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Problem 4: $(a_n)$ bdd. All $a_{n_j}$ converges to a common $\alpha$. Does $(a_n)$ converge??

\begin{proof}
Assume that $(c_n)$ doesn't converge to $\alpha$, the common limit of all convergent subsequences. Then there is an $\varepsilon _0 \textgreater 0 $ such that for all $N\in\mathbb{N}$, there is a corresponding term $n_N \textgreater N$ such that $|a_{n_N}-\alpha |\geq \varepsilon _0$. 

Now, we form a subsequence
\begin{align}
c_{n_1}, \quad c_{n_{n_1}}, \quad c_{n_{n_{n_1}}},...\notag
\end{align}
and denote it by $(b_j)_{j=1}^\infty$.  If $(b_j)$ contains infinitely many terms such that 
\begin{align}
b_j\geq \alpha + \varepsilon _0 ,
\end{align}
then, we ignore the opposite terms and form a subsequence $(b_{j_k})_{k=1}^\infty$. If $(b_j)$ contains only finitely many terms such that (1) holds, then we ignore these terms and keep the opposite terms, forming the subsequence $(b_{j_k})_{k=1}^\infty$. In these cases, we have $b_{j_k} \geq \alpha +\varepsilon_0$ and $b_{j_k} \leq \alpha -\varepsilon_0$ respectively. Bolzano-Weierstrass Theorem  yields that we have a convergent subsequence 
\begin{align}
(b_{j_{k_l}})_{l=1}^\infty \notag
\end{align}
and its limit must $\geq \alpha +\varepsilon_0 \textgreater \alpha $ ($\leq \alpha -\varepsilon_0 \textless \alpha$ respectively). But condition says that the limit equals $\alpha$, a contradiction. 
\end{proof}

[Note] This exercise indicates that if a  bounded sequence $(a_n)$ has merely one accumulation point $a$, then $a_n\to a$. But if not bounded, for example, $a_{2k+1} = (-k)^k$ while $a_{2k} = (-\frac{1}{k})^k$, this statement is not necessarily true. 

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