Separation Axioms - Topology, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


2 SEPARATIONAXIOMS
De?nition 2.1 A space X is a T
0
space i? it satis?es the T
0
axiom, i.e. for each x,y ? X
such that x6=y there is an open set U ? X so that U contains one of x and y but not the other.
Obviously the property T
0
is a topological property. An arbitrary product of T
0
spaces is
T
0
. Discrete spaces are T
0
but indiscrete spaces of more than one point are not T
0
.
De?nition 2.2 A space X is a T
1
space or Frechet space i? it satis?es the T
1
axiom, i.e. for
each x,y?X such that x6=y there is an open set U ? X so that x?U but y / ?U.
T
1
is obviously a topological property and is product preserving. Every T
1
space is T
0
.
Example 2.3 The set {0,1} furnished with the topology {Ø,{0},{0,1}} is called Sierpinski
space. It is T
0
but not T
1
.
Proposition 2.4 X is a T
1
space i? for each x?X, the singleton set {x} is closed.
Proof. Easy.
De?nition 2.5 A space X is a T
2
space or Hausdor? space i? it satis?es the T
2
axiom, i.e.
for each x,y ? X such that x 6= y there are open sets U,V ? X so that x ? U, y ? V and
UnV =Ø.
T
2
is a product preserving topological property. Every T
2
space is T
1
.
Example 2.6 Recall the co?nite topology on a set X de?ned in Section 1, Exercise 3. If X
is ?nite it is merely the discrete topology. In any case X is T
1
, but if X is in?nite then the
co?nite topology is not T
2
.
Proposition 2.7 Let f,g : X ? Y be maps with Y Hausdor?. Then {x? X / f(x) = g(x)}
is closed.
Proof. Let A ={x? X / f(x)6= g(x)}, and suppose x? A. Since f(x)6= g(x), there are
open sets U,V ? Y so that f(x)? U, g(x)? V and U nV =Ø. Let W = f
- 1
(U)ng
- 1
(V).
Then W is open and x?W. Moreover, W ? A. Thus A is open, so{x?X / f(x) =g(x)} is
closed.
In particular if X is T
2
and f : X ? X is a map then the ?xed-point set of f [i.e. the set
of points x for which f(x) =x] is closed.
De?nition 2.8 A space X is regular i? for each x ? X and each closed C ? X such that
x / ?C there are open sets U,V ? X so that x?U, C ? V and UnV =Ø. A regular T
1
space
is called a T
3
space.
The properties T
3
and regular are both topological and product preserving. Every T
3
space
is T
2
.
Example 2.9 The slit disc topology onR
2
is T
2
but not regular, hence not T
3
.
Take X =R
2
and let P ={(x,y)?R
2
/ x6= 0} and L ={(0,y)?R
2
}. Topologise X as
follows (c.f. Theorem 1.12):
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
Page 2


2 SEPARATIONAXIOMS
De?nition 2.1 A space X is a T
0
space i? it satis?es the T
0
axiom, i.e. for each x,y ? X
such that x6=y there is an open set U ? X so that U contains one of x and y but not the other.
Obviously the property T
0
is a topological property. An arbitrary product of T
0
spaces is
T
0
. Discrete spaces are T
0
but indiscrete spaces of more than one point are not T
0
.
De?nition 2.2 A space X is a T
1
space or Frechet space i? it satis?es the T
1
axiom, i.e. for
each x,y?X such that x6=y there is an open set U ? X so that x?U but y / ?U.
T
1
is obviously a topological property and is product preserving. Every T
1
space is T
0
.
Example 2.3 The set {0,1} furnished with the topology {Ø,{0},{0,1}} is called Sierpinski
space. It is T
0
but not T
1
.
Proposition 2.4 X is a T
1
space i? for each x?X, the singleton set {x} is closed.
Proof. Easy.
De?nition 2.5 A space X is a T
2
space or Hausdor? space i? it satis?es the T
2
axiom, i.e.
for each x,y ? X such that x 6= y there are open sets U,V ? X so that x ? U, y ? V and
UnV =Ø.
T
2
is a product preserving topological property. Every T
2
space is T
1
.
Example 2.6 Recall the co?nite topology on a set X de?ned in Section 1, Exercise 3. If X
is ?nite it is merely the discrete topology. In any case X is T
1
, but if X is in?nite then the
co?nite topology is not T
2
.
Proposition 2.7 Let f,g : X ? Y be maps with Y Hausdor?. Then {x? X / f(x) = g(x)}
is closed.
Proof. Let A ={x? X / f(x)6= g(x)}, and suppose x? A. Since f(x)6= g(x), there are
open sets U,V ? Y so that f(x)? U, g(x)? V and U nV =Ø. Let W = f
- 1
(U)ng
- 1
(V).
Then W is open and x?W. Moreover, W ? A. Thus A is open, so{x?X / f(x) =g(x)} is
closed.
In particular if X is T
2
and f : X ? X is a map then the ?xed-point set of f [i.e. the set
of points x for which f(x) =x] is closed.
De?nition 2.8 A space X is regular i? for each x ? X and each closed C ? X such that
x / ?C there are open sets U,V ? X so that x?U, C ? V and UnV =Ø. A regular T
1
space
is called a T
3
space.
The properties T
3
and regular are both topological and product preserving. Every T
3
space
is T
2
.
Example 2.9 The slit disc topology onR
2
is T
2
but not regular, hence not T
3
.
Take X =R
2
and let P ={(x,y)?R
2
/ x6= 0} and L ={(0,y)?R
2
}. Topologise X as
follows (c.f. Theorem 1.12):
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form{z}?(PnD) form a basis of neighbourhoods of z where
D is an open disc in the plane centred at z.
Clearly X is T
2
. However, note that L is a closed subset of X and that as a subspace, L
is discrete. Thus any subset of L is closed in X; in particular, L-{ (0,0)} is closed and does
not contain (0,0), although every open set containing (0,0) meets every open set containing
L-{ (0,0)}. Thus X is not regular and hence not T
3
.
Example 2.10 Every indiscrete space is vacuously regular but no such space (of more than 1
point!) is T
0
and hence also no such space is T
2
.
