Download Separation axioms

CHAPTER 1
Separation axioms
1. Degrees of separation
In chapter ?? we already encountered two measures for the size of topology T on a
set X, namely those of being separable (definition ??) and of being second countable
(definition ??). These two measures however, dont’ take the size of X as a set into
account at all.
The attributes of a topology T on a set X that we shall encounter in this section,
in a way compare the size to T to the size of X. Not quite literally so but rather
we shall ask whether T contains enough sets to separate distinct points by disjoint
neighborhoods. Likewise, we shall ask whether points can be separated from closed
sets by disjoint neighborhoods and, indeed, if disjoint closed sets themselves possess
disjoint neighborhoods. While in Euclidean space the answers to all these questions
are affirmative, (see example 2.7 below), in general topological spaces this need not be
so.
For our main definition below, we first state the following auxiliary definition.
Definition 1.1. We say that two subsets A and B of the topological space X are
separated if Ā ∩ B = ∅ = A ∩ B̄.
For example, closed and disjoint sets A, B ⊂ X are separated.
Definition 1.2. Let X be a topological space. Then
1. X is called 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. X is called 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. X is called 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 (as already stated in
definition ??).
4. X is called T3 if for any closed set A ⊂ X and any point x ∈ X − A there are
disjoint open sets UA , Ux ⊂ X such that A ⊂ UA and x ∈ Ux .
5. X is called 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. X is called T5 if for any two separated sets A, B ⊂ X (see definition 1.1 above)
there are disjoint open sets UA , UB ⊂ X with A ⊂ UA and B ⊂ UB .
7. X is called regular if it is both T0 and T3 .
8. X is called normal if it is both T1 and T4 .
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2
1. SEPARATION AXIOMS
The various properties Ti , i = 0, ..., 5 of a topological space X are not entirely
independent. To state their various dependencies, we first prove an easy lemma.
Lemma 1.3. Let X be a T1 space, then for any point x ∈ X, the set {x} is closed.
Proof. For any point y ∈ X − {x} let Uy be a neighborhood of y not containing
x. Then
X − {x} = ∪y∈X−{x} Uy
showing that X − {x} is a union of open sets and hence open.
Using this lemma and definition 1.2, the following implications are readily verified:
(1)
T1
T2
T1 + T3
T1 + T4
T5
=⇒
=⇒
=⇒
=⇒
=⇒
T0
T1
T2
T3
T4
The next theorem shows that the separation properties Ti are inherited by a subspace,
with some subtleties in the case of T4 and T5 .
Theorem 1.4. Let X be a topological space and let Y ⊂ X be a subspace of X.
(a) If X is Ti for some i ∈ {0, 1, 2, 3} then so is Y .
(b) If Y is closed and X is Ti for some i ∈ {4, 5}, then Y is also Ti .
Proof. (a) For the cases of i = 0, 1, 2 let x, y ∈ Y be two distinct points. If Ux
and Uy are the neighborhoods of x and y in X showing up in the definition of Ti for
X, then Ux ∩ Y and Uy ∩ Y are neighborhoods of x and y in Y that can be taken to
verify property Ti for Y itself.
For property T3 , let A ⊂ Y be a closed subset and let x ∈ Y − A be any point.
Then there is a closed subset A0 ⊂ X with A = A0 ∩ Y and clearly x ∈
/ A0 . Let UA0 and
0
Ux0 be disjoint neighborhoods of A and x in X, then UA = Y ∩ UA0 and Ux = Ux0 ∩ Y
are disjoint neighborhoods of A and x in Y .
(b) For property T4 , let A, B ⊂ Y be two disjoint closed subsets of Y . Since Y is
assumed to be closed in X, we see that A, B are also closed subsets of X. But then
they can be separated by neighborhoods U, V in X and are therefore separated by
U ∩ Y and V ∩ Y in Y . A similar argument works for T5 , the details are left to the
reader (exercise ??).
Remark 1.5. If Y ⊂ X is not a closed subspace, then the proof of part (b) of
theorem 1.4 fails in it present form. Namely, if A, B ⊂ Y are closed and disjoint
subsets of Y , it is certainly true that there must exist closed subsets A0 , B 0 ⊂ X with
A = A0 ∩ Y and B = B 0 ∩ Y . However, it is not in general possible to choose A0 and
B 0 so as to be disjoint. An example of this type is example 86 in [].
