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

jin8
Messages
24
Reaction score
0

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top