This is post #11 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.
In the previous post Perfect sets and Cantor sets, I, we show that every nonempty perfect set is uncountable. We now show that any perfect contains a Cantor set. Hence the cardinality of any perfect set is continuum.
Given a perfect set , we construct a Cantor set within
. Consider the following cases:
- Case 1.
contains some bounded closed interval
.
- Case 2.
does not any bounded closed interval.
If case 1 holds, then we can apply the middle third process on and produce a Cantor set. So in the remaining dicussion of this post, we assume case 2 holds. This means that for each closed interval
, there is some
such that
.
Background Discussion
Let and
. The point
is a right-sided limit point of
if for each open interval
containing
, the open interval
contains a point of
. The point
is a left-sided limit point of
if for each open interval
containing
, the open interval
contains a point of
. The point
is a two-sided limit point of
if it is both a right-sided limit point and a left-sided limit point of
. For the proof of the following lemma, see the post labeled #10 listed below.
In Lemma 2 below, we apply the least upper bound property and the greatest lower bound property. See the post labeled #4 listed below.
Key Lemmas for Construction
Lemma 1
Suppose that is an uncountable set. Then
contains a two-sided limit point.
As a corollary to the lemma 1, for the perfect set in question, all but countably many points of
are two-sided limit points of
.
Lemma 2
Suppose is a nonempty perfect set that satisfies Case 2 indicated above. Suppose that for the closed interval
, we have:
,
- the left endpoint
is a right-sided limit point of
,
- the right endpoint
is a left-sided limit point of
.
Then we have such that:
- there are no points of
in the open interval
,
- the point
is a left-sided limit point of
,
- the point
is a right-sided limit point of
.
Proof. Since Case 2 holds, for the closed interval in question, there is some
such that
. Then we can find an open interval
such that
and
and
.
Any point in is an upper bound of
. By the least upper bound property,
has a least upper bound
. Any point in
is an lower bound of
. By the greatest lower bound property,
has a greatest lower bound
. Then
and
satisfy the conclusion of the lemma.
Lemma 3
Suppose is a nonempty perfect set. Suppose we have a closed interval
such that the left endpoint
is a right-sided limit point of
and the right endpoint
is a left-sided limit points of
. Then we have
such that:
- the open interval
contains points of
,
- both endpoints
and
are two-sided limit points of
,
.
Proof. Suppose we have a closed interval as described in the lemma. Then
is a nonempty perfect set. Thus
is uncountable. So pick
such that
is a two-sided limit point of
.
Choose open interval such that
and
. Since
is a two-sided limit point of
, choose
and
such that
,
and both
and
are two-sided limit points of
. It follows that
and
satisfy the conclusion of the lemma.
Lemma 4
Suppose is a nonempty perfect set that satisfies Case 2 indicated above. Suppose we have a closed interval
such that the endpoints are two-sided limit points of
. Then we have disjoint closed intervals
and
such that
and
,
- the lengths of both
and
are less then
,
- for each of
and
, the endpoints are two-sided limit points of
.
Proof. This is the crux of the construction of a Cantor set from a perfect set and is the result of applying Lemma 2 and Lemma 3.
Applying Lemma 2 on and obtain the open interval
. We remove the open interval
from
and obtain two disjoint closed intervals
and
. Each of these two subintervals contains points of the perfect set
since the endpoints are limit points of
in the correct direction.
Now apply Lemm 3 to shrink to obtain a smaller subinterval
such that the length of
is less than
and is thus less than
. Likewise, apply Lemma 3 on
to obtain
such that the length of
is less than
and is thus less than
. Note that both
and
constain points of
, making both
and
perfect sets and compact sets.
Construction
Suppose is a nonempty perfect set that satisfies Case 2. Pick two two-sided limit
and
of
. Obtain
and
as a result of applying Lemma 4. Let
.
Then we apply Lemma 4 on the closed interval and obtain closed intervals
,
. Likewise we apply Lemma 4 on the closed interval
and obtain closed intervals
,
. Let
.
Continue this induction process. Let . The set
is a Cantor set and has all the properties discussed in the posts labled #6 and #7 lised at the end of this post.
We claim that . For any countably infinite sequence
of zeros and ones, let
be the first
terms in
. Let
. Then
for some countably infinite sequence
of zeros and ones (see post #6 listed below). Then every open interval
containing
would contain some closed interval
. Thus
is a limit point of
. Hence
.
Links to previous posts on the topology of the real line:
1. The Euclidean topology of the real line (1)
2. The Euclidean topology of the real line (2)
3. The Euclidean topology of the real line (3) – Completeness
4. The Euclidean topology of the real line (4) – Compactness
5. The Cantor bus tour
6. The Cantor set, I
7. The Cantor set, II
8. The Cantor set, III
9. Perfect sets and Cantor sets, I
10. The Lindelof property of the real line
Does anyone know how to prove the following: Any closed interval is the union of c pairwise disjoint perfect sets, where c is the power of the continuum? Would such sets need to be Cantor-like sets? This exercise was given in Natanson’s Volume 1 on Real Analysis.
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