and
are sets, there is some algebra associated with maps (
functions) from
to
If
and
are sets, there is some algebra associated with maps (
functions) from
to
Let
be
two subsets of
then
1.
2.
(see the text, Page 12 Lemma 2.8)
Let
be
two subsets of
then
3.
4.
To see that relationship 3.is inclusion rather than
equality consider the situation where
,and 
Then
and
4. is even more immediate.
Let
in
general
(
)
However, in general, the sets may not be equal. Let
and
be inclusion then
Lemma: If
is
onto then for all
(
)
Lemma: If
is
onto then for all
Which implies that if
then
Lemma:
The finite union of closed sets is closed and the arbitrary intersection of closed sets is closed.
Proof:
Use the boolean algebra of sets. A finite union of closed sets is the complement of a finite intersection of open sets and so on.
Definition:
Let
be
a topological space and
some subset.A point
is
called a limit point of
if for every open set
such
that
there is at least one
and
(
contains at least one point of
other than
)
Lemma:
Let
be a least upper bound for the set
.
Then either
or
is a limit point of
,
perhaps both.
Proof:
Assume
.
for
every
, which is the defining property of a limit point. Suppose not. Choose
such that
and choose
such that
is an upper bound for
.
since if there was an
with
then
contradicting the choice of
.
But if such a
exists then
is
not the least upper bound of
Theorem
A subset
of a topological space
is closed if and only if it contains all its limit points.
Proof:
Suppose
closed that is,
is open. Then any point
in
is not a limit point, since
itself is an open set, containing the point and not
intersecting
.
Suppose a set
contains all its limit points. Choose any
Since
is not a limit point of
we can find an open set
such that
and
. In particular
. So
is open.
Definition:
In
a
closed interval
is
said to be nested in
if
.
Note that this implies
More generally, if
is
nested in
,and
is
nested in
then
and so on.
Theorem:
Let
be an countable sequence of nested closed intervals, then
Proof:
Consider the set
. First note , because the intervals are nested, that
.
Next, since any
is
an upper bound,
has a least upper bound
.
Finally since
is the. least upper bound either
or
is a limit point for
and
hence
since
is
closed. But, again by definition
and
is the least upper bound for
and hence
By the same argument
Thus
Corollary:
In the setting of the previous Theorem, suppose
then
Proof:
Suppose the is a second point
Since for all
,
and
one concludes that
Hence
But then
To be turned in Thursday March 9:
Prove as an exercise. Begin by
defining
in our sense of the term "limit"
Also Due March 9: page 53 of Text, Problems 4.3, 4.4, 4.5, 4.8,and 4.10