# Difference between revisions of "Metrizable implies monotonically normal"

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## Latest revision as of 14:48, 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 (i.e., metrizable space) must also satisfy the second topological space property (i.e., monotonically normal space)

View all topological space property implications | View all topological space property non-implications

Get more facts about metrizable space|Get more facts about monotonically normal space

## 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.