Theorem 2.11 A space X is regular i? for each x? X, the closed neighbourhoods of x form
a basis of neighbourhoods of x.
Proof. ?: given x ? X, and a neighbourhood N of x, there is an open set O ? X such
that x?O? N. Consider the point x and the closed set X- O, which does not contain x. By
regularity, there are open sets U and V such that x? U, X- O ? V and U nV =Ø. Thus
x ? U ? X- V ? O ? N, so X- V is a closed neighbourhood of X contained in the given
neighbourhood N of x.
?: given x?X and the closed set C ? X-{ x}, since X- C is open and contains x, there
is a closed neighbourhood N of x so that N ? X- C. Let V = X- N. Then V is open and
C ? V. Since N is a neighbourhood of x, there is an open set U such that x? U ? N. Then
UnV ? Nn(X- N) =Ø, so X is regular.
De?nition 2.12 A space X is normal i? for each pair A,B of disjoint closed subsets of X,
there is a pair U,V of disjoint open subsets of X so that A? U, B ? V and U nV =Ø. A
normal T
1
space is called a T
4
space.
The properties T
4
and normal are both topological properties but, perhaps surprisingly, are
not product preserving. Every T
4
space is clearly a T
3
space, but it should not be surprising
that normal spaces need not be regular.
Example 2.13 Sierpinski’s space is vacuously normal but is not regular since 0 and{1} cannot
be separated.
Example 2.14 The tangent-discs topology onR
2
is T
3
but not normal, hence not T
4
.
Let X, P and L be as in Example 2.9 but topologise X as follows:
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form {z}?D form a basis of neighbourhoods of z where D is
the union of two open discs in P tangent to L at z, one to the left of L and the other to
the right of L.
Clearly X is T
1
.
X is regular, hence T
3
. Indeed, suppose z ? X and N ?N(z). If z ? P, then there is an
open disc D centred at z so that z?D? N. The closed disc of half the radius also centred at
z is a closed neighbourhood of z and lies in N. If z ? L, then there is D as in the de?nition
of the neighbourhood basis of z so that {z}?D ? N. The union of the closed discs of half
the radii of those of D also tangent to L at z is a closed neighbourhood of z contained in N.
Thus in either case, closed neighbourhoods form a neighbourhood basis so by Theorem 2.11, X
is regular.
Page 3


2 SEPARATIONAXIOMS
De?nition 2.1 A space X is a T
0
space i? it satis?es the T
0
axiom, i.e. for each x,y ? X
such that x6=y there is an open set U ? X so that U contains one of x and y but not the other.
Obviously the property T
0
is a topological property. An arbitrary product of T
0
spaces is
T
0
. Discrete spaces are T
0
but indiscrete spaces of more than one point are not T
0
.
De?nition 2.2 A space X is a T
1
space or Frechet space i? it satis?es the T
1
axiom, i.e. for
each x,y?X such that x6=y there is an open set U ? X so that x?U but y / ?U.
T
1
is obviously a topological property and is product preserving. Every T
1
space is T
0
.
Example 2.3 The set {0,1} furnished with the topology {Ø,{0},{0,1}} is called Sierpinski
space. It is T
0
but not T
1
.
Proposition 2.4 X is a T
1
space i? for each x?X, the singleton set {x} is closed.
Proof. Easy.
De?nition 2.5 A space X is a T
2
space or Hausdor? space i? it satis?es the T
2
axiom, i.e.
for each x,y ? X such that x 6= y there are open sets U,V ? X so that x ? U, y ? V and
UnV =Ø.
T
2
is a product preserving topological property. Every T
2
space is T
1
.
Example 2.6 Recall the co?nite topology on a set X de?ned in Section 1, Exercise 3. If X
is ?nite it is merely the discrete topology. In any case X is T
1
, but if X is in?nite then the
co?nite topology is not T
2
.
Proposition 2.7 Let f,g : X ? Y be maps with Y Hausdor?. Then {x? X / f(x) = g(x)}
is closed.
Proof. Let A ={x? X / f(x)6= g(x)}, and suppose x? A. Since f(x)6= g(x), there are
open sets U,V ? Y so that f(x)? U, g(x)? V and U nV =Ø. Let W = f
- 1
(U)ng
- 1
(V).
Then W is open and x?W. Moreover, W ? A. Thus A is open, so{x?X / f(x) =g(x)} is
closed.
In particular if X is T
2
and f : X ? X is a map then the ?xed-point set of f [i.e. the set
of points x for which f(x) =x] is closed.
De?nition 2.8 A space X is regular i? for each x ? X and each closed C ? X such that
x / ?C there are open sets U,V ? X so that x?U, C ? V and UnV =Ø. A regular T
1
space
is called a T
3
space.
The properties T
3
and regular are both topological and product preserving. Every T
3
space
is T
2
.
Example 2.9 The slit disc topology onR
2
is T
2
but not regular, hence not T
3
.
Take X =R
2
and let P ={(x,y)?R
2
/ x6= 0} and L ={(0,y)?R
2
}. Topologise X as
follows (c.f. Theorem 1.12):
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form{z}?(PnD) form a basis of neighbourhoods of z where
D is an open disc in the plane centred at z.
Clearly X is T
2
. However, note that L is a closed subset of X and that as a subspace, L
is discrete. Thus any subset of L is closed in X; in particular, L-{ (0,0)} is closed and does
not contain (0,0), although every open set containing (0,0) meets every open set containing
L-{ (0,0)}. Thus X is not regular and hence not T
3
.
Example 2.10 Every indiscrete space is vacuously regular but no such space (of more than 1
point!) is T
0
and hence also no such space is T
2
.
Theorem 2.11 A space X is regular i? for each x? X, the closed neighbourhoods of x form
a basis of neighbourhoods of x.
Proof. ?: given x ? X, and a neighbourhood N of x, there is an open set O ? X such
that x?O? N. Consider the point x and the closed set X- O, which does not contain x. By
regularity, there are open sets U and V such that x? U, X- O ? V and U nV =Ø. Thus
x ? U ? X- V ? O ? N, so X- V is a closed neighbourhood of X contained in the given
neighbourhood N of x.
