Download 4. Topologies and Continuous Maps.

4. Topologies and Continuous Maps.
In this section we introduce the concept of a topological space. Recall that a graph was defined to be a set
together with a collection of subsets that are called
”edges”. A simplicial complex is a set together with a
collection of subsets called ”simplices”. The notion of
a topological space extends this idea vastly. A topological space is a set together with a collection of subsets called ”open subsets”. In this section we describe
some efficient ways of generating a topology, we give
some examples and we introduce some first properties
of topological spaces.
Topological Spaces.
We say that a subset of R is an open subset of R if
it is a union of open intervals (a, b) ⊂ R. Given this
notion the usual ǫ − δ-definition for continuous maps
takes a rather simple form.
4.1. Theorem. A map f : R → R is continuous if
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either one of the following two conditions hold:
(1) the preimage under f
open.
of every open subset is
(2) for every point x ∈ R and every number ǫ > 0,
there is a number δ = δ(x, ǫ) > 0 such that |z−x| < δ
implies |f (z) − f (x)| < ǫ, for every z ∈ R.
Proof. (1) ⇒ (2):
Let us be given x ∈ R and ǫ > 0. Then the set
Y := { y ∈ R | |y − f (x)| < ǫ } is an open interval and
therefore an open set. Because of (1), the preimage
f −1 (Y ) is also an open set. By definition, this means
that that preimage is the union of open intervals. In
particular, the point x must lie in some open interval
(a, b) ⊂ Y . Set δ := min { |x − a|, |x − b| }. Then
(x − δ, x + δ) ⊂ (a, b) and so (x − δ, x + δ) is mapped
under f into Y . This means that |z −x| < δ implies
f (z) ∈ Y , i.e., |f (z) − f (x)| < ǫ. In other words, f
satisfies (2).
(2) ⇒ (1):
Let us be given an open subset, say W , of Y . Set
V := f −1 W . Then we are asked to show that V
is an open subset, i.e., we are asked to show that V
is a union of open intervals. In other words we are
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asked to show that every point of V lies in some
open interval which in turn is contained in V . To
do this let x ∈ V . Then there is an open interval
in W which contains f (x) since W is supposed to
be an open subset. More precisely, there is an ǫ > 0
such that the open interval (f (x) − ǫ, f (x) + ǫ) ⊂ W .
Because of condition (1), there is a δ > 0 such that
|z − x| < δ implies |f (z) − f (x)| < ǫ. It follows that
the open interval (x − δ, x + δ) is mapped under f
into the open interval (f (x) − ǫ, f (x) + ǫ) ⊂ W . In
particular, this shows that the point x lies in the open
interval (x−δ, x+δ) which in turn lies in f −1 V = W .
♦
Observe that the collection of all open sets in R forms
a topology in the following sense.
Definition. Let X be any set (finite or not). Then
a collection X of subsets of X is called a topology
for X if the following holds:
(1) ∅ ∈ X and X ∈ X .
(2) X is closed under taking finite intersections.
(3) X is closed under taking arbitrary unions.
A set together with a topology is a topological space.
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The elements of the topology X of X are called
open subsets of X. A subset A ⊂ X is called
closed if its complement X − A is open.
Remark. Strictly speaking, a topological space X
is a pair (X, X ) of X with a topology X (and
in this way a topological space is a generalization of
a simplicial complex). However, one often does not
list the topology X explicitly. Instead one often simply writes X instead of the more correct (X, X ) especially if it is clear from the context what X is.
4.2. Definition. Two topologies X1 , X2 of a set
X are equal if X1 = X2 , i.e., if the open sets in
one topology are also open in the other topology. The
topology X1 is called finer (resp. coarser) than X2
if X1 ⊃ X2 (resp. X1 ⊂ X2 ).
We now take a quantum leap forward and generalize
the definition of continuous maps as follows.
