Page 217 - DMTH503_TOPOLOGY

Unit 25: The Nagata-Smirnov Metrization Theorem

Then the sets Notes

U = U and V = V n


n

n Z + n Z +

are disjoint open sets about C and D, respectively.
Lemma 2: Let X be normal, let A be a closed G set in X. Then there is a continuous function f : X

 [0, 1] such that f(x) = 0 for x  A and f(x) > 0 for x  A.
Proof: Write A as the intersection of the open sets U , for n  Z . For each n, choose a continuous
n +
function f : X  [0, 1] such that f(x) = 0 for x  A and f(x) = 1 for x  X – U . Define
n n
n
n
f(x) = f (x)/2 . The series converges uniformly, by comparison with1/2 , so that f is continuous.
n
Also, f vanishes on A and is positive on X – A.
25.1.2 Nagata–Smirnov Metrization Theorem
Statement: A space X is metrizable if and only if X is regular and has a basis that is countably
locally finite.
Proof: Step I: Assume X is regular with a countably locally finite basis . Then X is normal, and
every closed set in X is a G set in X. We shall show that X is metrizable by imbedding X in the

metric space ( ,  ) for some J.
J
Let  =  , where each collection  is locally finite. For each positive integer n, and each basis
n n
element    , choose a continuous function
n
 1 
f : X  0,
n,B   n  
such that f (x) > 0 for x  B and f (x) = 0 for x  B. The collection [f ] separates points from
n,B n,B n,B
closed sets in X: Given a point x and a neighbourhood U of x , there is basis element B such that
0 0
x  B  U. Then B   for some n, so that f (x ) > 0 and f vanishes outside U.
0 n n,B 0 n,B
Let J be the subset of Z ×  consisting of all pairs (n, B) such that B is an element of  .
+ n
Define F : X  [0, 1] J
by the equation F(x) = (f (x)) .
n,B (n,b)J
J
Relative to the product topology on [0, 1] , the map F is an imbedding.
Now we give [0, 1] the topology induced by the uniform metric and show that F is an imbedding
J
1
relative to this topology as well. Here is where the condition f < comes in. The uniform
n,B(x) n
topology is finer (larger) than the product topology. Therefore, relative to the uniform metric,
the map  is injective and carries open sets of X onto open sets of the image space  = F(x). We
must give a separate proof that F is continuous.

J
Note that on the subspace [0, 1] of  , the uniform metric equals the metric
J
((x ), (y )) = sup{|x – y |}
   
To prove continuity, we take a point x of X and a number  > 0, and find a neighbourhood W of
0
x such that
0
x  W  (F(x), F(x )) < 
0

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