?: given x?X and the closed set C ? X-{ x}, since X- C is open and contains x, there
is a closed neighbourhood N of x so that N ? X- C. Let V = X- N. Then V is open and
C ? V. Since N is a neighbourhood of x, there is an open set U such that x? U ? N. Then
UnV ? Nn(X- N) =Ø, so X is regular.
De?nition 2.12 A space X is normal i? for each pair A,B of disjoint closed subsets of X,
there is a pair U,V of disjoint open subsets of X so that A? U, B ? V and U nV =Ø. A
normal T
1
space is called a T
4
space.
The properties T
4
and normal are both topological properties but, perhaps surprisingly, are
not product preserving. Every T
4
space is clearly a T
3
space, but it should not be surprising
that normal spaces need not be regular.
Example 2.13 Sierpinski’s space is vacuously normal but is not regular since 0 and{1} cannot
be separated.
Example 2.14 The tangent-discs topology onR
2
is T
3
but not normal, hence not T
4
.
Let X, P and L be as in Example 2.9 but topologise X as follows:
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form {z}?D form a basis of neighbourhoods of z where D is
the union of two open discs in P tangent to L at z, one to the left of L and the other to
the right of L.
Clearly X is T
1
.
X is regular, hence T
3
. Indeed, suppose z ? X and N ?N(z). If z ? P, then there is an
open disc D centred at z so that z?D? N. The closed disc of half the radius also centred at
z is a closed neighbourhood of z and lies in N. If z ? L, then there is D as in the de?nition
of the neighbourhood basis of z so that {z}?D ? N. The union of the closed discs of half
the radii of those of D also tangent to L at z is a closed neighbourhood of z contained in N.
Thus in either case, closed neighbourhoods form a neighbourhood basis so by Theorem 2.11, X
is regular.
Suppose thatX is normal. LetM ={(x,y)/x,y?Q}. As in Example 2.9, every subset of
L is a closed subset ofX. Thus to eachA? L, we can assign open setsU
A
andV
A
ofX so that
A? U
A
,L- A? V
A
andU
A
nV
A
=Ø. LetM
A
=MnU
A
. We show that the function sending
A to M
A
is injective. This will lead to a contradiction because L is uncountable whereas M is
countable, so it is impossible to ?nd an injective function from the power set of L to that of
M. Loosely speaking, if X were normal then there would need to be more open subsets of X
than there are.
To show that the function constructed in the last paragraph really is injective, let A,B? L
be such that A6= B: to show that M
A
6= M
B
. Either A- B 6=Ø or B- A6=Ø: assume the
former. We haveØ6=A- B =An(L- B)? U
A
nV
B
. SinceU
B
andV
B
are disjoint and open,
we must have U
B
nV
B
=Ø, so V
B
? X- U
B
, and hence Ø 6= U
A
nV
B
? U
A
- U
B
. Every
non-emptyopensubsetofX meetsM so, sinceU
A
- U
B
isopen, wehave: Ø6=Mn(U
A
- U
B
)? Mn(U
A
- U
B
) =M
A
- M
B
. Thus if A- B6=Ø then M
A
- M
B
6=Ø. Similarly, if B- A6=Ø
then M
B
- M
A
6=Ø, so the function is injective as promised.
Example 2.15 Let S be the real line with the right half-open interval topology of Example 1.4.
Then S is normal but S× S is not normal. The space S× S is sometimes called Sorgenfrey’s
square.
Suppose that A and B are disjoint subsets of S. Since B is closed, for each a ? A, there
is a
0
> a so that [a,a
0
)nB =Ø. Let U = ?
a?A
[a,a
0
). Then U is an open set containing A.
Similarly de?ne V =?
b?B
[b,b
0
), where for each b?B, b
0
>b is chosen so that [b,b
0
)nA =Ø.
V too is open and contains B. It is claimed that U nV = Ø, for suppose not. Then there
are a ? A, b ? B such that [a,a
0
)n[b,b
0
) 6=Ø. Thus either b ? [a,a
0
) or a ? [b,b
0
), so either
[a,a
0
)nB6=Ø or [b,b
0
)nA6=Ø, contrary to the choice of a
0
and b
0
. Hence UnV =Ø, so S is
normal.
To show that Sorgenfrey’s square is not normal, let X = S × S and set L = {(x,y) ?
X / x+y = 0} and M ={(x,y) / x,y?Q}.
Withoutasinglechangeinnotation,wecanapplytheproofofnon-normalityofthetangent-
discs topology in Example 2.14 to the present X, L and M to show that Sorgenfrey’s square is
not normal, hence not T
4
.
Although the de?nition of normality does not seem to be very di?erent from the previous
de?nitions, we have already seen in Example 2.15 that there is a rather signi?cant di?erence
between this property and the previous separation properties. There are other important dis-
tinctionstoointhatnormalityisequivalenttosomeseeminglyverydi?erentconditions, making
the property very rich and important. We now consider some such properties. The ?rst char-
acterisation is not very di?erent but the remaining ones, Urysohn’s lemma, Tietze’s extension
theorem, the cover shrinking theorem and the partition of unity theorem are all so di?erent
from the de?nition (and even from each other!!) that this is the source of their importance in
topology and also in other branches of mathematics.
Proposition 2.16 The space X is normal i? for each A,U ? X with A closed, U open and
A? U, there is an open set V such that A? V ? ¯V ? U.
Proof. Easy.
De?nition 2.17 Say that two subsets A and B of X are completely separated in X i? there
is a continuous function f : X ? [0,1] with f(A) = 0 and f(B) = 1. [0,1] may be replaced by
[a,b] with a<b. Let X be a space. By a Urysohn family in X we mean a family{U
r
/ r?D}
of open subsets of X satisfying:
Page 4


2 SEPARATIONAXIOMS
De?nition 2.1 A space X is a T
0
space i? it satis?es the T
0
axiom, i.e. for each x,y ? X
such that x6=y there is an open set U ? X so that U contains one of x and y but not the other.
Obviously the property T
0
is a topological property. An arbitrary product of T
0
spaces is
T
0
. Discrete spaces are T
0
but indiscrete spaces of more than one point are not T
0
.
