# The discrete Metric- Completeness and Compactness

• Oct 9th 2010, 11:46 PM
Scopur
The discrete Metric- Completeness and Compactness
Hello, I'm just working through this problem but I am a little shaky with some of the definitions

Let M be a set and let d be the discrete metric on
M.
1. Show that (M, d) is complete.
2. Describe all the compact subsets of M.

Since (M,d) is equipped with the discrete metric all sequences inside that metric converge to some pt that repeats. How is this cauchy?
• Oct 9th 2010, 11:53 PM
yeKciM
Quote:

Originally Posted by Scopur
Hello, I'm just working through this problem but I am a little shaky with some of the definitions

Let M be a set and let d be the discrete metric on
M.
1. Show that (M, d) is complete.
2. Describe all the compact subsets of M.

Since (M,d) is equipped with the discrete metric all sequences inside that metric converge to some pt that repeats. How is this cauchy?

hm... every converging sequence is Cauchy sequence :D

Th : "every compact metric space is complete .... "

if X is compact metric space and let $\displaystyle (x_n)$ be Cauchy sequence in X. because of compactness there is sub sequence $\displaystyle x_{nk}$ of our sequence.. and sub sequence is converging $\displaystyle x_{nk} \to x_0 \in X \;\; (k \to \infty )$ now we have

$\displaystyle d(x_n,x_0) <= d(x_n, x_{nk} ) + d ( x_{nk} , x_0)$

now first on the right side can be made very small because it is Cauchy , and second to because of convergence of the sub sequence :D so we have

$\displaystyle d(x_n,x_0) \to 0 \;\; (n\to \infty)$

meaning sequence $\displaystyle x_n$ is converging and space X is complete :D
• Oct 10th 2010, 01:50 AM
bubble86
for part b) use the fact that given any open cover of a compact set, there exists a finite subcover. Open balls of radius one about x, are denoted $\displaystyle \ B_1 (x)$ and $\displaystyle \forall x \in M$ , $\displaystyle \ B_1 (x) = \{x\}$
so only finite subsets are compact.