4.3. Definition. Let (X, X ) and (Y, Y) be topological spaces. Then a map f : X → Y is a continuous
map if the preimage of every open set is open, i.e., if
f −1 U ∈ X , for every U ∈ Y. A homeomorphism
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is a bijection h : X → Y such that h as well the
inverse h−1 is a continuous map. Two spaces X, Y
are homeomorphic (notation: X ∼
= Y ) if there is
a homeomorphism h : X → Y .
By what we have seen above this definition of continuity includes as special case the familiar continuous maps of the real line. However, we also have
to be aware that the definition is so general that it
has to be taken with a grain of salt. For instance,
it may be surprising that not every continuous bijection is a homeomorphism. But it is actually quite
easy to construct counterexamples. For this take any
set X and equip it with the discrete or the indiscrete topology. Then the identity but not its inverse
is continuous. Here the discrete topology of X is
the power set of X, i.e., the collection of all subsets of X. The indiscrete topology of X is the
topology which consists only of the empty set and the
set X. A slightly more typical example is the map
f : [0, 1) → S 1 , t 7→ (cos(2πt), sin(2πt)). It is clear
that this map is a continuous bijection but not a homeomorphism (why?). Extending the range, we get the
map f¯ : [0, 1) → R2 , t 7→ (cos(2πt), sin(2πt)). This
is an example of a map which is an embedding but not
a homeomorphism onto its image. Embeddings which
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are homeomorphisms onto their image are also known
as imbeddings.
To illustrate the power of the definition on this level
we give a topological proof of a well-known theorem
in elementary number theory. The proof is due to
Furstenberg [Amer. Math. Monthly 62 (1955)].
4.4. Theorem.
numbers.
There are infinitely many prime
Proof. We first introduce a topology P on the set
Z of integers (the so called profinite topology) by
setting
U ∈ P :⇔ for every a ∈ U there is a t ∈ Z − 0
with a + tZ ⊂ U
.
(here tZ denotes the set tZ := { tx ∈ Z | x ∈ Z }.
We claim P forms a topology of Z. For this we have
to verify the three conditions of a topology. Now, it
is clear that ∅, Z ∈ P. Moreover, if U1 , U2 ∈ P
and if a ∈ U1 ∩ U2 , then there are t1 , t2 ∈ Z with
a + t1 Z ⊂ U1 , a + t2 Z ⊂ U2 . It then follows that
a + (t1 t2 )Z ⊂ U1 ∩ U2 , and so U1 ∩ U2 ∈ P. Finally,
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it is obvious that arbitrary unions of sets from P are
sets in P.
To continue the proof observe that tZ, Z − tZ ∈ P,
i.e,, tZ is open and closed in the profinite topology.
Of course, tZ is open since (ta) + tZ S
⊂ tZ. To see
that tZ is closed, note that Z − tZ = s6=t sZ, i.e.,
the complement is a union of open sets and so open.
We are now ready to finish the proof. Assume the subset P ⊂ Z S
of all prime numbers in Z is finite.
S Then
the union
p∈P pZ
S is closed. Hence Z − p∈P pZ
is open. But Z − p∈P pZ = {±1} and {±1} is
not open in the profinite topology. This contradiction
finishes the proof. ♦
Remark. The construction used in the previous proof
is much more general. Indeed, using normal subgroups
N in groups, instead of tZ in Z, one can define a
profinite topology for every group.
Open and Closed Sets.
Recall that a subset Y ⊂ X of a topological space
(X, X ) is called an open subset iff it is an element
of the topology of X, i.e., iff Y ∈ X . It is called a
closed subset if its complement is an open subset. An
open subset U ⊂ X is called an open neighborhood
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for a set A ⊂ X if A ⊂ U (the subset A may be
just a single point).
Let Y ⊂ X be any subset of the topological space
(X, X ), Then we denote by
o
[
int(Y ) := Y := {U | U ∈ X and U ⊂ Y }
\
cl(Y ) := Ȳ := {X − U | U ∈ X and
Y ⊂ X − U }.
