Page 208 - DMTH503_TOPOLOGY

Topology

Notes is locally finite in (0, 1) but not in , as in the collection
 = {(1/(n+1), 1/n) | n  }.
+
Lemma 1: Let  be a locally finite collection of subsets of X. Then:
(a) Any sub collection of  is locally finite.

(b) The collection  {A} A  of the closures of the elements of  is locally finite.

(c)  A  A   A  A.
Proof: Statement (a) is trivial. To prove (b), note that any open set that intersects the set A
necessarily intersects A. Therefore, if is a neighbourhood of x that intersects only finitely
many elements A of , then can intersect at most the same number of sets of the collection .
(It might intersect fewer sets of , A and A can be equal even though A and A are not).
2
1
2
1
To prove (c), let Y denote the union of the elements of :
 A  Y.
A 
In general,  A  Y; we prove the reverse inclusion, under the assumption of local finiteness.
Let x Y; let be a neighbourhood of x that intersects only finitely many elements of , say
A ,..., A . We assert that x belongs to one of the sets A , ..., A and hence belongs to  A. For
1 k 1 k
otherwise, the set  A – ...– A would be a neighbourhood of x that intersect no element of 
k
1
and hence does not intersect Y, contrary to the assumption that x Y.
24.1.1 Countably Locally Finite

Definition: A collection  of subsets of X is said to be countably locally finite of  can be written
as the countable union of collections  , each of which is locally finite.
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24.1.2 Open Refinement and Closed Refinement

Definition: Let  be a collection of subsets of the space X. A collection  of subsets of X is said to
be a refinemet of  (or is said to refine ) if for each element B of , there is an element A of 
containing B. If the elements of  are open sets, we call  an open refinement of ; if they are
closed sets, we call  a closed refinement.
Lemma 2: Let X be a metrizable space. If  is an open covering of X, then there is an open
covering E of X refining  that is countably locally finite.
Proof: We shall use the well-ordering theorem in proving this theorem. Choose a well-ordering,
< for collection . Let us denote the elements of  generically by the letters U, V, W,.... .
Choose a metric for X. Let n be a positive integer, fixed for the moment. Given an element of
, let us define S ( ) to be the subset of obtained by “shrinking” a distance of 1/n. More
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precisely, let
S ( ) = {x|B(x, 1/n)  )}.
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(It happens that S ( ) is a closed set, but that is not important for our purposes.) Now we use the
n
well-ordering < of  to pass to a still smaller set. For each in , define
T () = S ( ) V.
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n
V

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