2. EXAMPLES
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2. Examples
Example 2.1. A space that is T3 − T5 but not T0 − T2 . 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.
To see this, let x = 0 and y = 1. The smallest open set containing x is [0, 2i which
incidentally also contains y. Thus (R, TP ) is not T0 , T1 nor T2 .
Observe that in this example the closed subsets and R agree with the open subsets.
Thus, given any closed subset A ⊂ R and a point x ∈ R − A, the open sets UA and Ux
from separation axiom T3 can be taken to be UA = A and Ux = X − A. Accordingly,
(R, TP ) is T3 .
Let A, B ⊂ R be two separated sets. Then Ā ∩ B = ∅ implies that B ⊂ R − Ā. But
X − Ā is open and thus also closed in the partition topology and thus B̄ =⊂ R − Ā.
Given this, the two open sets UA and UB from the separation axiom for T5 can be taken
to be UA = Ā and UB = R − Ā.
Example 2.2. A space that is T0 − T1 but not T2 − T5 . Let X = R have the finite
complement topology and let x, y ∈ R be any two distinct points. Then the two sets
Ux = X −{y} and Uy = X −{x} are neighborhoods of x and y respectively with y ∈
/ Ux
and x ∈
/ Uy . Thus (R, Tf c ) is T1 and T0 . It is not however T2 − T5 since no disjoint
non-empty open subsets of R even exist in this topology.
Example 2.3. A space that is T0 but not T1 − T5 . Consider the included point
topology Tp on R with p = 0. There are no disjoint non-empty open sets in this
topology and so (R, Tp ) cannot be T2 − T5 . It is also not T1 since every non-empty open
set has to contain p. It is however T0 since given any two points x, y, with x 6= p, the
set {y, p} (or simply {p} if y = p) is an open set containing y but not x.
Example 2.4. A space that is T0 and T4 −T5 but not T1 −T3 . Consider X = R with
the excluded point topology T p with p = 0. Given any two distinct points x, y ∈ R,
one of them, say x, is not equal to 0. But then the set Ux = {x} is a neighborhood of
x not containing y. Thus (R, T p ) is T0 . However, it is not T1 (and thus also not T2 )
since taking y = 0, the only open set containing y is R which of course also contains x.
The closed subsets of R in this topology are those containing p = 0. Thus, given
any closed set A ⊂ R, the only neighborhood of A is R itself. This immediately shows
that (R, T p ) cannot be T3 . It also shows that it is T4 since no pair of non-empty disjoint
closed subsets of R exist.
Finally, let A, B ⊂ R be two separated sets. Note that Ā = A∪{0} and B̄ = B∪{0}.
Then Ā ∩ B = ∅ implies that 0 ∈
/ B while A ∩ B̄ = ∅ implies that 0 ∈
/ A But then A
and B are both already open sets and so we can set UA = A and UB = B showing that
(R, T p ) is T5 .
Example 2.5. A space that is T0 −T5 . 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 since according to (1), all the other separation axioms follow from these.
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1. SEPARATION AXIOMS
To see that X is Hausdorff, pick two points x, y ∈ X with x < y. Define
Ux = [x, yi
and
Uy = [y, y + 1i
These are disjoint neighborhoods of x and y respectively showing that (R, Tll ) is T 2.
To verify separation axiom T5 , let A, B ⊆ X be two separated sets. Then X − B̄
is an open set and so we can, 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
the open set UA as UA = ∪a∈A [a, xa i and in a similar vein define also UB . These
are open sets (being a union of open sets) and contain A and B respectively. If we
had UA ∩ UB 6= ∅, then there would have to be some a ∈ A and some b ∈ B so that
[a, xa i ∩ [b, xb i 6= ∅. Suppose that 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 would yield a contradiction
since b ∈ B and [a, xa i ⊆ X − B̄. Therefore UA and UB must be disjoint, showing that
(R, Tll ) is T5 .
Example 2.6. Metric spaces are always T0 − T5 . Let (X, d) be a metric space and
consider X as a topological space by giving it the metric topology Td .
(X, Td ) is always T2 (and hence also T0 and T1 ) since, given any two distinct points
x, y ∈ X, the sets Bx (r/2) and By (r/2) are disjoint neighborhoods of x and y. Here r
is taken to be r = d(x, y).
