Heine-borel related proof

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.

 

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.

 

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.