# separability, the countable chain condition and the lindelof property

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```SEPARABILITY, THE COUNTABLE CHAIN CONDITION
AND THE LINDELOF PROPERTY IN
LINEARLY ORDERABLE SPACES
D. J. LUTZER AND H. R. BENNETT
1. Introduction.
It is known that, in metric spaces, separability,
the Lindelof property and the countable chain condition1 (abbreviated CCC) are each equivalent to second countability.
Furthermore,
any space which is separable or hereditarily
Lindelof satisfies the
CCC (though not conversely).
In this paper, we examine separability, the Lindelof property and the CCC in linearly orderable topological spaces (abbreviated LOTS), i.e., spaces whose topology is the
open interval topology of some linear ordering of the underlying set.
We shall prove that if X is a LOTS, then
(A) X satisfies the CCC if and only if X is hereditarily Lindelof;
(B) X is separable if and only if X is hereditarily
separable.
Result (A) was obtained for connected LOTS by T. Inagaki ([5]
and [6]) and somewhat
later by G. Kurepa
[9]. The methods of
Inagaki and Kurepa rely heavily on connectedness;
our proofs avoid
this assumption.
(Also, see Remark 4.3, below.) Result (B) was
announced without proof by G. Kurepa in [9]; L. Skula provided a
proof in [14], but our proof is shorter.
The following notation will be convenient.
If (X, ^) is a linearly
ordered set and if a£X, then ]— =o, a[= {x£X|x<a}
and ]— co, a]
= {x£X|x^a|.
The sets ]a, °° [ and [a, co [ are defined analogously.
2. The countable chain condition in LOTS.2 The following lemma
is easily proved by transfinite
induction.
(2.1) Lemma. Suppose X is a LOTS satisfying the CCC and suppose
FCJ,
Let zEY. Then there are countable subsets P and Q of ] — oo , z]
r\ Y and [z, oo [f~\ Y respectively such that if yEY, then there are points
pEP and qEQ such that p^y^q.
(2.2) Theorem. A LOTS satisfying the CCC is hereditarily Lindelof.
Proof. Suppose that X is a LOTS satisfying the CCC. To show
that X is hereditarily Lindelof, it will suffice to show that if V is any
collection of open intervals in X, then there is a countable subcollecPresented to the Society, January 25, 1969 under the title Covering properties of
linearly ordered spaces and Souslin's problem; received by the editors April 18,1969.
1 A space X satisfies the countable chain condition if any disjoint collection of open
sets is countable.
2 The authors are indebted to Mary Ellen Rudin for suggesting the approach used
in this section.
664
SEPARABILITY,THE COUNTABLECHAIN CONDITION
tion llCTj
665
which covers the set UT) (cf. [4, p. 141, exercise B]).
Let F=Ul3. For each x£ F, let J(x) = {y£ Y\ the set of all points
of F lying between x and y can be covered by a countable subcollection of V}. It is clear that each set I(x) is open in X and that for any
two points x,yEY, either I(x) = I(y) or else I(x)(~\I(y) = 0. Because
A satisfies the CCC, the collecion \$= {/(x)|x£
Y} is countable, say
3 = {7(x)|x£C}
where C is some countable subset of Y.
Fix x£C. Using (2.1), find countable subsets P(x) and Q(x) of
I(x)C\]— oo, x] and 7(x)P\[x,
oo [ respectively
such that if y£/(x),
then there are points pEP(x)
and qEQ(x) such that p^y^q.
By
definition of I(x), ior each pEP(x)
and g£<2(x) there are countable
subcollections V(x, p) and V(x, q) of V which cover the sets YC\ [p, x]
and Yf\[x, q] respectively.
Let 1l = U{u(x, r)\ rEP(x)\jQ(x)
and
x£C}. Then It is a countable subcollection of V which covers Y.
3. Separability in LOTS. Lemma (3.1) follows from (2.2) above;
alternatively,
it may be deduced from Lemma (3.2) below (which
can be proved by modifying the proof that a LOTS is normal given
in [3, p. 39]).
(3.1) Lemma. Suppose that X is a LOTS satisfying the CCC and
that Y is a discrete subspace of X (discrete in the relative topology). Then
Y is countable.
(3.2) Lemma. Any LOTS is hereditarily
(3.3) Theorem.
Proof.
collectionwise normal.3
A separable LOTS is hereditarily
separable.
Suppose that A is a separable LOTS and that iCJ.