To see that it is also T3 − T5 , it suffices to show that it is T5 (according to (1)).
Thus let A, B ⊂ X be two separated set. We claim that for every point a ∈ A there is
some ra > 0 such that Ba (ra ) ∩ B = ∅. For if not, part (c) of lemma ?? would imply
that a ∈ B̄ contradicting the assumption A ∩ B̄ = ∅. Let Ua = Ba (ra /2) and define
UA = ∪a∈A Ua . In a similar vein define also rb , Ub and UB . It is clear that UA and
UB b are open set containing A and B respectively. It remains to show that they are
disjoint. If not, we could find a point p ∈ UA ∩ UB . But there there would be points
a ∈ A and b ∈ B such that p ∈ Ua ∩ Ub showing further that
d(a, b) ≤ d(a, p) + d(p, b) < ra /2 + rb /2
For concreteness, suppose that ra ≤ rb . Then the above equality implies that d(a, b) <
rb showing that a ∈ Ub , a contradiction. Thus we arrive at UA ∩ UB = ∅ verifying
property T5 for (X, Td ).
Example 2.7. (Rn , TEu ) is T0 − T5 . Recall that the Euclidean topology TEu on
Rn is a metric topology associated to the Euclidean metric d. Thus example 2.6 shows
that (Rn , TEu ) is T0 − T5 .
3. Topological invariants
We have thus far expanded most of our efforts towards the study of properties
of topological spaces and continuous functions between them. These are necessary
prerequisites for the more involved concepts studied in later chapters and will be used
throughout the remainder of the book. Before delving into these, we pause here to
reflect on a question that lies at the heart of most efforts in topology overall.
3. TOPOLOGICAL INVARIANTS
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Question 3.1. Given a pair of topological spaces X and Y , are they homeomorphic?
This is a very difficult question in general and will never be answered in full generality. Indeed, even if one restricts the choices of X and Y to a particularly nice class
of topological spaces, namely manifolds, this question is undecidable. It follows from
the stunning work of Kurt Gödel [] that any sufficiently complicated system of logic is
riddled with so called “undecidable statements”, i.e. statements that can neither be
proved nor disproved. A consequence of Gödel’s work is that we can never fully answer
questions 3.1 above.
Nevertheless, we can try to obtain partial answers to question 3.1. The main tool
in doing so is the use of topological invariants which we now introduce and then use
to contribute, if only infinitesimally, toward unravelling question 3.1.
Definition 3.2. A topological invariant is a function O that assigns to every
topological space an “object”O(X) such that if X and Y are homeomorphic than
O(X) = O(Y ). The nature of the objects O(X) can range from being a simple object,
such as a number, to intricate objects such as a group (for instance the fundamental
group of a topological space as discussed in chapter ??).
Remark 3.3. It is important to note that the implication
X∼
=⇒
O(X) = O(Y )
=Y
for a topological invariant O, only goes one direction. It will generally not be possible to
conclude that X and Y are homeomorphic if O(X) = O(Y ) (with very few exceptions,
see chapter ?? for one such exception). Our typical use for topological invariants shall
be such that if X and Y are two topological spaces for which O(X) 6= O(Y ) then we
can conclude that X Y .
We have actually already encountered a number of topological invariants prior to
this section. We start by giving one example with full details and then state the others
with fewer specificities.
Example 3.4. Consider the set
V={
Is separable. ,
Is not separable. }
Thus V is a set of two elements, each being a phrase. We use V to define the function
O : { All topological spaces } → V
by setting
O(X) =

 Is separable.
;
If X is a separable topological space.
 Is not separable.
;
If X is not a separable topological space.
We claim that O is a topological invariant. To see this, suppose that X is a separable
space with A ⊂ X being a countable dense subset of X. Let f : X → Y be a
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1. SEPARATION AXIOMS
homeomorphism and let B = f (A) ⊂ Y . Since f is a bijection, it preserves cardinality
of sets and so B is also countable. To see that B is also dense, let V ⊂ Y be any open
set. Then U = f −1 (V ) is an open subset of X and so A ∩ U 6= ∅ according to corollary
??. But then B ∩ V is also non-empty and so, again by corollary ??, B must be dense.
Similarly, if Y is separable then the same argument shows that X is also separable.
In conclusion, given two homeomorphic spaces X and Y , either both or neither is
separable and therefore O is a topological invariant.
