Convergent Subsequences in Compact Metric Space

[Chipz]

Homework Statement

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

Homework 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.

I'll add more.

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.

 

[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

 

[Chipz]

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

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.

 

[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.)