Proving the Existence of a Supremum in a Compact Subset of R^n

  • Thread starter Thread starter bxn4
  • Start date Start date
  • Tags Tags
    Compact
bxn4
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
bxn4 said:
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

<br /> \Delta := \sup\{d(x,y)\;|\;x,y\in E\}\; ?<br />

It is convenient to consider a function d:E\times E\to\mathbb{R}, 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!
 
jostpuur said:
Am I correct to guess that you meant the supremum

<br /> \Delta := \sup\{d(x,y)\;|\;x,y\in E\}\; ?<br />

It is convenient to consider a function d:E\times E\to\mathbb{R}, 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
 
Back
Top