Page 198 - DMTH503_TOPOLOGY

Topology

Notes Proof: As you might expect, we construct D by using Zorn’s lemma. It states that, given a set A
that is strictly partially ordered, in which every simply ordered subset has an upper bound, A
itself has a maximal element.
The set A to which we shall apply Zorn’s lemma is not a subset of X, nor even a collection of
subsets of X, but a set whose elements are collections of subsets of X. For purpose of this proof,
we shall call a set whose elements are collections of subsets of X a “superset” and shall denote it
by an outline letter. To summarize the notation:
c is an element of X.
C is a subset of X.
 is collection of subset of X.
is a superset whose elements are collections of subsets of X.
Now by hypothesis, we have a collection  of subsets of X that has the finite intersection
property. Let  denote the superset consisting of all collections  of subsets of X such that  
and  has the finite intersection property. We use proper inclusion  as our strict partial order
of . To prove our lemma, we need to show that  has a maximal element D.
In order to apply Zorn’s lemma, we must show that if  is a “sub-superset” of  that is simply
ordered by proper inclusion, then  has a upper bound in . We shall show in fact that the
collection
  U  ,
  B
which is the union of the collections belonging to , is an element of ; the it is the required
upper bound on .
To show that  is an element of , we must show that   and the  has the finite intersection
property. Certainly  contains , since each element of B contains . To show that  has the finite
intersection property, let C , ..., C be elements of . Because  is the union of the elements of ,
1 n
there is, for each i, an element  of  such that C  . The superset { ,....,  } is contained in .
i i i i n
So it has a largest element; that is, there is an index K such that B B for i = 1, ..., n. then all the
i K
sets C ...., C are elements of  . Since  has the finite intersection property, the intersection of
1 n k k
the sets C , ..., C is non-empty, as desired.
1 n
Lemma 2: Let X be a set; Let D be a collection of subsets of X that is maximal with respect to the
finite intersection property. Then:
(a) Any finite intersection of elements of D is a element of D.
(b) If A is a subset of X that intersects every element of D, then A is an element of D.
Proof:
(a) Let B equal the intersection of finitely many elements of D. Define a collection of E by
adjoining B to D, so that E = D {B}. We show that Ehas the finite intersection property;
then maximality of D implied that E = D, so that B D as desired.
Take finitely many elements of E. If none of them is the set B, then their intersection is
non-empty because D has the finite intersection property. If one of them is the set B, then
their intersection is of the form
D ,... D B.
1 m
Since B equals a finite intersection of elements of D, this set is non-empty.
(b) Given A, define E = D {A}. We show that E has the finite intersection property from
which we conclude that A belongs to D. Take finitely many elements of E. If none of them
is the set A, their intersection is automatically non-empty. Otherwise, it is of the form
D ... D A.
1 n

192 LOVELY PROFESSIONAL UNIVERSITY