Show that the sequence {$a_n$} converges to $a \in \mathbb {R}$ $\leftrightarrow$ all real subsequence {$a_{n_k}$} of {$a_n$} converge to $a$ I need help with the following two tasks:
a) Show that the sequence {$a_n$] converges to a $\in \mathbb {R}$ $\leftrightarrow$ all proper subsequence {$a_{n_k}$} of {$a_n$} converge to a.
Well the right direction $\rightarrow$ is easy to proof. If {$a_n$} converges to $a$, it is bounded. After the "Bolzano-Weierstraß-theorem" each bounded sequence has a convergent subsequence. Obviously, it converges to a too (right?). But now the problem is to proof that each real subsequence converge to $a$.
$\leftarrow$ Well thats the biggest problem for me. We are talking about proper subsequences, so we can't use {$a_n$} as a  subsequence of {$a_n$}.The only thing that I could imagine is to say: Let {$a_n$} be cauchy. If the subsequence of a couchy-sequence converges to $a$ for $n_k$ $\rightarrow$ $\infty$. Then {$a_n$} converges to $a$ too.
b) Let {$b_n$} be a sequence with the following property: Each subsequence of {$b_n$} has another subsequence, that converges to $b$. Show that {$b_n$} converges to $b$.
Well I guess if we prove a), we are able to conclude b).
I am thankful for any advice.
 A: For one of the directions in (a), if $a_n$ does not converge to $a\in\mathbb R$, then there exists $\epsilon_0 > 0$ such that $|a_n-a|>\epsilon_0$ for infinitely many bad $n$. Enumerating an increasing subsequence $n_1<n_2<\dots$ of such $n$, what can you say about the subsequence $\{a_{n_k}\}_{k=1}^\infty$?
It seems like you've got a strategy for (b).
A: Concerning a):
For the "$\implies$" direction, you do not need use Bolzano-Weierstraß to get a converging subsequence. Instead, you have to take any (!) subsequence of $(a_n)$ and prove that it converges to $a$.
Hint: $a_n \to a$ means that the difference $\lvert a_n - a \rvert$ converges to $0$. (That is the definition of convergence, right?) Can you show that $\lvert a_{n_k} - a \rvert$ converges to $0$ for $k \to \infty$?
For the "$\impliedby$" direction, I think that you didn't see a subtlety which makes the proof very easy.
Hint: $(a_n)$ itself is a subsequence of $(a_n)$.
Concerning b):
As far as I'm concerned, this does not follow directly from a), but is harder. I suggest you try a contraposition: Suppose $(b_n)$ does not converge to $b$. (What does this mean, by definition of convergence?) Can you construct a subsequence of $(b_n)$ from this, which has no sub-sub-sequence converging to $b$? Then you have proven the contraposition of the statement.
