Why Can't a Sequence Have Two Distinct Limits in a Metric Space?

[bedi]

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

 

[micromass]

The conditions $|x_n-x_0|<1/n$ imply that $(x_n)_n$ converges to $x_0$. Thus $x_0$ is a limit point of the $(x_n)_n$ (which is the set S). Since limits of sequences are unique in $\mathbb{R}^k$, the limit point $x_0$ is unique.

 

[bedi]

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...

 

[micromass]

If y is a limit point of $(x_n)_n$, then there exists a subsequence $(x_{k_n})_n$ that converges to y. But since the original sequence $(x_n)_n$ converges to x, the subsequence must also converge to x.

So $(x_{k_n})_n$ converges to both x and y. Thus x must equal y.

 

[Bacle2]

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.