Download 3. Topological spaces.

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Manifold wikipedia , lookup

Orientability wikipedia , lookup

Surface (topology) wikipedia , lookup

3-manifold wikipedia , lookup

Sheaf (mathematics) wikipedia , lookup

Fundamental group wikipedia , lookup

Covering space wikipedia , lookup

Continuous function wikipedia , lookup

General topology wikipedia , lookup

Grothendieck topology wikipedia , lookup

Transcript
ALGBOOK
TOPOLOGY
3.1
19. januar 2002
3. Topological spaces.
n
n
n
n
(3.1) Definition. A topological space is a set !!X together with a collection !!{Uα }α∈I
of subsets Uα of X such that
(1) The set X and the empty set ∅ are in the collection {Uα }α∈I .
(2) For every subset !!J of I the union ∪β∈J Uβ of the sets in {Uβ }β∈J is in the
collection {Uα }α∈I .
(3) For every finite subset J of I the intersection ∩β∈J Uβ of the sets in {Uβ }β∈J
is in the collection {Uα }α∈I .
The sets Uα are called the open sets of X and the complement !!X \ Uα of the open
sets are called closed. We say that the collection of sets {Uα }α∈I is a topology on X.
Often we simply say that X is a topological space.
Let !!x be a point of X. A subset !!Y of X that contains x is a neighbourhood of
x if there exists an open subset U of X such that x ∈ U ⊆ Y . A collection {Uβ }β∈J
of open sets in X is called an open covering of X if the union of the sets is X, that
is X = ∪β∈J Uβ .
(3.2) Example. Let X be a set. The set X with the collection {∅, X} consisting of
the empty set and X itself is a topological space. This topology is called the trivial
topology on X.
(3.3) Example. Let X be a set. The set X with the collection of all subsets of X
is a topological space. This topology is called the discrete topology.
n
(3.4) Example. Let X be a set. The set X with the collection of sets consisting of
∅ and all the subsets U of X whose complement !!X \ U is a finite set is a topological
space. We call this topology the finite complement topology.
(3.5) Remark. Let X be a topological space with open sets {Uα }α∈I . For every
subset Y of X we have that the collection of sets {Uα ∩ Y }α∈I are the open subsets
of a topology on Y . We call this topology on Y the topology induced by the topology
on X, and we say that Y is a subspace of X.
n
n
(3.6) Definition. Let X be a topological space and let x be a point of X. A
collection of sets !!B = {Uβ }β∈J consisting of open neighbourhoods Uβ of x is a basis
for the neighbourhoods of x if there, for every open neighbourhood U of x, is an open
set Uβ belonging to B such that x ∈ Uβ ⊆ U .
A collection of subsets B = {Uγ }γ∈K of X consisting of open sets Uγ of X is a
basis for the topology if the members !!Bx = {U ∈ B : x ∈ U } containing x is a basis
for the neighbourhoods of x for every point x ∈ X.
(3.7) Example. The collection of all open sets is a basis for the topology on X.
n
(3.8) Definition. For every subset !!Y of X we denote by !!Y the intersection of
all the closed sets that contain Y . Equivalently Y is the set consisting of all points x
in X such that every open neighbourhood of x contains at least one point of Y . We
call Y the closure of the set Y .
topology3
ALGBOOK
TOPOLOGY
3.2
19. januar 2002
n
n
(3.9) Definition. Let X and Y be topological spaces. A map !!ψ : X → Y is called
continuous if the inverse image ψ −1 (V ) of every open subset V of Y is open in X.
The map is an isomorphism if there is a continuous homomorphism !!ω : Y → X
which is inverse to ψ. That is ωψ = idX and ψω = idY .
(3.10) Example. Let X be a topological space and Y a subset considered as a
topological space with the induced topology. Then the inclusion map Y → X is
continuous.
(3.11) Example. The set theoretic inverse of a bijecive continuous map ψ : X → Y
is not necessarily bijective. For example the identity map idX : X 0 → X 00 from
the topological space X 0 with X as underlying set and the discrete topology to
the topological space X 00 with X as underlying set and trivial topology is always
continous. However, the inverse, which is also idX is not continous if X has more
than one point.
(3.12) Remark. For every topological space X the map idX is continuous. When
ψ : X → Y and ω : Y → Z are continuous maps of topological spaces we have that
ωψ : X → Z is continuous. In other words the topological spaces with continuous
