamples in Topology” by L. A. Steen and J. A. Seebach, Jr published
by Springer Verlag (1970) and Dover (1995).
2. Examples
(1) Let X = R be given the partition topology TP associated to the partition P = {[2a, 2a+2i | a ∈
Z}. This space is not a Ti space for any i = 0, ..., 5. Taking x = 0 and y = 1, there is no open
set U which contains x but not y or contains y but not x. Every open set of X either contains
both x and y or contains neither of the two.
(2) Let X = R have the finite complement topology. This space is T1 (but not T2 ). Let x, y ∈ X
be arbitrary points and set Ux = X − {y} and set Uy = X − {x}. These are clearly open sets
and x ∈ Ux − Uy and y ∈ Uy − Ux . The space is not T2 since every two non-empty open sets
intersect. The same remains true for the countable complement topology.
1
2
SEPARATION AXIOMS
(3) Consider X = R with the excluded point topology T p with p = 0. This space is T0 but fails
to be T1 since the only open set containing p is X.
(4) Let X = R be given the lower limit topology Tll . Then X is Ti for every i = 0, 1, ..., 5. To
see this is suffices to show that it is T2 and T5 (all the others follows from these two). It
is clear that X is Hausdorff: given two points x, y ∈ X with x < y, define Ux = [x, yi and
Uy = [y, y + 1i. These are open and disjoint sets containing x and y respectively.
To see that X is T5 , let A, B ⊆ X be two separated sets. Then X − B̄ is an open set. We
can therefore for each a ∈ A ⊆ X − B̄ find an xa ∈ X such that [a, xa i ⊆ X − B̄ (since the
half-open intervals are a basis for the topology). Define UA = ∪a∈A [a, xa i. In the same vein
define UB . These are open sets (since they are a union of open sets) and clearly A ⊆ UA and
B ⊆ UB . It remains to see that they are disjoint. Suppose not, then UA ∩ UB 6= ∅. Thus there
is some a ∈ A and some b ∈ B so that [a, xa i ∩ [b, xb i =
6 ∅. Suppose a < b (the case b < a is
treated analogously), then b ∈ [a, xa i (since b is the smallest element of [b, xb i) but this is a
contradiction since b ∈ B and [a, xa i ⊆ X − B̄. Thus UA and UB must be disjoint. x
(5) Convince yourself that the Euclidean spaces (Rn , TE ) are all Ti for i = 0, 1, 2, 3, 4, 5.