Given that this was our first example of a topological invariant, we took extra care
with defining O. Since definition 3.2 stipulated that O(X) be an object, we went trough
the trouble of introducing the set V and let O take values in V. In the future we will
allow ourselves to be a bit more sloppy. Namely, we will simply say that the “property
of being separable”is a topological invariant or, even shorter, that “separability”is a
topological invariant.
Here is a typical application of this very simple topological invariant.
Example 3.5. Consider the two topological spaces: the Euclidean topology (R, TEu )
and the excluded point topology (R, T p ) (from example ??). Then
O(R, TEu ) = Is separable.
O(R, T p ) = Is not separable.
The first line above follows from the fact that Q ⊂ R is a dense countable subset of R
when R is given the Euclidean topology. The second line follows from the observation
that in the excluded point topology on R, the closure of a subset A ⊂ R is found as
Ā = A ∪ {p}. Thus if A is a countable set then so is Ā and can therefore not equal to
R.
We conclude that (R, TEu ) (R, T p ).
As we progress in our study of topological spaces we shall encounter numerous,
more and more intricate topological invariants. For now however we contend ourselves
with the following result:
Theorem 3.6. Each of the properties of a topological space listed in the table below
is a topological invariant:
• Separability (definition ??)
• T0 (definition 1.2)
• Second countability (definition ??) • T1 (definition 1.2)
• First countability (definition ??). • T2 (definition 1.2)
• T3 (definition 1.2)
• T4 (definition 1.2)
• T5 (definition 1.2)
Proof. The proofs of the topological invariance of each of the nine properties
listed are very similar to each other (and the proof of separability was already given in
example 3.4). We thus only give one of them, the one for property T2 .
Thus let X be a Hausdorff space and let f : X → Y be a homeomorphism. We’d
like to show that Y too must be Hausdorff. To see that, let y1 , y2 ∈ Y be any two
distinct points and let xi = f −1 (yi ) for i = 1, 2. Then x1 and x2 must be distinct points
also since f is a bijection. Since X is Hausdorff, there are disjoint neighborhoods U1
4. AN APPLICATION
7
and U2 of x1 and x2 respectively. Set Vi = f (Ui ) for i = 1, 2. Clearly then yi ∈ Vi and
moreover, each Vi is an open set. The latter claim follows from the continuity of f −1 .
Thus, Y too must be Hausdorff and we arrive at the conclusion that if X ∼
= Y then
either both or neither of X and Y is Hausdorff.
Example 3.7. The two spaces (R, TP ) and (R, Tf c ) cannot be homeomorphic. Here
TP is the partition topology on R associated to the partition P = {[2a, 2a + 2i | a ∈ Z}
while Tf c is the finite complement topology on R. These two spaces are not homeomorphic because (R, TP ) is not T1 but (R, Tf c ) is T1 (as demonstrated in examples 2.1
and 2.2 respectively).
Example 3.8. The two spaces (R, TEu ) (the Euclidean topology on R) and (R, T p )
(the excluded point topology on R) cannot be homeomorphic because only the first of
these is T5 (as exhibited in examples 2.4 and 2.7).
4. An application
We conclude this chapter with an application of the topological invariants from
section 3 to address question 3.1 in a (very) limited capacity. We will consider the
topological spaces obtained by endowing R with one of the following nine choices of a
topology:
1. TEu - The Euclidean topology from example ??.
2. Tdis - The discrete topology from example ??.
3. Tp - The included point topology from example ??.
4. T p - The excluded point topology from example ??.
5. Tf c - The finite complement topology from example ??.
6. Tcc - The countable complement topology from example ??.
7. TF,p - The Fort topology from example ??.
8. Tll - The lower limit topology from example ??.
9. Tul - The upper limit topology from example ??.
We organize our results in the table below. Given the row containing the topology T1
and the column containing T2 , the corresponding entry in the table is either X or ×
indicating whether (R, T1 ) and (R, T2 ) are or are not homeomorphic. The table entries
are clearly symmetric about the subdiagonal and the subdiagonal itself consists of X’s.
We explain some of the table entries and leave the rest for the reader to work out.
This is an instructive exercise that we highly recommend.
To get started, consider second countability as a topological invariant. With respect
to this invariant, the nine topologies above are divided as follows:
Lemma 4.1. Of the nine choices of topologies TEu , Tdis , Tp , T p , Tf c , Tcc , TF,p , Tll
and Tul on R, only the Euclidean topology TEu is second countable.
