# Analysis (convergent sequences) help!

1. Mar 14, 2007

### Pseudo Statistic

I'm having a bit of trouble with two analysis questions, they are:
1) a_n -> a iff every subsequence of {a_n} converges to a

2) a_n->a iff {a_n} is bounded, and a is its only cluster point.

For the first, I was thinking of doing something along the lines of saying that a subsequence of a_n would be a_(f(n)); that is, f being a function that maps the natural numbers onto itself, and has the properly that whenever n>= m, f(n) >= f(m) (nondecreasing) and from there, saying that |a_n - a| < eps given some eps(ilon), for all n> N (Where N is a natural number)... and then arguing that, since f(n) spits out a natural number, then for some f(n) >= N, a_f(n) thus has the limit....

For the second (and probably the first too, but I can't see it?), I'm supposed to make use of the theorem:
i) x is a cluster point of {x_n} <---> for all eps>0, N natural number, there exists an n > N such that |x_n - x| < eps, x real number
ii) x is a cluster point of {x_n} iff there exists a converging subsequence {x_(n_k)} of {x_n} that converges to x.

I'm unsure as to how to approach the second one.. but my idea is..
First, work my way from a_n->a to a_n is bounded, and a is a cluster point, and work backward from there.
However, I'm unsure as to how cluster points and limits are related...
I'd really appreciate it if someone can point out where I could begin, and possibly clarify some things I'm confused about... if possible. :D
Thanks for any replies!

Last edited: Mar 14, 2007
2. Mar 14, 2007

### Dick

I think you know everything you need to prove your questions but you are just sort of muddling everything together. To prove a iff b, you need to prove two things, a implies b and b implies a. So clearly state each separate implication and get started. For example 2) becomes:

a) a_n->a implies a_n bounded.
b) a_n->a implies a is the only cluster point.
c) a_n bounded and a is the only cluster point of a_n implies a_n->a.

Treat each one separately and remember proof by contradiction can be useful.