If
and
are sets
and
is a function:
Let
be
two subsets of
then
1.
2.
More generally
let be
a set of subsets
of
then
1.
2.
Given a point set
,
a topology on is a subset of
,
the set of subsets of
that
satisfy the 4 topology axioms.
We will sometimes use the notation
for a topology
There are two extreme topologies on any set
.
the
indiscrete topology
on
, and
the discrete topology
on
.
Given two topologies
and
,
on
,
if
then we say
is
finer than
.
and
is
coarser than
.
Given two topologies
and
,
(as sets) is also a topology. Note that
may
not be a topology.
Lemma 1:
Given two topologies
and
,
on
and
,the identity function, suppose we consider the left-most
"
"
with the topology
and
the right-most with the topology
then
is continuous if and only if
Proof:
being
continuous amounts to saying that every
open
set is also
open
Given a topological space
,
a point set
,
and a function
,
the collection of sets
, all open
forms a topology on
, and such that
is
continuous with respect to this topology. If
is
the given topology on
we
write
for this "pulled back" topology. That it is a topology follows quickly from
the Boolean Algebra above. Moreover,
Lemma
Given a topological space
,
and a function (set map)
,
then
is
continuous with respect to the topology
if and only if
is
continuous.
If
is
a second topology on such that
is continuous then
Proof:
1.Since
the composition of continuous functions is continuous, what is left to show is
that
is
continuous implies that
is continuous. In particular, we need to show that
is
open for any open set
.
By definition ,
for some open
But then
an open set.
2. Apply 1. to the case
and
apply Lemma 1. above.
Given a topological space
,
a point set
,
and a function
which
is onto, we define a topology on
, called the quotient topology under
,
to be the collection of sets
is
open in
If
is
the given topology on
we
write
for this "pushed out " topology. That it is a topology, again, follows quickly
from the Boolean Algebra above. Moreover,
Lemma
Given a topological space
,
and a function (set map)
,
then
is
continuous with respect to the topology
if and only if
is
continuous.
If
is
a second topology on
such
that
is continuous then
Proof:
1.Since
the composition of continuous functions is continuous, what is left to show is
that
is
continuous implies that
is continuous. In particular, we need to show that
is
open for any open set
.
But then
an open set so , by definition
is
open
2. Again, apply Lemma 1. to
the case