{Sn} is convergent -> {|Sn|} is convergent

[Jamin2112]

{Sn} is convergent ---> {|Sn|} is convergent

Homework Statement

I need to prove that if {s_n} is convergent, then {|s_n|} is convergent.

Homework Equations

s_n is convergent if for some s and all ε > 0 there exists a positive integer N such that |s_n - s| < ε whenever n ≥ N.

The Attempt at a Solution

Proof. By contrapositive. Suppose {|s_n|} is not convergent. Then for all s there exists an ε > 0 such that ||s_n| - s|| ≥ ε for all n.

... I need to somehow show that this implies that {s_n} does not converge.

Maybe some fancy triangle inequality thing like

ε ≤ ||s_n| - s| ≤ ||s_n| - s_n| + |s_n - s|

Wat do, PF?

 

[Jamin2112]

Ah, wait. I think I figured out something.

ε ≤ ||s_n| - s| ≤ ||s_n| - s_n| + |s_n - s| ≤ 2|s_n| + |s_n - s|

Am I any closer?

 

[MaxManus]

You are supposed to show that |Sn| converges to |S| not to S.
The trick is to show that ||Sn|-|S|| <= |Sn-S|

Hint to showing it:
|S| = |Sn + (S-Sn)|
And use the triangle inequality.

 

[Jamin2112]

You are supposed to show that |Sn| converges to |S| not to S.
The trick is to show that ||Sn|-|S|| <= |Sn-S|

Hint to showing it:
|S| = |Sn + (S-Sn)|
And use the triangle inequality.

I knew the triangle inequality would show his face somewhere.

 

[Jamin2112]

http://images.icanhascheezburger.com/completestore/2009/3/5/128807888821558038.jpg

 

[Jamin2112]

Hey, I need a hint for another problem. (You guys give stellar hints. Pat yourselves on the back.)

Suppose {p_n} is a Cauchy sequence in a metric space X, and some subsequence {p_ni} converges to a point p in X. Prove that the full sequence {p_n} converges to p.

As always, I'd love to use contradiction.

Suppose {p_n} does not converge to p. Then for some ε > 0, d(p,p_n) ≥ ε for all n. Then d(p_m,p_n)+d(p_m,p) ≥ ε for all m, n. But {pm} is a Cauchy sequence, so with the proper choice of m, n we'll have d(p_m,p_n) < ε. ... blah blah I hope this is going somewhere. A little hint, maybe?