De?nition 2.2 A space X is a T
1
space or Frechet space i? it satis?es the T
1
axiom, i.e. for
each x,y?X such that x6=y there is an open set U ? X so that x?U but y / ?U.
T
1
is obviously a topological property and is product preserving. Every T
1
space is T
0
.
Example 2.3 The set {0,1} furnished with the topology {Ø,{0},{0,1}} is called Sierpinski
space. It is T
0
but not T
1
.
Proposition 2.4 X is a T
1
space i? for each x?X, the singleton set {x} is closed.
Proof. Easy.
De?nition 2.5 A space X is a T
2
space or Hausdor? space i? it satis?es the T
2
axiom, i.e.
for each x,y ? X such that x 6= y there are open sets U,V ? X so that x ? U, y ? V and
UnV =Ø.
T
2
is a product preserving topological property. Every T
2
space is T
1
.
Example 2.6 Recall the co?nite topology on a set X de?ned in Section 1, Exercise 3. If X
is ?nite it is merely the discrete topology. In any case X is T
1
, but if X is in?nite then the
co?nite topology is not T
2
.
Proposition 2.7 Let f,g : X ? Y be maps with Y Hausdor?. Then {x? X / f(x) = g(x)}
is closed.
Proof. Let A ={x? X / f(x)6= g(x)}, and suppose x? A. Since f(x)6= g(x), there are
open sets U,V ? Y so that f(x)? U, g(x)? V and U nV =Ø. Let W = f
- 1
(U)ng
- 1
(V).
Then W is open and x?W. Moreover, W ? A. Thus A is open, so{x?X / f(x) =g(x)} is
closed.
In particular if X is T
2
and f : X ? X is a map then the ?xed-point set of f [i.e. the set
of points x for which f(x) =x] is closed.
De?nition 2.8 A space X is regular i? for each x ? X and each closed C ? X such that
x / ?C there are open sets U,V ? X so that x?U, C ? V and UnV =Ø. A regular T
1
space
is called a T
3
space.
The properties T
3
and regular are both topological and product preserving. Every T
3
space
is T
2
.
Example 2.9 The slit disc topology onR
2
is T
2
but not regular, hence not T
3
.
Take X =R
2
and let P ={(x,y)?R
2
/ x6= 0} and L ={(0,y)?R
2
}. Topologise X as
follows (c.f. Theorem 1.12):
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form{z}?(PnD) form a basis of neighbourhoods of z where
D is an open disc in the plane centred at z.
Clearly X is T
2
. However, note that L is a closed subset of X and that as a subspace, L
is discrete. Thus any subset of L is closed in X; in particular, L-{ (0,0)} is closed and does
not contain (0,0), although every open set containing (0,0) meets every open set containing
L-{ (0,0)}. Thus X is not regular and hence not T
3
.
Example 2.10 Every indiscrete space is vacuously regular but no such space (of more than 1
point!) is T
0
and hence also no such space is T
2
.
Theorem 2.11 A space X is regular i? for each x? X, the closed neighbourhoods of x form
a basis of neighbourhoods of x.
Proof. ?: given x ? X, and a neighbourhood N of x, there is an open set O ? X such
that x?O? N. Consider the point x and the closed set X- O, which does not contain x. By
regularity, there are open sets U and V such that x? U, X- O ? V and U nV =Ø. Thus
x ? U ? X- V ? O ? N, so X- V is a closed neighbourhood of X contained in the given
neighbourhood N of x.
?: given x?X and the closed set C ? X-{ x}, since X- C is open and contains x, there
is a closed neighbourhood N of x so that N ? X- C. Let V = X- N. Then V is open and
C ? V. Since N is a neighbourhood of x, there is an open set U such that x? U ? N. Then
UnV ? Nn(X- N) =Ø, so X is regular.
De?nition 2.12 A space X is normal i? for each pair A,B of disjoint closed subsets of X,
there is a pair U,V of disjoint open subsets of X so that A? U, B ? V and U nV =Ø. A
normal T
1
space is called a T
4
space.
The properties T
4
and normal are both topological properties but, perhaps surprisingly, are
not product preserving. Every T
4
space is clearly a T
3
space, but it should not be surprising
that normal spaces need not be regular.
Example 2.13 Sierpinski’s space is vacuously normal but is not regular since 0 and{1} cannot
be separated.
Example 2.14 The tangent-discs topology onR
2
is T
3
but not normal, hence not T
4
.
Let X, P and L be as in Example 2.9 but topologise X as follows:
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form {z}?D form a basis of neighbourhoods of z where D is
the union of two open discs in P tangent to L at z, one to the left of L and the other to
the right of L.
Clearly X is T
1
.
X is regular, hence T
3
. Indeed, suppose z ? X and N ?N(z). If z ? P, then there is an
open disc D centred at z so that z?D? N. The closed disc of half the radius also centred at
z is a closed neighbourhood of z and lies in N. If z ? L, then there is D as in the de?nition
of the neighbourhood basis of z so that {z}?D ? N. The union of the closed discs of half
the radii of those of D also tangent to L at z is a closed neighbourhood of z contained in N.
Thus in either case, closed neighbourhoods form a neighbourhood basis so by Theorem 2.11, X
is regular.
Suppose thatX is normal. LetM ={(x,y)/x,y?Q}. As in Example 2.9, every subset of
L is a closed subset ofX. Thus to eachA? L, we can assign open setsU
A
andV
A
ofX so that
A? U
A
,L- A? V
A
andU
A
nV
A
=Ø. LetM
A
=MnU
A
. We show that the function sending
A to M
A
is injective. This will lead to a contradiction because L is uncountable whereas M is
countable, so it is impossible to ?nd an injective function from the power set of L to that of
M. Loosely speaking, if X were normal then there would need to be more open subsets of X
than there are.
