# Convergent Subsequences in Compact Metric Space

1. Aug 30, 2010

### Chipz

1. The problem statement, all variables and given/known data
Suppose that $$(x_n)$$ is a sequence in a compact metric space with the property that every convergent subsequence has the same limit $$x$$. Prove that $$x_n \to x$$ as $$n\to \infty$$

2. Relevant equations
Not sure, most of the relevant issues pertain to the definitions of the space. In this case I believe the following is relevant:
In a compact metric space every sequence must have a convergent subsequence, defining it as sequentially compact.

3. The attempt at a solution
My basic hang up is this: Does every subsequence have to be convergent? If so...you can.

Suppose not:
Assume there exists a subsequence $$s_n$$ in $$(x_n)$$ s.t. $$s_m =\displaystyle\sum\limits_{k=1}^{m} x_k \to y \neq x$$

Then there would exist an $$m>a>0$$

Under the assumption that $$s_a = \displaystyle\sum\limits_{k=a}^{\infty} x_k \to x$$

Where $$s_m = \displaystyle\sum\limits_{k=1}^{m+a} x_k$$ Would not be convergent.

Then the sequence $$(x_n)$$ is not Cauchy, which implies it's not a Complete Metric Space.

2. Aug 30, 2010

### lanedance

how about contradiction? may need some tightening but general idea is there

assume xn doesn't tend to x, then either there it does not converge or it tends to y not equal to x

if it tends to y its clearly a contradication

now if it does not converge it still visits x infinitely often, but it must also be in the neighbourhood of some other point infinitely often (compactness) which is also a contradiction

3. Aug 30, 2010

### Chipz

That's essentially the proof I gave. Although I've revised it a little to be more true for subsequences rather than partial sums.

Let $$(n_1>n_2>n_3...) \in \mathbb{N}$$

Suppose $$lim x_n = x$$

given any $$\epsilon > 0$$ we must find an $$N$$ s.t. $$j \ge N$$ then $$|x_n_j - x| < \epsilon$$ $$\forall n \ge N$$

$$n_j \ge j$$ $$\forall j$$ by induction and $$n_1 \ge 1$$ because $$n_1 \in \mathbb{N}$$

if $$n_j \ge j \to n_{j+1} \ge n_j \ge j \to n_{j+1} > j +1$$ so if $$j\ge N$$ $$n_j \ge N$$ then $$|x_n_j -x| > \epsilon$$

And thus every subsequence needs to converge to the same value.

4. Aug 30, 2010

### ystael

This appears to be a (confusingly written) proof that if $$x_n \to x$$ then every subsequence of $$(x_n)$$ also converges to $$x$$. Unfortunately that statement is the converse of what you are required to prove.

Begin with the hypothesis that every subsequence of $$(x_n)$$ converges to $$x$$, and suppose that $$x_n \not\to x$$. What does it mean for $$x_n$$ not to converge to $$x$$? (What is the negation of the statement that $$(x_n)$$ is eventually in every neighborhood of $$x$$?)

(Use words! If your argument isn't essentially correct, symbols and abbreviations don't make it any clearer; they just make it harder to understand what needs fixing.)