Lecture 3
3. Constructing topologies
In this section we discuss several methods for constructing topologies on a given
set.
Definition.
If
T
and
T
are two topologies on the same space
X
, such that
T
⊂
T
(as sets), then
T
is said to be
stronger than
T
. Equivalently, we will say
that
T
is
weaker than
T
.
Remark that this condition is equivalent to the continuity of the map
Id : (
X,
T
)
→
(
X,
T
)
.
Comment.
Given a (nonempty) set
X
, and a collection
S
of subsets of
X
, one
can ask the following:
Question 1
:
Is there a topology on
X
with respect to which all the sets in
S
are open?
Of course, this question has an affirmative answer, since we can take as the topology
the collection of
all
subsets of
X
. Therefore the above question is more meaningful
if stated as:
Question 2
:
Is there the weakest topology on
X
with respect to which all the
sets in
S
are open?
The answer to this question is again affirmative, and it is based on the following:
Remark 3.1.
If
X
is a nonempty set, and (
T
i
)
i
∈
I
is a family of topologies on
X
, then the intersection
i
∈
I
T
i
is again a topology on
X
.
In particular, if one starts with an arbitrary family
S
of subsets of
X
, and if we
take
Θ(
S
) =
T
:
T
topology on
X
with
T
⊃
S
,
then the intersection
top
(
S
) =
T
∈
Θ(
S
)
T
is the weakest (i.e. smallest) among all topologies with respect to which all sets in
S
are open.
The topology
top
(
S
) defined above cane also be described constructively as
follows.
Proposition 3.1.
Let
S
be a collection of subsets of
X
.
Then the sets in
top
(
S
)
, which are a proper subsets of
X
, are those which can be written a (arbitrary)
unions of finite intersections of sets in
S
.
13
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14
LECTURE 3
Proof.
It is useful to introduce the following notations. First we define
V
(
S
)
to be the collection of all sets which are finite intersections of sets in
S
. In other
words,
B
∈
V
(
S
)
⇐⇒ ∃
D
1
, . . . , D
n
∈
S
such that
D
1
∩ · · · ∩
D
n
=
B.
With the above notation, what we need to prove is that for a set
A
X
, we have
A
∈
top
(
S
)
⇐⇒ ∃
V
A
⊂
V
(
S
) such that
A
=
B
∈
V
A
B.
The implication “
⇐
” is pretty obvious. Since
top
(
S
) is a topology, and every set in
S
is open with respect to
top
(
S
), it follows that every finite intersection of sets in
S
is again in
top
(
S
), which means that every set in
V
(
S
) is again open with respect
to
top
(
S
). But then arbitrary unions of sets in
V
(
S
) are again open with respect
to
top
(
S
).
To prove the implication “
⇒
” we define
T
0
=
A
⊂
X
:
∃
V
A
⊂
V
(
S
) such that
A
=
B
∈
V
A
B ,
and we will show that
(1)
top
(
S
)
⊂ {
X
} ∪
T
0
.
By the definition of
top
(
S
) it suffices to prove the following
Claim
:
The collection
T
1
=
{
X
} ∪
T
0
is a topology on
X
, which contains all
the sets in
S
.
The fact that
T
1
⊃
S
is trivial.
The fact that
∅
, X
∈
T
1
is also clear.
The fact that arbitrary unions of sets in
T
1
again belong to
T
1
is again clear,
by construction.
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 Three '10
 Smith
 Topology, Sets, Continuous function, Yi, Empty set, Open set, Topological space

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