.
No attempt is made to give a thorough development. Consult any
advanced undergraduate textbook on analysis or topology for more
details. Our treatment of topology is from a persistently
set-theoretic viewpoint.One of the nice things about posting mathematics on the web is that you can use nifty symbols for things sometimes. Thus, whenever you see a yellow and black striped horizontal bar, it will denote the end of a proof.
We shall consider some elements of topology pertaining to.
No attempt is made to give a thorough development. Consult any
advanced undergraduate textbook on analysis or topology for more
details. Our treatment of topology is from a persistently
set-theoretic viewpoint.
An open set of real numbers is any union of a collection of open
intervals in
.
A closed set of real numbers is any set whose complement is open.
Since there exists a one-to-one function from any open interval onto
,
and since
,
it follows that every open interval has cardinality
.
Therefore, every open set of real numbers has cardinality
.
It is easy to check that any finite set is closed, and we may readily
exhibit countable closed sets, for example, the set is
closed. In order to assess the cardinality of uncountable closed
sets, we require the Cantor-Bendixson Theorem, which we shall develop
shortly. Note that if we assume the continuum hypothesis, then every
uncountable closed set of real numbers has cardinality
.
However, we can show that this is true even in the absence of the
continuum hypothesis.
The rational intervals, that is, open intervals with rational
endpoints, form a basis for the topology of
.
Hence, any open set can be expressed as the union of a collection of
rational intervals. It follows from this that the set of all open
sets of
has cardinality
.
Likewise, the set of all closed sets of
has cardinality
.
A set of real numbers is called compact if it is closed and
bounded. A function
is continuous if
is open for every open set
.
A dense set of reals is a set that is dense in
.
If
are dense open sets of reals, then the intersection
is dense in
.
Proof: In order to prove that
is dense in
,
we need to show that
is nonempty for every nonempty open set
in
.
For this, it suffices to prove that
is nonempty for every nonempty open interval
.
Clearly, for each
,
is dense and open. Thus let
be an enumeration of the rational intervals. Let
,
and for each
let
,
where
is the least such
such
that the closed interval
. Then
where
.
The Baire category theorem appears frequently in analysis. For example, the Banach-Steinhaus theorem and the open mapping theorem are results of Theorem 13. Also, we have the following
The set
of all rationals is not the intersection of a countable collection of
open sets.
Proof: If 
where
are open sets, then each 
is dense in 
.
Thus each 
,
where 
is the nth rational, is dense
and open, hence 
is nonempty.
In the language of analysis,
is not a
.
Our interest in perfect sets is motivated by the need to determine the cardinality of an uncountable closed set of reals. First, some definitions:
Let
be a set of real numbers. A real number
is
a limit point of
if every open set containing
contains some
distinct from
.
A point
is
called an isolated point of
if there exists an open set containing
which contains no other element of
besides
.
A set of real numbers is perfect if it is nonempty, closed, and has no isolated points.
Next we have the following lemma on the cardinality of perfect sets.
Exercise: Prove that the Cantor set is a perfect set
Every perfect set has cardinality
.
Proof: Let
be
a perfect set. We need to find a one-to-one function
.
We construct, by induction on length, for every finite
-sequence
,
elements
of
such that:
if length(s) = length(t)
and
,
then
;
if
,
then
exists;
if
,
then
.
Since
is closed, each
is in
and
is a one-to-one mapping of
into
.
Let
be arbitrary.
If
,
let
and
denote
and
.
To satisfy 4.2 and 4.3 it is enough to make sure that for each
,
and
are close to
.
For instance, it suffices to let
and
,
where
and
We can find
as follows:
Since
is a limit point of
and
is closed, there is some
that satisfies 4.6.
The point here is that we
are trying to show that every closed set of reals is either at most
countable or contains a subset which is perfect and therefore has
cardinality
.
Let
be a strictly descending sequence of closed sets. Then
is countable.
Proof: For each
,
there is a rational interval
that is disjoint from
but not from
.
However, there are only countably many rational intervals.
Every set of reals has at most countably many isolated points.
Proof: Let
.
Let
be an enumeration of the rational intervals. For every isolated point
of
,
let
be the least
such that
is the only element of
in
.
Clearly,
if
are distinct isolated points of
.
If
is
an uncountable closed set, then
,
where
is perfect and
is countable.
Proof: For every
,
let
A' = the set of all limit points of A
(the derivative of
A). It is easy to see that
is closed, and if
is closed then
.
Thus we let
if
is a limit
Since
,
there exists an ordinal
such that
for all
.
(In fact, the least
with this property is countable, by Lemma 4.3.) We let
.
If
,
then
and so it is perfect. Thus the proof is completed by showing that
is at most countable.
We have
;
hence if
,
then there is a unique
such that
is an isolated point of
.
As in Lemma 4.4, we let
denote the least
such that
is the only point of
in the interval
.
Note that if
and
,
then
,
and hence
.
Thus the correspondence
maps
to
is one-to-one, and it follows that
is at most countable.
Finally we have the following Corollary, which is where we have been heading all along. The proof is immediate.
If
is a closed set, then either
or
.