Difference between revisions of "Metrizable implies monotonically normal"
(→Proof of monotonicity) 
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The disjointness of <math>U</math> and <math>V</math> follows from the triangle inequality. It suffices to prove that for every <math>a \in A</math> and <math>b \in B</math>, <math>U_a \cap V_b = \emptyset</math>. To see this, note:  The disjointness of <math>U</math> and <math>V</math> follows from the triangle inequality. It suffices to prove that for every <math>a \in A</math> and <math>b \in B</math>, <math>U_a \cap V_b = \emptyset</math>. To see this, note:  
−  <math>d(a,b) \  +  <math>d(a,b) \ge d(a,B), \qquad d(a,b) \ge d(A,b)</math> 
Thus:  Thus:  
−  <math>d(a,b) \  +  <math>d(a,b) \ge d(a,B)/2 + d(A,b)/2</math> 
This, along with the triangle inequality yields that the balls <math>U_a</math> and <math>V_b</math> cannot intersect.  This, along with the triangle inequality yields that the balls <math>U_a</math> and <math>V_b</math> cannot intersect. 
Revision as of 14:47, 6 July 2019
This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property must also satisfy the second topological space property
View all topological space property implications  View all topological space property nonimplications

Contents
Statement
Any metrizable space is monotonically normal. In fact, we can construct an explicit monotone normality operator using the metric.
Definitions used
Metrizable space
Further information: Metrizable space
Monotonically normal space
Further information: Monotonically normal space
Proof
Construction of the operator
Suppose is a metric space. We construct a monotone normality operator on as follows. For any two closed subsets and of :
 For every , let be the open ball about of radius . Here, is the infimum of the distances for . Note that since is closed and , .
 Analogously, for every , define to be the open ball about of radius .
Then, the sets:
are the required disjoint open subsets.
Proof of disjointness
The disjointness of and follows from the triangle inequality. It suffices to prove that for every and , . To see this, note:
Thus:
This, along with the triangle inequality yields that the balls and cannot intersect.
Proof of monotonicity
We need to show that if and , then the open set about becomes smaller, and the open set about becomes larger. This is clear from the definitions, because has fewer balls, and the radii of the balls also become smaller. Similarly has more balls, and the radii of the balls also become larger.