Convergent Subsequences in Compact Metric Space

[Chipz]

Homework Statement

Suppose that [tex](x_n)[/tex] is a sequence in a compact metric space with the property that every convergent subsequence has the same limit [tex]x[/tex]. Prove that [tex]x_n \to x[/tex] as [tex]n\to \infty[/tex]

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 [tex]s_n[/tex] in [tex](x_n)[/tex] s.t. [tex]s_m =\displaystyle\sum\limits_{k=1}^{m} x_k \to y \neq x[/tex]

Then there would exist an [tex]m>a>0[/tex]

Under the assumption that [tex]s_a = \displaystyle\sum\limits_{k=a}^{\infty} x_k \to x[/tex]

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

Then the sequence [tex](x_n)[/tex] 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 [tex](n_1>n_2>n_3...) \in \mathbb{N}[/tex]

Suppose [tex]lim x_n = x[/tex]

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

[tex]n_j \ge j[/tex] [tex]\forall j[/tex] by induction and [tex]n_1 \ge 1[/tex] because [tex]n_1 \in \mathbb{N}[/tex]

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

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

 

[ystael]

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

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

(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.)