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Compact subset of R^n

by bxn4
Tags: compact, subset
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bxn4
#1
Mar5-08, 02:56 AM
P: 2
I am struggling to prove the following: Let E be a compact nonempty subset of R^k and let delta = {d(x,y): x,y in E}. Show E contains points x_0,y_0 such that d(x_0,y_0)=delta.
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jostpuur
#2
Mar5-08, 03:17 AM
P: 2,065
Quote Quote by bxn4 View Post
I am struggling to prove the following: Let E be a compact nonempty subset of R^k and let delta = {d(x,y): x,y in E}. Show E contains points x_0,y_0 such that d(x_0,y_0)=delta.
Am I correct to guess that you meant the supremum

[tex]
\Delta := \sup\{d(x,y)\;|\;x,y\in E\}\; ?
[/tex]

It is convenient to consider a function [itex]d:E\times E\to\mathbb{R}[/itex], and use some basic topological results, or their immediate consequences. For example: The Cartesian product of compact sets is a compact set. In metric spaces compact sets are sequentially compact. The distance function d is continuous. Continuous mappings map compact sets into compact sets. The Heine-Borel Theorem. Just put pieces together!
bxn4
#3
Mar5-08, 11:57 AM
P: 2
Quote Quote by jostpuur View Post
Am I correct to guess that you meant the supremum

[tex]
\Delta := \sup\{d(x,y)\;|\;x,y\in E\}\; ?
[/tex]

It is convenient to consider a function [itex]d:E\times E\to\mathbb{R}[/itex], and use some basic topological results, or their immediate consequences. For example: The Cartesian product of compact sets is a compact set. In metric spaces compact sets are sequentially compact. The distance function d is continuous. Continuous mappings map compact sets into compact sets. The Heine-Borel Theorem. Just put pieces together!
Yes, it is the supremum. I am using the fact that E is compact to show that there is a subsequence that converges in E. Then I'd want to say that the limit of d(x_n_j, y_n_j) is
\Delta but not sure how to show it. We have not talked about continuous functions. We have only studied sequences so far.

thanks


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