Proof. We have already examined two of these topologies with regards to their
second countability. The Euclidean topology TEu is second countable by example ??
while the included point topology Tp is not second countable according to example ??.
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1. SEPARATION AXIOMS
TEu
TEu X
Tdis ×
Tp
×
Tp
×
Tf c
×
Tcc
×
TF,p ×
Tll
×
Tul
×
Table
Tdis Tp T p Tf c Tcc TF,p Tll Tul
× × × × × × × ×
X × × × × × × ×
× X × × × × × ×
× × X × × × × ×
× × × X × × × ×
× × × × X × × ×
× × × × × X × ×
× × × × × × X X
× × × × × × X X
1. Homeomorphism relationships.
Example ?? shows that (R, Tdis ) is not separable and so by lemma ?? it cannot be
second countable either. Let’s address the remaining six cases:
T p Every set {x} ⊂ R with x 6= p is open and cannot be written as the union of
any other non-empty open sets. Thus every basis B for (R, T p ) must at least
contain {x}, x ∈ R − {p} which is an uncountable set. Consequently, every
basis is uncountable.
Tf c Suppose a countable basis B = {U1 , U2 , U3 , ...} existed. Then we could pick an
arbitrary point x ∈ R and consider the countable non-empty set Bx = {Ui ∈
B | x ∈ Ui }. Since for every y 6= x, the set R − {y} is open, it must be a union of
elements from B showing that for every y 6= x there exists an element Ui ∈ Bx
with y ∈
/ Ui . But then ∩Ui ∈Bx Ui = {x} so that
R − {x} = R − ∩Ui ∈Bx Ui = ∪Ui ∈Bx (R − Ui )
Tcc
TF,p
Tll
Tul
so that R−{x} is a countable union of finite sets, a contradiction. Thus (R, Tf c )
cannot be second countable.
Since Tf c ⊂ Tcc and Tf c is not second countable then neither is Tcc .
The Fort topology is finer than the excluded point topology which we already
saw was not second countable. Accordingly, neither is the Fort topology.
Suppose that there were a countable basis B = {U1 , U2 , U3 , ...} for (R, Tll ).
Without loss of generality we can assume that Ui = [ai , bi i for some ai , bi ∈ R,
ai < bi (explain this!). Since R is uncountable we can find a point x ∈ R −
{a1 , a2 , a3 , ...}. It is now easy to see that the open set [a, a+1i cannot be gotten
as a union of the basis elements. Thus (R, Tll ) is not second countable.
Similar argument as in the lower limit topology Tll .
The results of the preceding lemma suffice to fill out the first row and column of
table 1. We next turn to the Hausdorff (or T2 ) property.
4. AN APPLICATION
9
Lemma 4.2. Of the nine choices of topologies TEu , Tdis , Tp , T p , Tf c , Tcc , TF,p , Tll
and Tul on R, the ones that are Hausdorff are TEu , Tdis , TF,p , Tll and Tul
Proof. Here is what we already know: Examples 2.2, 2.3 and 2.4 show that
(R, Tf c ), (R, Tp ) and (R, T p ) are not Hausdorff while examples 2.5 and 2.7 show that
(R, Tll ) and (R, TEu ) are Hausdorff. A straightforward adaption of example 2.5 shows
that (R, Tul ) is also Hausdorff. Thus we have three more cases to address:
Tdis This topology is clearly Hausdorff since given any two distinct point x, y ∈ R,
the sets {x} and {y} are disjoint neighborhoods of x and y.
Tcc As in the finite complement topology, no disjoint non-empty open sets exist.
TF,p Let x, y ∈ R be two distinct points. We can assume that x 6= p. Then {x} and
R − {x} are two disjoint neighborhoods of x and y respectively. Thus (R, TF,p )
is Hausdorff.
Lemma 4.2 allows us to populate table 1 with an additional 32 marks of ×.
Lemma 4.3. The function f : R → R given by f (t) = −t is a homeomorphism from
(R, Tll ) to (R, Tul ).
Proof. This follows immediately from part (e) of theorem ?? and the observation
that f −1 (ha, b]) = [−b, −ai and f ([c, di) = h−d, −c].
The remaining 22 entries from table 1 are left as an exercise.