This might seem like a strange topic but right now our topology class is studying compact metric spaces. One of the more famous of these is the “Cantor set” or “Cantor space”. I discussed the basics of these here.
Now if you know the relationship between a countable product of two point discrete spaces (in the product topology) and Cantor spaces/Cantor Sets, this post is probably too elementary for you.
Construction: start with a two point set
and give it the discrete topology. The reason for choosing 0 and 2 to represent the elements will become clear later. Of course,
is a compact metric space (with the discrete metric:
if
.
Now consider the infinite product of such spaces with the product topology:
where each
is homeomorphic to
. It follows from the Tychonoff Theorem that
is compact, though we can prove this directly: Let
be any open cover for
. Then choose an arbitrary
from this open cover; because we are using the product topology
where each
is a one or two point set. This means that the cardinality of
is at most
which requires at most
elements of the open cover to cover.
Now let’s examine some properties.
Clearly the space is Hausdorff (
) and uncountable.
1. Every point of
is a limit point of
. To see this: denote
by the sequence
where
. Then any open set containing
is
and contains ALL points
where
for
. So all points of
are accumulation points of
; in fact they are condensation points (or perfect limit points ).
(refresher: accumulation points are those for which every open neighborhood contains an infinite number of points of the set in question; condensation points contain an uncountable number of points, and perfect limit points are those for which every open neighborhood contains as many points as the set in question has (same cardinality).
2.
is totally disconnected (the components are one point sets). Here is how we will show this: given
there exists disjoint open sets
. Proof of claim: if
there exists a first coordinate
for which
(that is, a first
for which the canonical projection maps disagree (
). Then
,
are the required disjoint open sets whose union is all of
.
3.
is countable, as basis elements for open sets consist of finite sequences of 0’s and 2’s followed by an infinite product of
.
4.
is metrizable as well;
. Note that is metric is well defined. Suppose
. Then there is a first
. Then note
which is impossible.
5. By construction
is uncountable, though this follows from the fact that
is compact, Haudorff and dense in itself.
6.
is homeomorphic to
. The homeomorphism is given by
. It follows that
is homeomorphic to a finite product with itself (product topology). Here we use the fact that if
is a continuous bijection with
compact and
Hausdorff then
is a homeomorphism.
Now we can say a bit more: if
is a copy of
then
is homeomorphic to
. This will follow from subsequent work, but we can prove this right now, provided we review some basic facts about countable products and counting.
First lets show that there is a bijection between
and
. A bijection is suggested by this diagram:

which has the following formula (coming up with it is fun; it uses the fact that
:
for
even
for
odd
for
odd, 
for
even, 
Here is a different bijection; it is a fun exercise to come up with the relevant formulas:

Now lets give the map between
and
. Let
and denote the elements of
by
where
.
We now describe a map
by

Example: 
That this is a bijection between compact Hausdorff spaces is immediate. If we show that
is continuous, we will have shown that
is a homeomorphism.
But that isn’t hard to do. Let
be open;
. Then there is some
for which
. Then if
denotes the
component of
we wee that for all
(these are entries on or below the diagonal containing
depending on whether
is even or odd.
So
is of the form
where each
is open in
. This is an open set in the product topology of
so this shows that
is continuous. Therefore
is a homeomorphism, therefore so is
.
Ok, what does this have to do with Cantor Sets and Cantor Spaces?
If you know what the “middle thirds” Cantor Set is I urge you stop reading now and prove that that Cantor set is indeed homeomorphic to
as we have described it. I’ll give this quote from Willard, page 121 (Hardback edition), section 17.9 in Chapter 6:
The proof is left as an exercise. You should do it if you think you can’t, since it will teach you a lot about product spaces.
What I will do I’ll give a geometric description of a Cantor set and show that this description, which easily includes the “deleted interval” Cantor sets that are used in analysis courses, is homeomorphic to
.
Set up
I’ll call this set
and describe it as follows:
(for those interested in the topology of manifolds this poses no restrictions since any manifold embeds in
for sufficiently high
).
Reminder: the diameter of a set
will be 
Let
be a strictly decreasing sequence of positive real numbers such that
.
Let
be some closed n-ball in
(that is,
is a subset homeomorphic to a closed n-ball; we will use that convention throughout)
Let
be two disjoint closed n-balls in the interior of
, each of which has diameter less than
.

Let
be disjoint closed n-balls in the interior
and
be disjoint closed n-balls in the interior of
, each of which (all 4 balls) have diameter less that
. Let 

To describe the construction inductively we will use a bit of notation:
for all
and
will represent an infinite sequence of such
.
Now if
has been defined, we let
and
be disjoint closed n-balls of diameter less than
which lie in the interior of
. Note that
consists of
disjoint closed n-balls.
Now let
. Since these are compact sets with the finite intersection property (
for all
),
is non empty and compact. Now for any choice of sequence
we have
is nonempty by the finite intersection property. On the other hand, if
then
so choose
such that
. Then
lie in different components of
since the diameters of these components are less than
.
Then we can say that the
uniquely define the points of
. We can call such points 
Note: in the subspace topology, the
are open sets, as well as being closed.
Finding a homeomorphism from
to
.
Let
be defined by
. This is a bijection. To show continuity: consider the open set
. Under
this pulls back to the open set (in the subspace topology)
hence
is continuous. Because
is compact and
is Hausdorff,
is a homeomorphism.
This ends part I.
We have shown that the Cantor sets defined geometrically and defined via “deleted intervals” are homeomorphic to
. What we have not shown is the following:
Let
be a compact Hausdorff space which is dense in itself (every point is a limit point) and totally disconnected (components are one point sets). Then
is homeomorphic to
. That will be part II.