Proving the Discreteness of a Metric Space with Open Closure Property"

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SUMMARY

The discussion centers on proving that a metric space M, where the closure of every open set is open, is discrete. The key conclusion is that every singleton set in M must be open. The approach involves demonstrating that if a point x in M is not open, then a converging sequence in M-{x} would converge to x, which contradicts the property of the metric space. The solution emphasizes that under the given hypothesis, open sets are equal to their closures, leading to the conclusion that all points in M are open.

PREREQUISITES
  • Understanding of metric spaces and their properties
  • Familiarity with the concepts of open and closed sets in topology
  • Knowledge of convergence in metric spaces
  • Basic principles of set theory
NEXT STEPS
  • Study the properties of discrete metric spaces
  • Learn about the relationship between open sets and their closures in topology
  • Explore examples of converging sequences in metric spaces
  • Investigate the implications of the closure of open sets being open in various metric spaces
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Mathematics students, particularly those studying topology and metric spaces, as well as educators looking for examples of discrete metric spaces and their properties.

jin8
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Homework Statement


the problem:
Let M be a metric in which the closure of every open set is open. Prove that M is discrete


The Attempt at a Solution


To show M is discrete, it's enough to show every singleton set in M is open.
For any x in M, assume it's not open,
then there exist a converging sequence in M-{x} converges to x

I want to show such sequence does not exist, but I really don't know how to use the original statement that the closure of open set is open

Thank for help
 
Physics news on Phys.org
First show that under this hypothesis, an open set is in fact equal to its own closure. Then with the right choice of open set it's not hard to see that points are open.
 

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