To show that the function constructed in the last paragraph really is injective, let A,B? L
be such that A6= B: to show that M
A
6= M
B
. Either A- B 6=Ø or B- A6=Ø: assume the
former. We haveØ6=A- B =An(L- B)? U
A
nV
B
. SinceU
B
andV
B
are disjoint and open,
we must have U
B
nV
B
=Ø, so V
B
? X- U
B
, and hence Ø 6= U
A
nV
B
? U
A
- U
B
. Every
non-emptyopensubsetofX meetsM so, sinceU
A
- U
B
isopen, wehave: Ø6=Mn(U
A
- U
B
)? Mn(U
A
- U
B
) =M
A
- M
B
. Thus if A- B6=Ø then M
A
- M
B
6=Ø. Similarly, if B- A6=Ø
then M
B
- M
A
6=Ø, so the function is injective as promised.
Example 2.15 Let S be the real line with the right half-open interval topology of Example 1.4.
Then S is normal but S× S is not normal. The space S× S is sometimes called Sorgenfrey’s
square.
Suppose that A and B are disjoint subsets of S. Since B is closed, for each a ? A, there
is a
0
> a so that [a,a
0
)nB =Ø. Let U = ?
a?A
[a,a
0
). Then U is an open set containing A.
Similarly de?ne V =?
b?B
[b,b
0
), where for each b?B, b
0
>b is chosen so that [b,b
0
)nA =Ø.
V too is open and contains B. It is claimed that U nV = Ø, for suppose not. Then there
are a ? A, b ? B such that [a,a
0
)n[b,b
0
) 6=Ø. Thus either b ? [a,a
0
) or a ? [b,b
0
), so either
[a,a
0
)nB6=Ø or [b,b
0
)nA6=Ø, contrary to the choice of a
0
and b
0
. Hence UnV =Ø, so S is
normal.
To show that Sorgenfrey’s square is not normal, let X = S × S and set L = {(x,y) ?
X / x+y = 0} and M ={(x,y) / x,y?Q}.
Withoutasinglechangeinnotation,wecanapplytheproofofnon-normalityofthetangent-
discs topology in Example 2.14 to the present X, L and M to show that Sorgenfrey’s square is
not normal, hence not T
4
.
Although the de?nition of normality does not seem to be very di?erent from the previous
de?nitions, we have already seen in Example 2.15 that there is a rather signi?cant di?erence
between this property and the previous separation properties. There are other important dis-
tinctionstoointhatnormalityisequivalenttosomeseeminglyverydi?erentconditions, making
the property very rich and important. We now consider some such properties. The ?rst char-
acterisation is not very di?erent but the remaining ones, Urysohn’s lemma, Tietze’s extension
theorem, the cover shrinking theorem and the partition of unity theorem are all so di?erent
from the de?nition (and even from each other!!) that this is the source of their importance in
topology and also in other branches of mathematics.
Proposition 2.16 The space X is normal i? for each A,U ? X with A closed, U open and
A? U, there is an open set V such that A? V ? ¯V ? U.
Proof. Easy.
De?nition 2.17 Say that two subsets A and B of X are completely separated in X i? there
is a continuous function f : X ? [0,1] with f(A) = 0 and f(B) = 1. [0,1] may be replaced by
[a,b] with a<b. Let X be a space. By a Urysohn family in X we mean a family{U
r
/ r?D}
of open subsets of X satisfying:
(i)
¯ D =R;
(ii) ?
r?D
U
r
=X and n
r?D
U
r
=Ø;
(iii) if r <s then U
r
? U
s
.
Lemma 2.18 LetU ={U
r
/ r?D} be a Urysohn family in the space X and de?ne a function
?
U
:X ?R by ?
U
(x) = inf{r?D / x?U
r
}. Then ?
U
is continuous.
Proof. Notethat?
U
iswell-de?nedbecauseby(ii),forany x,{r?D/x?U
r
}isnon-empty
and bounded below.
Let ? = ?
U
and let (a,b)? R. By Theorem 1.19(iii) it is enough to show that ?
- 1
((a,b))
is open. Let x ? ?
- 1
((a,b)). Then a < ?(x) < b so by (i), there are r,s,t ? D so that
a < r < t < ?(x) < s < b. Then x / ? U
t
so by (iii), x / ? U
r
; also x? U
s
. Thus U
s
- U
r
is an
open neighbourhood of x: note that U
s
- U
r
? ?
- 1
(a,b), so ? is continuous.
Lemma 2.19 Let A and B be subsets of a space X. Then A and B are completely separated
in X i? there is a Urysohn family {U
r
/ r?D} with A? U
0
and B? X- U
1
.
Proof. ?: Let f : X ? [0,1] be continuous with f(A) = 0 and f(B) = 1. For each r?R
let U
r
={x?X / f(x)<
r+1
2
}. Then{U
r
/ r?R} is a Urysohn family of the desired form.
?: Let U = {U
r
/ r ? D} be a Urysohn family with A ? U
0
and B ? X- U
1
, and let
? =?
U
:X ?R be as in Lemma 2.18. De?ne f :X ? [0,1] by
f(x) =
?
?
?
0 : if ?(x)= 0
?(x) : if ?(x)? [0,1]
1 : if ?(x)= 1.
Bytheorem1.22f iscontinuous. Furthermore, ifx?Athenforeachr > 0,x?U
r
; so?(x)= 0
and hence f(x) = 0. On the other hand, if x?B then for each r < 1, x / ?U
r
; so ?(x)= 1 and
hence f(x) = 1.
Theorem 2.20 (Urysohn’s lemma) A space X is normal i? any two disjoint closed subsets
are completely separated in X.
Proof. ?: easy.
?: LetAandB betwodisjointclosedsubsetsofX. We?ndaUrysohnfamilyasinLemma
2.19.
For each non-negative integer n, let
D
n
={0,1}?
n
m
2
n
/ m is a positive integer and m= 2n
o
,
and set D = (-8 ,0)?(1,8)?[?
n= 0
D
n
]. Then
¯ D =R.
For r < 0 set U
r
=Ø and for r >1 set U
r
=X. We de?ne U
r
for r?Dn[0,1] inductively
as follows: induction is on n with U
r
de?ned for r?D
n
. By Proposition 2.16, there is an open
set U
0
such that A? U
0
? U
0
? X- B and an open set U
1
such that U
0
? U
1
? U
1
? X- B.
