Page 211 - DMTH503_TOPOLOGY

Unit 24: Local Finiteness and Paracompactness

3. A closed covering of X and locally finite. Notes
4. An open covering of X and locally finite.
Proof: It is trivial that (4) (1).
What we need to prove our theorem is the converse. In order to prove the converse, we must go
through the steps (1) (2) (3) (4)
anyway, so we have for convenience listed there conditions in the statement of the lemma.
(1) (2).
Let  be an open covering of X. Let  be an open refinement of  that covers X and is countably
locally finite; let
 = 
n
where each  is locally finite.
n
Now we apply essentially the same sort of shrinking trick, we have used before to make sets
from different   disjoint. Given i, let
n
V i 
 i 
Then for each n Z and each element of  , define
+ n
S ( )   V i
n

i n
[Note that S ( ) is not necessarily open, nor closed.]
n
Let  = {S ( ) :  }
n n n
Then  =  . We assert that  is the required locally finite refinement of , covering X.
n n
Let x be a point of X. We wish to prove that x lies in an element of , and that x has a neighbourhood
intersecting only finitely many elements of . Consider the covering  =  ; let N be the
n
smallest integer such that x lies in an element of  . Let be an element of  containing x. First,
N N
note that since x lies in no element of  for i < N, the point x lies in the element S () of . Second,
i N
note that since each collection  is locally finite, we can choose for each n = 1, ..., N a
n
neighbourhood W of x that intersects only finitely many elements of  . Now if W intersects
n n n
the element S (V) of  , it must intersect the element V of  , since S (V) V. Therefore, W
n n n n n
intersects only finitely many elements of  . Furthermore, because is in  , intersects no
n N
element of  for n > N. As a result, the neighbourhood
n
W W ...W 
1 2 n
of x intersects only finitely many elements of .
(2) (3). Let  be an open covering of X. Let  be the collection of all open sets of X such that
is contained in an element of . By regularity,  covers X. Using (2), we can find a refinement
 of  that covers X and is locally finite. Let
  {C : C C }

Then  also covers X; it is locally finite by lemma (1) and it refines .
(3) (4): Let  be an open covering of X. Using (3), choose  to be a refinement of  that covers
X and is locally finite. (We can take  to be closed refinement if we like, but that is irrelevant.)
We seek to expand each element B of  slightly to an open set, making the expansion slight
enough that the resulting collection of open sets will still be locally finite and will still refine .

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