maps form a category, called the category of topological spaces.
(3.13) Remark. Let X be a topological space and let Obj(K) be the collection of
open sets of X. For each pair of open sets U, V in X we let Hom(U, V ) consist of
the inclusion map of U in V if U is contained in V , and otherwise let Hom(U, V ) be
empty. Then Obj(K) with these morphisms form a category.
→
(3.14) Exercises.
1. Let X be a set and let X = U0 ⊃ U1 ⊃ U2 ⊃ · · · be a sequence of subsets.
(1) Show that the sets ∅ and {Un }n∈N are the open sets of a topology of X.
(2) Show that if ∩n∈N Un 6= 0 the set ∩n∈N Un is not open in X.
2. Let X be a set and let x0 be an elements of X.
(1) Show that X with the collection of all subsets of X that contain x0 is a
topological space.
(2) Show that X has a basis for the topology with open sets consisting of 1 or 2
elements.
(3) Find the closed points of X.
3. Let Y = {y, X} be the disjoint union of a point y and the underlying set X of
a topological space with open sets {Uα }α∈I . Show that Y with the family of sets
{y, Uα }α∈I is a topological space.
4. Let X and Y be topological spaces and ψ : X → Y a map.
(1) Show that when X has the discrete topology then ψ is continuous.
(2) Show that when Y has the trivial topology then ψ is continuous.
5. Give another example than (?) of a continuous bijective homomorphism ψ : X →
Y of topological spaces which is not an isomorphism.
ALGBOOK
TOPOLOGY
3.3
19. januar 2002
6. Let X and Y be topological spaces and B a basis for the topology on Y . Show
that a map ψ : X → Y is continuous if and only if the inverse image of every open
set belonging to B is open in X.
7. Let X be a set and let B = {Uα }α∈I be a family of subsets Uα with the property
that for every pair of sets Uα , Uβ in the family B and every point x ∈ Uα ∩ Uβ there
is a Uγ in B such that x ∈ Uγ ⊆ Uα ∩ Uβ . Let U be the family of all subsets U of X
such that for every point x ∈ U there is a Uα in B such that x ∈ Uα ⊆ U .
→
(1) Show that X with the family of sets U is a topological spaces.
(2) Show that the sets of B form a basis for the topological space of part (1).
n
8. Let X and Y be topological spaces and let !!V be the collection of subsets of the
cartesian product X × Y of the form U × V , where U is open in the X and V is open
in Y .
n
(1) Show that X × Y with the sets !!U which consists of all the unions of the sets
in V form a topological space. We call this topology the product topology on
X ×Y.
(2) Show that the projection !!π : X ×Y → X defined by π(x, y) = x is continuous
when X × Y has the product topology.
(3) Assume that X and Y have the finite complement topology. Show that in
most cases the finite complement topology on X × Y is different from the
product topology.
n
n
→
→
→
9. Let X = Z. An arithmetic progression consists of numbers of the form !!Vp,q =
{pn + q : n ∈ Z} where p and q are integers, and p 6= 0.
(1) Show that for every integer m we have that Vp,q = Vp,mp+q .
(2) Let p0 , p00 , q 0 , q 00 be natural numbers. Show that for every number n in Vp0 ,q0 ∩
Vp00 ,q00 there are natural numbers p, q such that n ∈ Vp,q ⊆ Vp0 ,q0 ∩ Vp00 ,q00 .
(3) Show that the collection of all subsets of Z that are arithmetic progressions
satisfy the conditions of Exercise (6), and consequently is the basis for a
topology on X.
(4) Show that all the arithmetic progressions Vp,q are closed in the topology of
part (3).
(5) Let Y be the union of all the sets Vp,0 where p is a prime number. Show that
X \ Y = {−1, 1} and that {−1, 1} is not open in X.
(6) Use part (4) and (5) to prove that there exists infinitely many prime numbers.
10. Let X be a set with a metric, that is, for each pair of points x, y of X there is
a real number d(x, y) such that for all elements x, y, z of X we have:
(1)
(2)
(3)
(4)
d(x, y) ≥ 0.
d(x, y) = 0 if and only if x = y.
d(x, y) = d(y, x).
d(x, z) ≤ d(x, y) + d(y, z).
Let U consist of all sets U with the property that for every point x of U there is a
ALGBOOK
TOPOLOGY
3.4
19. januar 2002
real number εx such that the set {y ∈ X : d(x, y) < εx } is contained in U .
(1) Show that X with the family U is a topological space.
(2) Show that for each point x of X the sets Ux,n = {y ∈ X : d(y, x) < 1/n} for
all natural numbers n form a basis for the neighbourhoods of x.