Now suppose n is such that U
r
has been de?ned for r ? D
n
so that if r,s ? D
n
are such
that r < s, then U
r
? U
s
. Let t ? D
n+1
- D
n
, and let r = max{p ? D
n
/ p < t} and
s = min{p?D
n
/ p>t}. Then we have U
r
? U
s
so by Proposition 2.16, there is an open set
U
t
such that U
r
? U
t
? U
t
? U
s
. Thus the induction continues and hence we obtain a Urysohn
family as needed by Lemma 2.19.
Page 5


2 SEPARATIONAXIOMS
De?nition 2.1 A space X is a T
0
space i? it satis?es the T
0
axiom, i.e. for each x,y ? X
such that x6=y there is an open set U ? X so that U contains one of x and y but not the other.
Obviously the property T
0
is a topological property. An arbitrary product of T
0
spaces is
T
0
. Discrete spaces are T
0
but indiscrete spaces of more than one point are not T
0
.
De?nition 2.2 A space X is a T
1
space or Frechet space i? it satis?es the T
1
axiom, i.e. for
each x,y?X such that x6=y there is an open set U ? X so that x?U but y / ?U.
T
1
is obviously a topological property and is product preserving. Every T
1
space is T
0
.
Example 2.3 The set {0,1} furnished with the topology {Ø,{0},{0,1}} is called Sierpinski
space. It is T
0
but not T
1
.
Proposition 2.4 X is a T
1
space i? for each x?X, the singleton set {x} is closed.
Proof. Easy.
De?nition 2.5 A space X is a T
2
space or Hausdor? space i? it satis?es the T
2
axiom, i.e.
for each x,y ? X such that x 6= y there are open sets U,V ? X so that x ? U, y ? V and
UnV =Ø.
T
2
is a product preserving topological property. Every T
2
space is T
1
.
Example 2.6 Recall the co?nite topology on a set X de?ned in Section 1, Exercise 3. If X
is ?nite it is merely the discrete topology. In any case X is T
1
, but if X is in?nite then the
co?nite topology is not T
2
.
Proposition 2.7 Let f,g : X ? Y be maps with Y Hausdor?. Then {x? X / f(x) = g(x)}
is closed.
Proof. Let A ={x? X / f(x)6= g(x)}, and suppose x? A. Since f(x)6= g(x), there are
open sets U,V ? Y so that f(x)? U, g(x)? V and U nV =Ø. Let W = f
- 1
(U)ng
- 1
(V).
Then W is open and x?W. Moreover, W ? A. Thus A is open, so{x?X / f(x) =g(x)} is
closed.
In particular if X is T
2
and f : X ? X is a map then the ?xed-point set of f [i.e. the set
of points x for which f(x) =x] is closed.
De?nition 2.8 A space X is regular i? for each x ? X and each closed C ? X such that
x / ?C there are open sets U,V ? X so that x?U, C ? V and UnV =Ø. A regular T
1
space
is called a T
3
space.
The properties T
3
and regular are both topological and product preserving. Every T
3
space
is T
2
.
Example 2.9 The slit disc topology onR
2
is T
2
but not regular, hence not T
3
.
Take X =R
2
and let P ={(x,y)?R
2
/ x6= 0} and L ={(0,y)?R
2
}. Topologise X as
follows (c.f. Theorem 1.12):
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form{z}?(PnD) form a basis of neighbourhoods of z where
D is an open disc in the plane centred at z.
Clearly X is T
2
. However, note that L is a closed subset of X and that as a subspace, L
is discrete. Thus any subset of L is closed in X; in particular, L-{ (0,0)} is closed and does
not contain (0,0), although every open set containing (0,0) meets every open set containing
L-{ (0,0)}. Thus X is not regular and hence not T
3
.
Example 2.10 Every indiscrete space is vacuously regular but no such space (of more than 1
point!) is T
0
and hence also no such space is T
2
.
Theorem 2.11 A space X is regular i? for each x? X, the closed neighbourhoods of x form
a basis of neighbourhoods of x.
Proof. ?: given x ? X, and a neighbourhood N of x, there is an open set O ? X such
that x?O? N. Consider the point x and the closed set X- O, which does not contain x. By
regularity, there are open sets U and V such that x? U, X- O ? V and U nV =Ø. Thus
x ? U ? X- V ? O ? N, so X- V is a closed neighbourhood of X contained in the given
neighbourhood N of x.
?: given x?X and the closed set C ? X-{ x}, since X- C is open and contains x, there
is a closed neighbourhood N of x so that N ? X- C. Let V = X- N. Then V is open and
C ? V. Since N is a neighbourhood of x, there is an open set U such that x? U ? N. Then
UnV ? Nn(X- N) =Ø, so X is regular.
De?nition 2.12 A space X is normal i? for each pair A,B of disjoint closed subsets of X,
there is a pair U,V of disjoint open subsets of X so that A? U, B ? V and U nV =Ø. A
normal T
1
space is called a T
4
space.
The properties T
4
and normal are both topological properties but, perhaps surprisingly, are
not product preserving. Every T
4
space is clearly a T
3
space, but it should not be surprising
that normal spaces need not be regular.
Example 2.13 Sierpinski’s space is vacuously normal but is not regular since 0 and{1} cannot
be separated.
Example 2.14 The tangent-discs topology onR
2
is T
3
but not normal, hence not T
4
.
Let X, P and L be as in Example 2.9 but topologise X as follows:
• if z?P, let the open discs inR
2
centred at z form a basis of neighbourhoods of z;
• if z?L let the sets of the form {z}?D form a basis of neighbourhoods of z where D is
the union of two open discs in P tangent to L at z, one to the left of L and the other to
the right of L.
Clearly X is T
1
.
X is regular, hence T
3
. Indeed, suppose z ? X and N ?N(z). If z ? P, then there is an
open disc D centred at z so that z?D? N. The closed disc of half the radius also centred at
z is a closed neighbourhood of z and lies in N. If z ? L, then there is D as in the de?nition
of the neighbourhood basis of z so that {z}?D ? N. The union of the closed discs of half
the radii of those of D also tangent to L at z is a closed neighbourhood of z contained in N.
Thus in either case, closed neighbourhoods form a neighbourhood basis so by Theorem 2.11, X
is regular.