∂Y := Ȳ − Y o
We say Y o is the interior, Ȳ is the closure and
∂Y is the boundary of Y . Thus the interior of a
set Y is the union of all open subsets contained in Y
and the closure of Y is the intersection of all closed
subsets containing Y .
We say a point x ∈ X is a limit point (or an accumulation point) for Y if, for every open neighborhood U of x, the intersection
U ∩ ( Y − {x} )
is non-empty. We denote
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Y ′ := the set of all limit points of Y .
Remark. Keep in mind that a point x need not be
a limit point of a set Y if all its open neighborhoods
meet Y . For this we have to make sure that they
meet Y in points other than x !
4.5. Theorem. Y o is an open set and Ȳ is a closed
set. Moreover, Ȳ = Y ∪ Y ′ .
Proof. Y o is an open set since any union of open
sets is open. To show that ȲT is closed, it sufficies to
show that any intersection
A∈A A of closed
T sets is
closed. But this follows from the fact X − A∈A A =
S
A∈A (X − A) is open.
Next, we show that Y ∪ Y ′ ⊂ Ȳ . So let x ∈ Y ∪ Y ′ .
If x ∈ Y , then of course x ∈ Ȳ . Hence let x ∈ Y ′ .
Then every open neighborhood U (x) of x intersects
Y (in a point other than x). In particular, x cannot
lie in X − Ȳ . For otherwise, there is at least one closed
set A ⊂ X with Y ⊂ A and x 6∈ A. But X − A
is open since A is closed. Hence there is an open
neighborhood of x which does not meet A and so
not Y . Contradiction. This proves Y ∪ Y ′ ⊂ Ȳ .
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For the other direction let x ∈ Ȳ . We wish to show
that x ∈ Y ∪ Y ′ . If x ∈ Y , then there is nothing to
show. So let x ∈ Ȳ − Y and let U (x) be an open
neighborhood of x in X. Then U (x) must meet
Y , for otherwise the complement of U (x) is a closed
set containing Y but not x. This is impossible since
x ∈ Ȳ and since Ȳ is the intersection of all closed
sets containing Y . It therefore follows that every
x ∈ Ȳ − Y has an open neighborhood that meets Y
in a different point than x. ♦
Remark. Note that Y o ⊂ Y ⊂ Ȳ . Note further that
Y is open iff Y o = Y and it is closed iff Ȳ = Y .
The closure of sets gives us the following useful characterization of continuous maps.
4.6. Theorem. Let f : X → Y be a map between
topological spaces X, Y . Then f is continuous if
and only if
f (Ā) ⊂ f (A),
for all subsets A ⊂ X.
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Proof. Suppose f is continuous. Let x ∈ Ā. If
f (x) ∈ f (A), we are done. So assume f (x) ∈ Y −
f (A). Let U be an open neighborhood of f (x).
Then f −1 U is an open neighborhood of x. Hence
it contains a point, say z, of A different from x.
Then f (z) is a point in f (A), different from f (x)
(recall f (x) ∈ Y − f (A)), which lies in U ∩ f (A).
This implies that f (x) lies in the closure of f (A).
Suppose f (Ā) ⊂ f (A), for all subsets A ⊂ X. Let
B ⊂ Y be any closed subset of Y . To show that
f is continuous it suffices to show that the preimage
A := f −1 B is closed. For this it is enough to show
that Ā ⊂ A (because then Ā = A). Now, x ∈ Ā
implies f (x) ∈ f (Ā) ⊂ f (A) ⊂ B̄ = B. Hence
x ∈ f −1 B = A. ♦
The notion of a limit point allows some more terminology (and in this way some better questions that can
be asked about topological spaces).
