Heine-borel related proof

  • Thread starter bedi
  • Start date
  • #1
81
0

Main Question or Discussion Point

I'm trying to understand that proof with little success. Particularly I don't understand why "S has x_0 as a limit point, and S has no other limit point in R^k". Please help
 

Attachments

Answers and Replies

  • #2
22,097
3,277
The conditions [itex]|x_n-x_0|<1/n[/itex] imply that [itex](x_n)_n[/itex] converges to [itex]x_0[/itex]. Thus [itex]x_0[/itex] is a limit point of the [itex](x_n)_n[/itex] (which is the set S). Since limits of sequences are unique in [itex]\mathbb{R}^k[/itex], the limit point [itex]x_0[/itex] is unique.
 
  • #3
81
0
Alright, but can you explain it with basic topology, like neighbourhoods and the definition of a limit point? I don't see why any other interior point of S couldn't be a also a limit point in that case. Rudin says that it's because it would violate the triangular inequality theorem, but still it's not clear enough...
 
  • #4
22,097
3,277
If y is a limit point of [itex](x_n)_n[/itex], then there exists a subsequence [itex](x_{k_n})_n[/itex] that converges to y. But since the original sequence [itex](x_n)_n[/itex] converges to x, the subsequence must also converge to x.

So [itex](x_{k_n})_n[/itex] converges to both x and y. Thus x must equal y.
 
  • #5
Bacle2
Science Advisor
1,089
10
By basic properties of the (standard) real numbers, any two real numbers that are

indefinitely-close to each other, e.g., d(x,y)<1/n for all n , then x=y by,e.g., the

Archimedean Property. Use the triangle inequality to show that, in a metric space,

if a sequence has two limits L1, L2, then L1 is indefinitely-close to L2.
 

Related Threads for: Heine-borel related proof

Replies
2
Views
6K
Replies
9
Views
1K
Replies
2
Views
946
Top