Suppose thatX is normal. LetM ={(x,y)/x,y?Q}. As in Example 2.9, every subset of
L is a closed subset ofX. Thus to eachA? L, we can assign open setsU
A
andV
A
ofX so that
A? U
A
,L- A? V
A
andU
A
nV
A
=Ø. LetM
A
=MnU
A
. We show that the function sending
A to M
A
is injective. This will lead to a contradiction because L is uncountable whereas M is
countable, so it is impossible to ?nd an injective function from the power set of L to that of
M. Loosely speaking, if X were normal then there would need to be more open subsets of X
than there are.
To show that the function constructed in the last paragraph really is injective, let A,B? L
be such that A6= B: to show that M
A
6= M
B
. Either A- B 6=Ø or B- A6=Ø: assume the
former. We haveØ6=A- B =An(L- B)? U
A
nV
B
. SinceU
B
andV
B
are disjoint and open,
we must have U
B
nV
B
=Ø, so V
B
? X- U
B
, and hence Ø 6= U
A
nV
B
? U
A
- U
B
. Every
non-emptyopensubsetofX meetsM so, sinceU
A
- U
B
isopen, wehave: Ø6=Mn(U
A
- U
B
)? Mn(U
A
- U
B
) =M
A
- M
B
. Thus if A- B6=Ø then M
A
- M
B
6=Ø. Similarly, if B- A6=Ø
then M
B
- M
A
6=Ø, so the function is injective as promised.
Example 2.15 Let S be the real line with the right half-open interval topology of Example 1.4.
Then S is normal but S× S is not normal. The space S× S is sometimes called Sorgenfrey’s
square.
Suppose that A and B are disjoint subsets of S. Since B is closed, for each a ? A, there
is a
0
> a so that [a,a
0
)nB =Ø. Let U = ?
a?A
[a,a
0
). Then U is an open set containing A.
Similarly de?ne V =?
b?B
[b,b
0
), where for each b?B, b
0
>b is chosen so that [b,b
0
)nA =Ø.
V too is open and contains B. It is claimed that U nV = Ø, for suppose not. Then there
are a ? A, b ? B such that [a,a
0
)n[b,b
0
) 6=Ø. Thus either b ? [a,a
0
) or a ? [b,b
0
), so either
[a,a
0
)nB6=Ø or [b,b
0
)nA6=Ø, contrary to the choice of a
0
and b
0
. Hence UnV =Ø, so S is
normal.
To show that Sorgenfrey’s square is not normal, let X = S × S and set L = {(x,y) ?
X / x+y = 0} and M ={(x,y) / x,y?Q}.
Withoutasinglechangeinnotation,wecanapplytheproofofnon-normalityofthetangent-
discs topology in Example 2.14 to the present X, L and M to show that Sorgenfrey’s square is
not normal, hence not T
4
.
Although the de?nition of normality does not seem to be very di?erent from the previous
de?nitions, we have already seen in Example 2.15 that there is a rather signi?cant di?erence
between this property and the previous separation properties. There are other important dis-
tinctionstoointhatnormalityisequivalenttosomeseeminglyverydi?erentconditions, making
the property very rich and important. We now consider some such properties. The ?rst char-
acterisation is not very di?erent but the remaining ones, Urysohn’s lemma, Tietze’s extension
theorem, the cover shrinking theorem and the partition of unity theorem are all so di?erent
from the de?nition (and even from each other!!) that this is the source of their importance in
topology and also in other branches of mathematics.
Proposition 2.16 The space X is normal i? for each A,U ? X with A closed, U open and
A? U, there is an open set V such that A? V ? ¯V ? U.
Proof. Easy.
De?nition 2.17 Say that two subsets A and B of X are completely separated in X i? there
is a continuous function f : X ? [0,1] with f(A) = 0 and f(B) = 1. [0,1] may be replaced by
[a,b] with a<b. Let X be a space. By a Urysohn family in X we mean a family{U
r
/ r?D}
of open subsets of X satisfying:
(i)
¯ D =R;
(ii) ?
r?D
U
r
=X and n
r?D
U
r
=Ø;
(iii) if r <s then U
r
? U
s
.
Lemma 2.18 LetU ={U
r
/ r?D} be a Urysohn family in the space X and de?ne a function
?
U
:X ?R by ?
U
(x) = inf{r?D / x?U
r
}. Then ?
U
is continuous.
Proof. Notethat?
U
iswell-de?nedbecauseby(ii),forany x,{r?D/x?U
r
}isnon-empty
and bounded below.
Let ? = ?
U
and let (a,b)? R. By Theorem 1.19(iii) it is enough to show that ?
- 1
((a,b))
is open. Let x ? ?
- 1
((a,b)). Then a < ?(x) < b so by (i), there are r,s,t ? D so that
a < r < t < ?(x) < s < b. Then x / ? U
t
so by (iii), x / ? U
r
; also x? U
s
. Thus U
s
- U
r
is an
open neighbourhood of x: note that U
s
- U
r
? ?
- 1
(a,b), so ? is continuous.
Lemma 2.19 Let A and B be subsets of a space X. Then A and B are completely separated
in X i? there is a Urysohn family {U
r
/ r?D} with A? U
0
and B? X- U
1
.
Proof. ?: Let f : X ? [0,1] be continuous with f(A) = 0 and f(B) = 1. For each r?R
let U
r
={x?X / f(x)<
r+1
2
}. Then{U
r
/ r?R} is a Urysohn family of the desired form.
?: Let U = {U
r
/ r ? D} be a Urysohn family with A ? U
0
and B ? X- U
1
, and let
? =?
U
:X ?R be as in Lemma 2.18. De?ne f :X ? [0,1] by
f(x) =
?
?
?
0 : if ?(x)= 0
?(x) : if ?(x)? [0,1]
1 : if ?(x)= 1.
Bytheorem1.22f iscontinuous. Furthermore, ifx?Athenforeachr > 0,x?U
r
; so?(x)= 0
and hence f(x) = 0. On the other hand, if x?B then for each r < 1, x / ?U
r
; so ?(x)= 1 and
hence f(x) = 1.
Theorem 2.20 (Urysohn’s lemma) A space X is normal i? any two disjoint closed subsets
are completely separated in X.