4.7. Definition. Let X be a topological space and
let A ⊂ X be a subset. Then
(1) A is dense in X if Ā = X,
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(2) A is nowhere dense if (Ā)o = ∅,
complement of Ā is dense in X,
i.e., the
(3) A is perfect if A = A′ ,
(4) a point x ∈ A is isolated if it is open, i.e.,
there is an open neighborhood U of x in X with
U ∩ {x} = {x}.
Remark. To see the equivalence in (2) recall that
C( int(A) ) = cl( C(A) ) and C( cl(A) ) = int( C(A) ),
where C(A) denotes the complement. Hence
C(∅) = X = cl( C(cl(A)) ) = C( int( cl(A)) ) )
and the claim follows.
Remark. The word ”limit point” is perhaps slightly
misleading. One says that a sequence {xn }n∈N of
points in a topological space X converges to a point
x ∈ X if every open neighborhood U of x contains
all but finitely many xn . But it is not true that every
limit point x has a sequence {xn }n∈N converging
to it. As an example consider R with the topology
given by ∅ and all sets opne in R and containing 0.
Then every point x ∈ R − {0} is a limit point of the
set A = {0}, but no sequence in A converges to x.
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Construction of Topologies.
The definition of a topological space is rather abstract
and not very easy to work with. In general, it is out
of the question to list all open subsets of a topology
although this is sometimes possible (e.g., for topologies of finite sets). In praxis, however, one describes
a topology differently, namely through a basis or even
a subbasis. In fact, we have given the topology of R
(see above) through a basis.
4.8. Definiton. A basis of a topology is a collection
B of subsets of a set X with
(1) every open set of the topology is a union of basis
elements.
(2) if B1 , B2 are basis elements and if x ∈ B1 ∩ B2 ,
then there is a basis element B3 such that x ∈ B3 ⊂
B1 ∩ B2 .
A collection of subsets of a set X is a subbasis if
its union equals X.
We turn a subbasis into a basis by taking all finite
intersections. We generate a topology from a basis by
taking all unions. So, for instance, the collection of all
open intervals is a basis as well as a subbasis.
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Remark. Note that the concept of subbasis makes it
now a trivial matter to construct topologies for a set
X. For this we simply give ourselves any collection
A of subsets of X whose union is X. This already
defines a unique topology, given by first taking all intersections of elements of A and then all unions of the
resulting subsets. The topology obtained in this way is
called the topology generated by the subbasis A.
Here are some less obvious ways of constructing topologies.
Product Topology. Let (X, X ), (Y, Y) be two topological spaces. Then notice that
B := { U × V | U ∈ X , V ∈ Y }
is not a topology. It satisfies all axioms of a topology
except that it is not closed under taking unions. So
it is a basis. The topology generated by this basis is
called the product topology for X × Y . A set in
the product topology is open if every point of it has
an open neighborhood of the form U × V . Given
two topological spaces X, Y , it is understood that
X × Y denotes the topological space with the product
topology.
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Quotient Topology. Given a surjective map ϕ :
X → Y between a topological space (X, X ) and a
set Y , the quotient topology for Y (with respect
to ϕ) is the topology given by
Y := { U | ϕ−1 (U ) ∈ X }.
The topological space (Y, Y) with the quotient topology is often called the quotient space and denoted
by X/ϕ.
Remark. Note that the map ϕ defines an equivalence relation on X by setting
x ∼ y ⇔ ϕ(x) = ϕ(y).
Vice versa, an equivalence relation ∼ on the space
X gives rise to a surjective map
ϕ : X → Y = X/ ∼,
x 7→ [x]
which maps every element x to its equivalence class.
It is the quotient map induced by the equivalence
relation ∼. The notation ”quotient space” is due to
the fact that quotient spaces are usually defined via
equivalence relations.
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The quotient topology reflects the fact that points are
being identified. Here are two typical examples.
Example. Let R be the set of real numbers with the
Euclidean topology. Define an equivelance relation ∼
on the unit interval
I := [0, 1] = { x ∈ R | 0 ≤ x ≤ 1 }
by setting
x ∼ y ⇔ x = y, or x = 0 and y = 1.
