The Lindelof property of the real line
May 29, 2010 at 10:38 pm | Posted in Basic Topology, Introductory topology, Lindelof space, The real line, Well known examples | Leave a commentTags: General topology, Hereditarily lindelof, Introductory topology, Lindelof space, Point-set topology, The real line, Topology
This is post #10 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 real line, the set of all integers (positive, negative and zero) is a closed set as well as a discrete set. It is closed because the complement is an open set. It is discrete since no point of
is a limit point of
. It turns out that in the real line, only countable sets can be both closed and discrete. Equivalently, the real line satisfies the following property:
Every uncountable subset has a limit point.
A set is discrete if no point of
is a limit point of
. The set
is a discrete set, as is the set
. However, unlike
, the set
is not a closed set. It turns out that in the real line, only countable sets can be discrete. Equivalently, the real line satisfies the following property:
Every uncountable subset contains one of its limit points.
It follows that for any uncountable subset of the real line, all but countably many points of
are limit points of
.
We show that both the statements and
are intimately tied to the Lindelof property. Specifically, both
and
hold for the real line because the real line itself and every subset of the real line has the Lindelof property.
We also prove that for every uncountable subset of the real line, there is at least one
such that
is a two-side limit point of
. It follows that for every uncountable subset
of the real line, all but countably many points of
are two-sided limit points of
.
Definitions
Let be a topological space. A collection
of subsets of
is said to be a cover of
if every point of
belongs to some set in
. The collection
is said to be an open cover if it is a cover of
and it consists entirely of open subsets of
. If
is an open cover of
and if
also covers
, then
is said to be a subcover of
.
The space is said to be a Lindelof space (or has the Lindelof property) if every open cover of
has a countable subcover. The space
is said to be hereditarily Lindelof if every subspace of
is a Lindelof space.
Theorem 1
Let be a topological space.
- If
is Lindelof, then
satisfies the property
.
- If
is hereditarily Lindelof, then
satisfies the property
.
Proof. We prove the contrapositive, that is, if has an uncountable subset that has no limit point, then
is not Lindelof. Let
be uncountable such that
has no limit point. Then
is a closed set. For each
, there is an open set
such that
and
contains no point of
other than
. Then the following is an open cover of
:
Note that only countably many cannot cover
. Thus
is an open cover of
that has no countable subcover, showing that
is not Lindelof.
To show 2, we also prove the contrapositive. Let be some uncountable set such that no point of
is a limit point of
. Then for each
,
is an open set relative to
. Thus the subspace
is not Lindelof, showing that
is not hereditarily Lindelof.
Theorem 2
Every subset of the real line has the Lindelof property, i.e.
is hereditarily Lindelof.
Remark. This follows from Lemma 4 in the post labeled #4 listed below. This theorem follows from the fact that the Euclidean topology of the real line is generated by a countable base (specifically, the open intervals with rational numbers as endpoints).
Corollary 3
The real line satisfies properties and
. Thus we have:
- Every uncountable subset of the real line has a limit point.
- Every uncountable subset of the real line contains one of its limit points.
Remark
This follows from Theorem 1 and Theorem 2. It also follows that if is an uncountable subset of the real line, all but countably many points of
are limit points of
.
Let and
. The point
is said to be right-sided limit point of
if for each open interval
such that
, we have
. The point
is said to be left-sided limit point of
if for each open interval
such that
, we have
. The point
is said to be a two-sided limit point of
if it is both a right-sided and left-sided limit point.
Based on the discussion in the previous post on completness (the post labeled #3 below), we can assume the least upper property whenever we work with the real line. This property states that for every subset of the real line, if
is bounded above, then
has a least upper bound (the upper bound that is the smallest among all the upper bounds). This assumption simplies the proof of the next theorem. The least upper bound property is equivalent to the greatest lower bound property, which states that for every subset
of the real line, if
is bounded below, then
has a greatest lower bound (the lower bound that is the greatest among all the lower bounds).
Theorem 4
Every uncountable subset of the real line has at least one point that is a two-sided limit point.
Proof. Suppose that is an uncountable set such that none of its points is a two-sided limit point. By corollary 3, all but countably many points of
is a limit point of
. So by removing these countably many points if necessary, assume that every point of
is a limit point of
.
For each ,
is either a right-sided limit point of
or a left-sided limit point of
but not both. So either
there are uncountably many points of
that are left-sided limit points of
or
there are uncountably many points of
that are right-sided limit points of
. Assume
and use the least upper bound property. The proof assuming
is similar and uses the greatest lower bound property.
Let be the set of all points of
that are left-sided limit points of
. By assumption
,
is uncountable. For each
, choose a rational number
such that
contains no point of
. Then there must be some rational number
such that
for uncountably many
. Thus the following set
is uncountable:
Note that the rational number is an upper bound of
. By the least upper bound property, let
be the least upper bound of
. Choose some
such that
. This is possible since
for all
and
is uncountable. There must be some
such that
. Otherwise,
would be an upper bound. Since
is a left-sided limit point of
,
contains infinitely many points of
. But
is supposed to contain no points of
, a contradiction.
Under assumption , the proof is similar and uses the greatest lower bound property. Thus we establish that in each uncountable subset of the real line, one of its points must be a two-sided limit point. As a corollary to this fact, all but countably many points of an uncountable set of real numbers must be two-sided limit points of the set.
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
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