Proof. ?: easy.
?: LetAandB betwodisjointclosedsubsetsofX. We?ndaUrysohnfamilyasinLemma
2.19.
For each non-negative integer n, let
D
n
={0,1}?
n
m
2
n
/ m is a positive integer and m= 2n
o
,
and set D = (-8 ,0)?(1,8)?[?
n= 0
D
n
]. Then
¯ D =R.
For r < 0 set U
r
=Ø and for r >1 set U
r
=X. We de?ne U
r
for r?Dn[0,1] inductively
as follows: induction is on n with U
r
de?ned for r?D
n
. By Proposition 2.16, there is an open
set U
0
such that A? U
0
? U
0
? X- B and an open set U
1
such that U
0
? U
1
? U
1
? X- B.
Now suppose n is such that U
r
has been de?ned for r ? D
n
so that if r,s ? D
n
are such
that r < s, then U
r
? U
s
. Let t ? D
n+1
- D
n
, and let r = max{p ? D
n
/ p < t} and
s = min{p?D
n
/ p>t}. Then we have U
r
? U
s
so by Proposition 2.16, there is an open set
U
t
such that U
r
? U
t
? U
t
? U
s
. Thus the induction continues and hence we obtain a Urysohn
family as needed by Lemma 2.19.
Theorem 2.21 (Tietze’s extension theorem) A space X is normal i? every bounded con-
tinuous functionf :C ?R, whereC is a closed subset ofX, has a bounded continuous extension
to X, i.e. there is a continuous function
ˆ
f :X ?R so that for each x?C,
ˆ
f(x) =f(x).
Proof. ?: Given f, let ß be an upper bound for|f(C)|, and for each n?N let r
n
=ß 
2
3


n- 1
.
We construct a sequencehg
n
:X ?Ri of continuous functions so that for each n,|g
n
(X)|= r
n
and|f- P
n
i=1
(g
i
|C)|= 2r
n
.
De?ne g
1
:X ?R to be the function sending X to 0. Now suppose g
1
,...,g
n- 1
have been
constructed. Thefunctionf- P
n- 1
i=1
(g
i
|C) :C ? [- 2r
n- 1
,2r
n- 1
] = [- 3r
n
,3r
n
]iscontinuous,so
thesetsA ={x?C /f(x)- P
n- 1
i=1
(g
i
(x))=- r
n
}andB ={x?C /f(x)- P
n- 1
i=1
(g
i
(x))= r
n
}
are closed and disjoint. Hence there is a continuous function g
n
: X ? [- r
n
,r
n
] ? R with
g
n
(A) =- r
n
and g
n
(B) =r
n
.
Suppose thatx?C: we show that|f(x)- P
n
i=1
(g
i
(x))|= 2r
n
. There are three possibilities:
either x ? A or x ? B or x ? C - (A? B). If x ? A then g
n
(x) = - r
n
and - 3r
n
= f(x)- P
n- 1
i=1
(g
i
(x)) = - r
n
, so - 2r
n
= f(x)- P
n
i=1
(g
i
(x)) = 0. Similarly if x ? B then
0= f(x)- P
n
i=1
(g
i
(x))= 2r
n
. Ifx?C- (A?B) then both off(x)- P
n- 1
i=1
(g
i
(x)) andg
n
(x) lie
between- r
n
andr
n
sothattheyarewithin2r
n
ofeachother, i.e. - 2r
n
= f(x)- P
n
i=1
(g
i
(x))= 2r
n
. Thus the induction proceeds.
Wenowhaveasequencehg
n
i,withitscorrespondingseries
P
8
n=1
g
n
. Thisseriesisuniformly
convergentsince
P
8
n=1
r
n
=ß P
8
n=1

2
3


n- 1
= 3ß , andso, setting
ˆ
f =
P
8
n=1
g
n
, itisseenthat
ˆ
f
is continuous and for eachx?X,|
ˆ
f(x)|= 3ß . Ifx?C then for eachn,|f(x)- P
n
i=1
g
i
|= 2r
n
,
so, letting n?8, we obtain|f(x)- ˆ
f(x)|= 2.lim
n?8
r
n
= 0, i.e.,
ˆ
f(x) =f(x).
?: Suppose that A and B are disjoint closed subsets of X. De?ne f : A?B ? [0,1] by
f(A) = 0 and f(B) = 1. By Theorem 1.22(ii), f is continuous so has an extension
ˆ
f. Then
ˆ
f
- 1
((-8 ,
1
2
)) and
ˆ
f
- 1
((
1
2
,8)) are disjoint open sets containing A andB. ThusX is normal.
Corollary 2.22 There is a continuous surjection [0,1]? [0,1]× [0,1].
Proof. Let C be Cantor’s ternary set as de?ned in Exercise 12 of Section 1. De?ne f :
C ? [0,1]× [0,1] as follows. Let x? C. Then there is a sequence ha
n
i of 0’s and 2’s so that
x =
P
8
n=1
an
3
n
. If we set f(x) = (
P
8
n=1
a
2n- 1
2
n+1
,
P
8
n=1
a
2n
2
n+1
), then f is continuous and surjective.
By Tietze’s extension theorem, f extends to a continuous function [0,1]? [0,1]× [0,1] which
must be a surjection.
De?nition 2.23 Let X be a space. A cover of X is a family of subsets whose union is all of
X. An (open) [closed] cover is a cover in which all subsets are (open) [closed]. A family of
subsets is (point-?nite) [locally ?nite] i? each point of X (lies in) [has a neighbourhood which
meets] only ?nitely many members of the family. By a shrinkage of a cover {U
a / a ? A} of
the space X is meant another cover {V
a / a ?A} so that for each a ?A, V
a ? U
a .
Theorem 2.24 Let X be a space. Then the following three conditions are equivalent:
(i) X is normal;
(ii) every point-?nite open cover of X has an open shrinkage;
(iii) every locally ?nite open cover of X has an open shrinkage.
Proof. (i)?(ii): let{U
a / a ?A} be a point-?nite open cover of X, and letS be the set of
all functions V :B?T, whereT is the topology of X, satisfying:
(a) B? A;