§ 15. Weak (initial) topology with respect to a family of maps.
15.1. Definition. Let fα : X → Yα, α ∈ A, be a family of maps of a set X into
spaces Yα. Then the family of sets
f
−1
α (U) : U open in Yα, for some α ∈ A
is a cover of the set X and, by Proposition 8.8 of Lecture 2, is a subbase of some
topology T on X. We call T the weak or initial topology with respect to the family of
maps {fα : α ∈ A}.
With respect to this topology all the maps fα are continuous by virtue of Proposition 13.2 of Lecture 3. Moreover, T is the smallest (weakest) topology on X having this property.
It is easy to see that in the definition it suffices to restrict attention to open sets U ⊂ Yα
from some subbase of the space Yα.
15.2. Examples.
1. Let M be a subset of the space (X, T ), iM : M → X the identity
embedding map. Then the weak topology on M with respect to the embedding iM coincides with the
subspace topology T |M.
2. Let Tα, α ∈ A, be a family of topologies on the set X, idα : X → Xα, α ∈ A, the
family of identity maps of the set X into the spaces Xα. Then on X there is defined
the least upper bound of the family of topologies A as the weak topology with respect to the family of
maps {idα : α ∈ A}.
15.3. Proposition. Let fα : X → Yα, α ∈ A, be a family of maps of the space
(X, T ) into the spaces Yα. Then the following conditions are equivalent:
1) T is the weak topology on X with respect to the family of maps fα, α ∈ A,
2) for any space Z, its map f : Z → X is continuous if and only
if the maps fα ◦ f : X → Yα, α ∈ A, are continuous.
Proof. 1) ⇒ 2). Necessity of condition 2) follows from the continuity of the maps fα, α ∈ A, and their composition with f. Sufficiency. By Proposition 13.3 of Lecture 3 it suffices
to show that the preimage of a set O from the subbase of the topology of X defined in Definition
15.1 is open in Z. Let O = f
−1
α (Oα) for some α ∈ A. Then
f
−1O = f
−1
(f
−1
α (Oα)) = (fα ◦ f)
−1
(Oα),
and f
−1O is open.
2) ⇒ 1). Since the identity map id of the space (X, T ) into (X, T ) is a homeomorphism, then by condition 2) the compositions fα ◦ id are continuous, α ∈ A. Hence the maps fα, α ∈ A,
are continuous, since for any open subset Oα ⊂ Yα we have
f
−1
α (Oα) = id((fα ◦ id)−1
(Oα)),
and the set (fα ◦ id)−1
(Oα) is open in X, α ∈ A.
Let T
0 be the weak topology on X with respect to the family of maps fα, α ∈ A. Then
T ≥ T 0
, and for any α ∈ A the composition fα ◦ id : X → Yα of the identity map (X, T
0
)
into (X, T ) and fα : X → Yα is continuous (T
0 being the weak topology on X with respect to the family of
maps fα, α ∈ A, and on the set X the maps fα and fα ◦ id coincide). By condition 2)
the identity map id : (X, T
0
) → (X, T ) is continuous, i.e. any set open in
(X, T ) is open in (X, T
0
). Hence T
0 ≥ T . Consequently, T = T
0
.
§ 16. Cartesian product of a family of sets.
16.1. Definition. Let a family of sets {Aα : α ∈ J} be given. Denote
by Q
α∈J
Aα the Cartesian product of these sets, i.e. the set of all maps
x : J →
S
α∈J Aα, such that x(α) ∈ Aα for any α ∈ J.
26
For finite J = {1, . . . , k} the product of the family of sets A1, . . . Ak is denoted by
A1 × . . . × Ak, and for J = N by Q∞
i=1
Ai
.
If S ⊂ J, then there is defined the natural projection
prS :
Y
α∈J
Aα →
Y
α∈S
Aα,
assigning to a point x of the product Q
α∈J Aα (a map x : J →
S
α∈J Aα) the point
prS(x) = x|S (the restriction x|S of the map x to the set S). If S consists of a single index
α, then prS will be denoted by prα.
The point x(α) ∈ Aα will be called the α-coordinate of the point x ∈
Q
α∈J
Aα and, as a rule, will be
denoted xα, and the point itself x will be denoted (xα). In particular, an element x of the product
Q∞
i=1
Ai will be denoted (xi).