Let
1(A) = {a£yl|
{a} is relatively open in A }. By (3.1), 1(A) is countable. Let D he a countable dense subset of X. Let
S>= []r,s[\r,sE
For each interval
D,r < s,
/£SD,
I(A)\J{a(J)\JE'£>}-
and
An]r,s[^0}.
choose a point a(J)EA(~\J.
Let D(A) =
Clearly, D(A) is a countable subset of A. To
prove that D(A) is dense in A, it will suffice to show that if U is open
in A and if Af\U^0,
a£/(,4),
then D(A)C\U^0.
then aED(A)C\U.
If a\$I(A),
Choose a£.4rw.
If
then either
(i) ]— oo, a[^0
and ]x, a[P\yl 7^0 whenever x<a, or else
(ii) ]a, oo [^0
and ]a, y[C\A 7^0 whenever y>a.
We consider only the first case. Since U is a neighborhood of a in X,
there is a point z<a such that ]z, a]QU. Applying (i), we choose
* Added in proof. A detailed proof of this result by Professor L. A. Steen will appear
in these Proceedings.
666
D. J. LUTZER AND H. R. BENNETT
points b £]z, a[f~\A;cE]b, a[(~\A; anddEJc,
in X and since ]z, c[^09^}c,
sE]c, a[f\D.
Letting J=]r,
[December
a[C\A. Since D is dense
a[, there are points r£]z, c[C\D and
s[, we obtain an element of £>. Then
a(J)E]z, a[nD(A)QUr\D(A).
4. Examples. In this section, we will attempt to show how our
results can be applied in general topology.
(4.1) Example.
Souslin spaces. Recall that a Souslin space is a
nonseparable
LOTS which satisfies the CCC. It was recently proved
that the existence of a Souslin space is independent of the axioms of
set theory ([lO] and [15]). It follows easily from a result of Mary
Ellen Rudin in [13] that if there is a Souslin space, then there is a
compact, connected Souslin space. Together with our Theorem (2.2),
this shows that Problem 4.7 of [l] cannot be answered affirmatively.
Similarly, Problem 4.6 of [l] cannot be answered affirmatively, since
Ponomarev
[12] and Bennett
[2] have proved that if there is a
Souslin space, then there is a Souslin space with a point-countable
base.
(4.2) Example.
Let X be the space i?x{0,
l} with the lexico-
graphic order topology: see [8] for details. (X is called the double
line.) It is clear that X is a separable LOTS. Hence, X is hereditarily
separable and hereditarily
Lindelof.
Let £ be the topology on R obtained
by taking the collection
{[a, b[\a, bER and a<b} as a base. The space (R, £), called the
Sorgenfrey line, is homeomorphic
to a subspace (the "top line of
points") of the double line. It follows from (2.2) and (3.3) that (R, £)
is hereditarily
separable,
hereditarily
Lindelof
and perfectly
normal
(cf. [8, p. 134, Example K]). It should be pointed out that (R, £) is
not itself a LOTS [ll].
(4.3) Remark.
We observe that the Souslin space with a pointcountable base mentioned in (4.1) cannot be (locally) connected by
[2, Theorem 15]. Also, the double line of (4.2) cannot be embedded in
a (locally) connected LOTS satisfying
show that the results of Inagaki and
certain applications.
(4.4) Remark
on a paper
the CCC. These two facts
for
of F. B. Jones.
Professor F. Burton
Jones has asked the authors to point out an error which appeared in
one of his papers [7]. The five conditions listed on p. 627 are not
equivalent, even in a connected LOTS, as our examples and the example given in [8, p. 164, Example K] show. In any LOTS, Jones'
property (2) is equivalent to the CCC and to the hereditary Lindelof
property. In a connected LOTS, Jones' properties (1), (4) and (5) are
equivalent.
1969]
SEPARABILITY,THE COUNTABLECHAIN CONDITION
5. Generalizations.
number.
667
In this section, m denotes an infinite cardinal
(5.1) Definition.
A space X is m-separable ii X contains a dense
subset having cardinality ^ m; X satisfies the m-chain condition if
any disjoint collection of open subsets of X has cardinality
^ m; X is
m-Lindelbf if every open cover of X has a subcover with cardinality =w.
Using the methods of §2 and §3, we can prove
(5.2) Theorem.
A LOTS satisfies the m-chain condition if and only
if it is hereditarily m-Lindeldf.
(5.3) Theorem.
An m-separable
LOTS is hereditarily
m-separable.
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University of Washington and
Texas Technological
College