Let ϕ be the induced quotient map. Then I/ϕ is
homeomorphic to the unit circle S 1 .
Example. Let again R be the set of real numbers
with the Euclidean topology. Let d : R → R, x 7→
x + 1, be the translation by 1. Then d induces an
equivalence relation ∼ by setting
x ∼ y ⇔ y = dn (x), for some n ∈ N
Let ϕ be the induced quotient map. Then R/ϕ is
again the unit circle S 1 .
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Subspace Topology. Last but not least we define
the subspace topology for a subset Y ⊂ X of a
topological space (X, X ) to be the topology
Y := { U ∩ Y | U ∈ X }.
Note that Y is an open set in the topological space
(Y, Y) even if Y is a closed set in (X, X ).
Some Examples of Topologies.
The mere definition of a topology is not very restrictive
and therefore it covers a lot of territory. One basically
only needs to check that it is closed under taking intersections and unions. So before one goes overboard
inventing new topologies one should be aware of the
fact that most of them are pretty useless unless we can
say what we need them for. One aspect of a topology
that one may want to keep in mind is that a topology
decides which sequences of elements in a set converge.
More precisely, we define
4.9. Definition. Let (X, X ) be a topological space.
Let x ∈ X be a point. We say a sequence xn , xn ∈
X, converges to x if, for every open neighborhood
U (x) of x there is an index N > 0 such that
xn ∈ U (x), for all n > N.
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Remark. Note that convergence is defined here without the use of any ǫ and δ.
Example. In the indiscrete topology every sequence
converges to every point. In the discrete topology only
the eventually constant sequences converge.
Here are some typical examples for topological spaces.
Example. As a first example consider the finite product Rn = R×. . .×R. If n ≥ 2, there are actually two
different ways of putting a topology on Rn . The first
is to start with the Euclidean toplogy of R and iterate
the construction of the product topology. More precisely, one defines recursively the topology for Rn+1
to be the product topology of Rn × R. The second
way, uses the fact that an open interval cannot only
be described by the order relation <, but also by a
distance function. The Euclidean distance is given
by
p
d(x, y) := (x1 − y1 )2 + . . . + (xn − yn )2 .
Given the Euclidean distance function, one sets
U (x, ǫ) := { y ∈ Rn | d(x, y) < ǫ },
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for every x ∈ Rn and every ǫ > 0. One says that
U (x, ǫ) is the ǫ-neighborhood for x. It is easily
verified that the set
B := { U (x, ǫ) | x ∈ Rn , ǫ > 0 }
of all ǫ-neighborhoods defines a basis. This basis in
turn defines a topology for Rn . Which of the two
topologies defined so far should be called the Euclidean
topology of Rn ? Fortunately, we do not have to make
a decision here. Both topologies are the same. For
this one only has to show that every point in every
box (a1 , b1 ) × . . . × (an , bn ) has an ǫ-neighborhood
contained in the box, and that every point in every
epsilon neighborhood lies in a box (a1 , b1 ) × . . . ×
(an , bn ) contained in the the ǫ-neighborhood. But
this is easy to verify.
Example. The Euclidean topology of Rn induces a
topology on the set Matm (R) of all m × m-matrices
with m2 = n. Indeed, the entries of an m×m matrix
define a unique point in Rn and vice versa. So there
is a bijection ϕ : Matm (R) → Rn if m2 = n. Define
the topology X of Matm (R) by setting
X := { ϕ−1 (U ) | U open in Rn }.
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In this topology two matrices are ”close” if all their
entries are close. Thus a sequence An ∈ Matm (R)
of matrices in Matm (R) converge to a matrix A ∈
Matm (R) if and only if (aij )n → aij for the entries
of the respective matrices.