16.2. Example. Let 2 = {0, 1}, {Ai = 2 : i ∈ N}. Then 2
N =
Q∞
i=1
Ai is the set of all
sequences of 0's and 1's (maps of the natural numbers into a two-element set).
2
N is also called the countable power of the two-point discrete space.
§ 17. Tychonoff product topology.
17.1. Definition. Let the factors Xα of the Cartesian product X =
Q
{Xα : α ∈ A
be topological spaces. Then on the set X one can consider the weak
topology with respect to the family of projections prα : X → Xα, α ∈ A. This topology is called
the Tychonoff product topology. The set X with this topology is called the topological, or Tychonoff, or simply the product of the spaces Xα.
By Definition 15.1, a subbase of the space X is formed by all sets
of the form pr−1
α (U), where U is taken from some base of the space Xα, and a base, consequently, by all their finite intersections
pr−1
α1
(Uα1
) ∩ . . . ∩ pr−1
αk
(Uαk
).
17.2. Examples.
1. The plane R
2
is homeomorphic to the product R × R of number lines.
2. Consider in Euclidean space R
3 a rectangular coordinate system Oxyz and define the torus T
2 as the surface of revolution of the circle
(x − 2)2 + z
2 = 1
y = 0
about the axis Oz.
The torus T
2
is homeomorphic to the product S
1 × S
1 of circles.
From Proposition 15.3 it follows.
17.3. Corollary. Let X be the product of the spaces Xα, α ∈ A. A map f : Z →
X is continuous if and only if all the compositions prα ◦ f : Z → Xα are continuous.
17.4. Proposition. The product of the subspaces Yα ⊂ Xα, α ∈ A, coincides with the subspace T
pr−1
α (Yα) : α ∈ A
of the product X =
Q
Xα : α ∈ A
.
Proof. We have
Y =
Y
Yα : α ∈ A
= {x ∈ X : x(α) ∈ Yα, α ∈ A} =
=
\
pr−1
α (Yα) : α ∈ A
⊂
Y{Xα : α ∈ A}.
Let qα = prα|Y : Y → Yα. A subbase of the subspace topology T
pr−1
α (Yα) : α ∈ A
is formed by
the sets Y ∩ pr−1
α (U), where U is open in Xα. But Y ∩ pr−1
α (U) = q
−1
α (U ∩ Yα), and the sets
q
−1
α (U ∩ Yα) form a subbase of the topology of Q
Yα : α ∈ A
.
17.5. Definition (products of maps). Let fα : Xα → Yα, α ∈ A, be maps. Then the map f :
Q
Xα →
Q
Yα, defined by the equalities f(x)(α) = fα
x(α)
, α ∈
A, is called the product of the maps fα and is denoted by Q
fα : α ∈ A
.
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If the index set is finite: A = {1, . . . , k}, then the product of the maps is denoted
by f1 × . . . × fk.
17.6. Proposition. The product f of continuous maps fα : Xα → Yα, α ∈ A,
is continuous.
Proof. Denote by qα the projection of the product Y =
Q
Yα0 : α
0 ∈ A
onto
the factor Yα. From Definition 17.5 the equality follows
qα ◦ f = fα ◦ prα. (17.1)
The map fα ◦ prα is continuous as a composition of continuous maps. Then the continuity of the map f follows from (17.1) and Proposition 15.3.
17.7. Definition (diagonal product of maps). Let fα : X →
Yα, α ∈ A, be maps. Then the map f : X →
Q
Yα : α ∈ A
, defined by the equalities f(x)(α) = fα(x), α ∈ A, is called the diagonal product of the maps fα and is
denoted by ∆{fα : α ∈ A}.
In the case of a finite index set we write f = f1∆ . . . ∆fk.
17.8. Proposition. The diagonal product f of continuous maps is continuous.
Proof. As in the case of the product of maps, we apply Proposition 15.3,
since qα ◦ f = fα.
17.9. Example. The Cantor perfect set. Take the segment I = [0, 1] of the number
line and call it the segment of rank zero. On it take two segments I0 = [0,
1
3
] and I1 = [ 2
3
, 1]
and call them segments of rank one. Call the interval J = ( 1
3
,
2
3
) the interval of rank one.
With each of the segments I0 and I1 we proceed as with the segment I, namely on I0 and I1 we take
two segments of rank two, i.e.
I00 = [0,
1
9
], I01 = [ 2
9
,
1
3
],
I10 = [ 2
3
,
7
9
], I11 = [ 8
9
, 1].
