Download Notes 3

Separation axioms
1. The axioms
The following categorization of topological spaces below is a measure of how “fine”or
how “coarse”the topology of the space is.
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 . Said differently, for any pair of points in X there is an open
set which contains one but not the other.
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 =⇒ T0
T2 =⇒ T1
T4 + T2 =⇒ T3
T5 =⇒ 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
2
2.
3.
4.
5.
but not x. Every open set of X either contains both x and y or contains neither
of the two.
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.
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.
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
follow 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
Convince yourself that the Euclidean spaces (Rn , TE ) are all Ti for i = 0, 1, 2, 3, 4, 5.