Example. The Euclidean topology of Rn induces
a topology for finitely generated matrix groups. For
instance, consider
SLm (R) = { A ∈ Matm (R) | det(A) = 1 }.
Then SLm (R) is not only a set of matrices but a space
of matrices with respect to the topology induced by
the topology of Matm (R). Moreover, SLm (R) is a
group with respect to matrix multiplication, because
of the product formula for determinants. Now, let G ⊂
SLm (R) be a finitely generated group of matrices. Let
A1 , . . . , Ag ∈ G be the finitely many matrices that
generate G. By definition, this means that every
matrix in G is a product of the matrices A1 , . . . , Ag .
The set
Hom(G, SLm (R))
:= { ϕ : G → SLm (R) | ϕ is a homomorphism }
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can be viewed as a subset of R(ng) where n =
m2 , because ϕ is determined by the g matrices
ϕ(A1 ), . . . , ϕ(Ag ) which in turn can be identified with
points in Rn . Thus Hom(G, SLm (R)) is a subspace
of some Euclidean space and therefore is a topological
space itself with respect to the subspace topology.
Example. Let Map(X, Y ) be the set of all maps
f : X → Y from X to Y . Given that Y is a
topological space, the sets
S(x, U ) := { f : X → Y | f (x) ∈ U },
for all open sets U ⊂ Y and all points x ∈ X, defines
a subbasis. The topology of Map(X, Y ), generated
by this subbasis, is called the topology of pointwise
convergence. It is one of several useful topologies for
function spaces.
Example. Let R be a commutative ring (e.g., a
polynomial ring). Then denote
spec(R) := the set of all prime ideals in R.
The Zariski topology of spec(R) is given by specifying closed sets (rather than open sets). More precisely, a set C ⊂ spec(R) is closed in the Zariski
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topology if there is an ideal A ⊂ R such that
C = {P | P is prime ideal with A ⊂ P }. The complements of the closed sets are the open sets in the Zariski
topology.
Properties of Topological Spaces.
For future reference we need to list some more properties of topological spaces.
4.10. Definition. Let X
Then
be a topological space.
(1) X is disconnected if it is the disjoint union
X = A ∪ B of two open and non-empty subspaces
A, B ⊂ X,
(2) X is connected, if it is not disconnected.
(3) X is totally disconnected, if the only connected
subspaces are the one-point sets,
(4) X is path-connected if, for any two points
x, y ∈ X, there is a continuous map (path) w :
[0, 1] → X with w(0) = x and w(1) = y.
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A subspace C ⊂ X is a component of X if it is
connected and if it equals every connected subspace
containing it.
Remark. Components are closed since the closure
of a component is a component. However, in general,
a component is not open. Components are of course
open if X has only finitely many components since
then the complement is a finite union of closed sets.
4.11. Example. Consider the following subspace K
in R2 with the subspace topology.
a
b
c
The intervals [a, c) and (c, b] are components. They
are closed but not open. Moreover, there is no separation K = U ∪ V with a ∈ U and b ∈ V and yet a, b
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do not lie in a component of K. Here a separation
of K is a disjoint union K = U ∪ V of non-empty,
open subsets U, V .
Next we list some separation axioms. We say that
two disjoint subsets A, B ⊂ X in a topological space
X can be separated if there are open neighborhoods
U (A), U (B) of A, B, respectively, with U (A) ∩
U (B) = ∅. Given this notation we next define:
T1: A topological space X is a T1-space if every
subset with a single element is closed (short: all points
are closed).
T2: A topological space X is a T2-space if any two
distinct points in X can be separated. A T2-space is
also called a Hausdorff space.
T3: A topological space X is a T3-space if any
point and any disjoint, closed set can be separated. A
T3-space is also called a regular space.
T4: A topological space X is a T4-space if any two
disjoint, closed sets can be separated. A T4-space is
also called a normal space.
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regular
79
normal
4.12. Theorem. Let X be a T1-space. Then
(1) X is a Hausdorff space if and only if the diagonal
∆ := {(x, x) | x ∈ X} is closed in the product space
X × X.