Between them lie the intervals of rank two
J0 = ( 1
9
,
2
9
), J1 = ( 7
9
,
8
9
).
We continue this construction. Suppose 2
n segments of rank n, Ii1...in
(each of the indices
taking the value 0 or 1), have been constructed. Each segment Ii1...in we divide into three equal parts:
the two outer segments Ii1...in0 and Ii1...in1 (the first and third thirds of the segment Ii1...in
) and the interval Ji1...in
lying between
them.
The union of all segments of rank n we denote by Cn. This is a subset of the segment I,
the complement of which consists of all intervals of rank ≤ n. Hence the intersection
C =
T
{Cn : n = 1, 2 . . .}
has as its complement (in the segment I) the union of all intervals Ji1...in
of every rank. Different intervals of this family are pairwise disjoint. Hence, in particular, it follows that the endpoints of all intervals Ji1...in belong to the set C. Moreover, C contains
the points 0 and 1. The set C is called the Cantor perfect set or the Cantor discontinuum.
The Cantor set C is homeomorphic to the Tychonoff product 2
N (the countable power
of the discrete two-point space).
Proof. Different segments of rank n are disjoint. Hence each point
x ∈ C belongs to a unique segment Ii1...in of rank n. Consequently, to each point
x ∈ C there corresponds uniquely a sequence of segments
Ii1 ⊃ Ii1i2 ⊃ . . . ⊃ Ii1i2...in ⊃ . . . ,
and hence a sequence of indices
i1, i2, . . . , in, . . . (17.2)
(of zeros and ones). Thus, assigning to each point x ∈ C the sequence (17.2),
we obtain a map g which, as just shown, is injective, and by the lemma on a sequence of nested contracting segments is surjective, i.e. is a map "onto"
(see Example 16.2).
Let us show that the map g of the Cantor set into the product 2
N with the Tychonoff topology is continuous. For any n ∈ N the composition of g and the projection prn onto the n-th factor is a map
28
of C into the two-point space {0, 1}. (The map g is the diagonal product of the maps g ◦prn, n ∈ N.)
The preimage of zero is the intersection of C with the union of segments of rank n whose indexing i1, i2, . . . , in has in = 0, the preimage of one is the intersection of C with the union of segments of rank n whose indexing
i1, i2, . . . , in has in = 1. Both sets are clopen subsets of C. Hence prn ◦ g
is continuous. To prove the continuity of g it remains to apply Corollary 17.3.
The Cantor set C is compact by the compactness criterion in R
n, and a continuous bijection
of a compactum is a homeomorphism (we will discuss this later). Hence g is a homeomorphism.
Assignment N 4
1. Verify that the following conditions on a bijective map f : X → Y are equivalent:
(a) the map f is a homeomorphism;
(b) a set O is open if and only if the set f(O) is open;
(c) a set F is closed if and only if the set f(F) is closed;
(d) a set O is open if and only if the set f
−1
(O) is open;
(e) a set F is closed if and only if the set f
−1
(F) is closed.
2. Verify that if f : X → Y is a homeomorphism, then for any A ⊂ X we have:
(a) f(ClA) = Cl(f(A));
(b) f(IntA) = Int(f(A));
(c) f(BdA) = Bd(f(A)).
3. Construct homeomorphisms:
(a) [0, 1] onto [a, b], a < b;
(b) (0, 1] onto [0, 1);
(c) (0, 1) onto R.
Prove that [0, 1], [0, 1) and (0, 1) are pairwise non-homeomorphic.
4. Prove that the following spaces are homeomorphic:
(a) R
2
;
(b) {(x, y) ∈ R
2
: x ∈ (0, 1), y ∈ (0, 1)};
(c) {(x, y) ∈ R
2
: x ∈ R, y > 0};
(d) {(x, y) ∈ R
2
: x
2 + y
2 < 1};
(e) {(x, y) ∈ R
2
: x > 0, y > 0};
(f) {(x, y) ∈ R
2
: |x| + y
2 > x};
(g) {(x, y, z) ∈ R
3
: x
2 + y
2 + z
2 = 1} \ (0, 0, 1) (the sphere S
2 without a point).
5. Prove that the following spaces are homeomorphic:
(a) R
2 \ (0, 0);
(b) {(x, y) ∈ R
2
: 0 < x2 + y
2 < 1};
(c) {(x, y) ∈ R
2
: x
2 + y
2 > 1};
(d) R
2 \ {(x, y) ∈ R
2
: x, y ∈ [0, 1]};
(e) R
2 \ {(x, y) ∈ R
2
: x ∈ [0, 1], y = 0}.
