Heine-borel related proof

1. Aug 18, 2012

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

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2. Aug 18, 2012

micromass

Staff Emeritus
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.

3. Aug 18, 2012

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...

4. Aug 18, 2012

micromass

Staff Emeritus
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.

5. Aug 19, 2012

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.