We also have the following sandwich properties:
(2) X is a regular space if and only if, for every
x ∈ X and every open neighborhood U (x), there is
an open subset V ⊂ X with x ∈ V ⊂ V̄ ⊂ U (x).
3) X is a normal space if and only if, for every closed
subset A ⊂ X and every open neighborhood U (A),
there is an open subset V ⊂ X with A ⊂ V ⊂ V̄ ⊂
U (A).
Remark. In general, V 6= V̄ . However, in the next
section we will see a sandwich property with V = V̄ .
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Proof. ad (1)
Let x, y be two distinct points in X. Then (x, y) is
a point in X × X which does not lie on the diagonal
∆ of X × X. If ∆ is closed, there is an open
box U × V ⊂ X × X (with U, V ⊂ X open) such
that (x, y) ∈ U × V and U × V ∩ ∆ = ∅. This
means that x ∈ U, y ∈ V and that there is no
point (u, v) ∈ U × V with u = v. Thus U, V are
disjoint and open neighborhoods of x, y, respectively.
Similarly, one shows that if U, V are disjoint and open
neighborhoods of x, y, respectively, then U × V is
an open neighborhood of (x, y) that does not meet
the diagonal ∆. This proves (1)
ad (2)
Suppose X is regular and let x ∈ X. Recall that X
is also supposed to be a T1-space. So the one-point
set {x} forms a closed set. Moreover, the complement X − U (x) is a closed set. Hence there are open
neighborhoods U, V with x ∈ U and X − U (x) ⊂ V
such that U ∩ V = ∅. It follows that X − V ⊂ U (x).
Hence we have x ∈ U ⊂ X − V ⊂ U (x). Moreover,
Ū ⊂ X − V since X − V is closed. Thus, altogether, we have x ∈ U ⊂ Ū ⊂ U (x). This proves one
direction of (2).
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For the other direction of (2), let x ∈ X be a point
and let A ⊂ X be a closed subset disjoint to x. Then
U (x) := X − A is an open neighborhood of x. Hence
there is an open set V ⊂ X with x ∈ V ⊂ V̄ ⊂ U (x).
In particular, U := X − V̄ is an open neighborhood
of A which does not meet the open neighborhood V
of X. This proves that X is regular.
ad (3)
The proof for (3) follows the same line of argument.
Simply replace the point x by the set A and argue
as before. ♦
Appendix: The Cantor Set.
The discussion of topological spaces has been rather
dry so far. It is time to add some meat. For this let
us consider a concrete example, the Cantor set. The
Cantor set has some strange properties that help us to
get a first idea of the intricacies of certain topological
spaces that come up quite naturally.
We begin with the definition of the Cantor set. Basically, the Cantor set is the set
C :=
Y
{0, 2} = {0, 2}N
n∈N
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of 02-sequences together with a certain topology. This
topology is the subspace topology given by the embedding
f : C → [0, 1],
c1 c2 c3 . . . 7→
X
cn
.
n∈N 3n
Thus the topology of C is given by the sets
f −1 (U ), U open in [0, 1].
Here is another view of the Cantor set which helps to
visualize its topology. Denote by
Cn := {c1 c2 . . . cn cn+1 . . . | c1 , . . . , cn ∈ {0, 2} and
ci ∈ {0, 1, 2}, for i ≥ n + 1 },
i.e., Cn is the set of all infinite words in the alphabet {0, 1, 2} with theSfirst n letters different from 2.
Then of course C = n Cn , i.e., the sets Cn ”converge” to C. The images of Cn under the embedding
f are the following intervals:
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C0
C1
C2
C3
4.13. Theorem. Let C be the Cantor space. Then
(1) C is perfect.
(2) C is totally disconnected.
(3) C is homogeneous, i.e., for any two points
x, y ∈ C there is a homeomorphism h : C → C with
h(x) = y.