6. Prove that the sphere S
n with a point removed is homeomorphic to R
n, n ∈ N.
7. Prove that the spaces Z, Q and R are pairwise non-homeomorphic.
8. A map f : X → Y is called an embedding if f is a homeomorphism of X onto the subspace f(X). Prove that Q does not embed in Z.
9. Prove that the line with the Euclidean topology, the line with the Zariski topology, and the Sorgenfrey
line are pairwise non-homeomorphic.
10. Give an example of a continuous bijective map f : X → Y of spaces X and Y
that is not a homeomorphism. Can one additionally require that X = Y ?
11. Prove that the n-th power of the line R (of the segment I = [0, 1]) is homeomorphic to Euclidean space R
n (the cube I
n to the subspace of Euclidean space R
n), n ∈ N.
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12. Consider in Euclidean space R
3 a rectangular coordinate system Oxyz and
define the torus T
2 as the surface of revolution of the circle
(x − 2)2 + z
2 = 1
y = 0
about the axis Oz.
Prove that the torus T
2
is homeomorphic to the product S
1 × S
1 of circles.
13. For a map f : X → Y the set {(x, f(x)) ⊂ X × Y : x ∈ X} is called the graph of the map. Prove the homeomorphism of the space X and the graph of any of its
continuous map f : X → Y . Is the statement true for an arbitrary map?
14. Embed S
1 × B2
, S
1 × S
1 × I, S
2 × I in R
3
, where B2 is the closed disk of unit radius
in R
2
, S
2 is the two-dimensional sphere.
15. Let A ⊂ X, B ⊂ Y . Verify the following equalities hold:
(a) Int(A × B) = IntA × IntB;
(b) Cl(A × B) = ClA × ClB;
(c) Bd(A × B) = BdA × BdB;
(d) Bd(A × B) = (BdA × B) ∪ (A × BdB);
(e) Bd(A × B) = (ClA × BdB) ∪ (BdA × ClB).
16. Prove that if the product f of maps fα : Xα → Yα, α ∈ A, is continuous, then
each map fα, α ∈ A, is continuous.
17. Prove that if the diagonal product f of maps is continuous, then each map fα, α ∈ A, is continuous.
18. Let X = {f ∈ R
N : the set {n ∈ N : f(n) 6= 0} is finite}. Find the closure of X in R
N.
19. Let (X, ρ) be a metric space, Tρ the metric topology on X. Prove
that the metric ρ is continuous on the square of the space (X, Tρ).
Prove that if the metric ρ (as a map) is continuous on the square of the space
(X, T ), then T ≥ Tρ.
20. Explain the continuity of the binary operations of addition and multiplication on R, of addition and
scalar multiplication in a normed space over the field R.
Additional problems of Assignment N 3
21. Do there exist non-homeomorphic spaces X and Y for which continuous bijections f : X → Y and g : Y → X are defined?
22. Prove that the square {(x, y) ∈ R
2
: x ∈ [0, 1], y ∈ [0, 1]} is homeomorphic to the disk {(x, y) ∈ R
2
:
x
2 + y
2 ≤ 1}.
23. Prove that the space R
n \ R is homeomorphic to S
n−2 × R
2
, n ∈ N, where S
m is the m-dimensional
sphere.
24. T
k = S
1 × · · · × S
1
| {z }
k
is the k-dimensional torus. Embed T
k in R
k+1
.
25. Is every continuous bijection of the line with the standard topology into the line (the Sorgenfrey line)
a homeomorphism?
Describe all homeomorphisms of the line with the standard topology.
Is every bijection of the line with the Zariski topology a homeomorphism?
26. Prove that any closed convex subset of the plane is homeomorphic to a point, or a segment, or a disk, or a ray, or a line, or a strip, or a half-plane, or the plane.
27. The subset of the space `
2
consisting of all points (x1, ..., xk, ...), for which 0 ≤
xk ≤ 1/2
k
, k ∈ N, is called the Hilbert cube in `
2
. Prove that the Hilbert cube is a closed subset of `
2
;
the interior of the Hilbert cube in `
2 is the empty set.
Prove that the Hilbert cube in `
2
is homeomorphic to the Hilbert cube [0, 1]N (the countable power
of the segment).
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