(4) C is not homeomorphic to [0, 1].
Proof. ad (1).
We are asked to show that C ′ = C.
Let c =
c1 c2 c3 c4 . . . ∈ C be a point in C. Then the sequence
c1 000 . . . , c1 c2 00 . . . , c1 c2 c3 0 . . . , c1 c2 c3 c4 . . .
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is a sequence of points in C different from c that
converges to c. It follows that C ⊂ C ′ . To prove
C ′ ⊂ C, let x be a limit point of C. Then x is a
limit point of Cn , for every n ≥ 1. But Cn is closed
as a finite
T union of closed intervals. Thus x ∈ Cn and
so x ∈ Cn = C.
ad (2).
Let U be a connected subset of C. Then U ⊂ [0, 1].
Let a = inf U and b = sup U be the greatest
lower bound resp. the least upper bound of U . Then
every point x ∈ (a, b) lies in U , for otherwise the
connected set U can be written as a disjoint union
(U ∩(a, x))∪(U ∩(x, b)) of non-trivial open sets which
is impossible. Thus U is an interval. But every
interval in C lies in Cn , for all n. Hence such an
interval has length 3−n , for all n. It follows that
U has length 0. This means that U is a set which
consists of a single point.
ad (3).
We follow an argument given in [Rosemann, Elementary Topology]. Here is the idea. Suppose x, y ∈ C
are two arbitrary points
x = x1 x2 x3 . . . and y = y1 y2 y3 . . . .
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Then define a homeomorphism
hxy : C → C, p = p1 p2 p3 . . . 7→ hxy (p) = c1 c2 c3 . . .
according to the rule
ci =
pi ,
if xi = yi
pi + 2 (mod2), otherwise
Here is an example
x = 220202222 . . .
y = 220022020 . . .
p = 020202020 . . .
hxy (p) = 020022222 . . .
Check that h(x) = y. For instance,
x = 220202222 . . .
y = 220022020 . . .
x = 220202222 . . .
hxy (x) = 220022020 . . .
For the general case note that hxy (X) = y, because
digits common to x and y are left unchanged under
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h, while digits of x different from the corresponding y-digit are changed (necessarily into the one for
y). It remains to show that f is a homeomorphism
(Exercise).
ad (4).
C cannot be an interval since C is totally disconnected and an interval is connected. ♦
We finish this section by showing that the Cantor set is
not only bijective but also homeomorphic to the product C × C. The product C 2 = C × C is also known
as the Sierpinski dust:
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4.14. Theorem. Let C be the Cantor set. Then
there is a homeomorphism h : C → C 2 .
Proof. (We here follow an argument from [Sieradski,
An introduction into topology and homotopy]). Define
a map
h : C → C 2,
c1 c2 c3 c4 . . . 7→ (h1 (x), h2 (x)) = (c1 c3 c5 . . . , c2 c4 c6 ..).
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We have the following for the distances of points x =
x1 x2 x3 . . . , y = y1 y2 y3 . . . in C and image points
h(x), h(y) in C 2 :
dist(x, y) < 3−2n
⇔ x, y ∈ C2n
(see above for the definition of Cn )
⇔ xi = yi , for all 1 ≤ i ≤ 2n
⇔ h1 (x), h1 (y) as well as h2 (x), h2 (y)
lie in a component of Cn
⇔ maximal distance of h(x) and h(y)
is smaller than ( 13 )n .
It follows that h and its inverse are continuous. ♦
Remark. In contrast to the previous theorem we have
that R is not homeomorphic to the product R × R =
R2 . To see this note that R−{0} is disconnected but
R2 −{p} is connected, for every p ∈ R2 . On the other
hand, we also have D ∼
= D × D, for every discrete
space D. So the Cantor set is more like a discrete
space than R. Is the Cantor set homeomorphic to a
discrete space? For the answer see the next lecture.
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