A case of Bolzano-Weierstrass Can anyone help me get started on this problem?
Suppose {$a_n$} is a bounded sequence. Let $L$ = sup{$a_n| n\in \mathbb{N}$}. 
a) Prove that if $L\notin$ {$a_n| n\in \mathbb{N}$} then there is a subsequence of {$a_n$} that converges to $L$.
b) Give an example to show that if if $L\in$ {$a_n| n\in \mathbb{N}$} there is not necessarily a subsequence of {$a_n$} which converges to $L$. 
 A: Lemma $\quad$ $L$ is finite.
Proof $\quad$ Since $(a_n)$ is bounded, there is some nonnegative $M$ such that $a_n\leq|a_n|\leq M$ for all $n$. Hence, $M$ is an upper bound for the sequence. Since $L$ is the least upper bound by definition, $L\leq M<\infty$. Also, $a_n\geq -|a_n|\geq-M$ and $L\geq a_n$, since $L$ is an upper bound. Hence, $L\geq -M>-\infty$. That is, $L$ is finite. $\blacksquare$
Now, by the definition of the supremum (the least upper bound), which is finite because of the lemma, for every positive integer $k$, there exists some integer $n_k$ such that $$(*)\quad L\geq a_{n_k}>L-\frac{1}{k}.$$
Claim $\quad$ If $L\notin\{a_n\,|\,n\in\mathbb{Z}_+\}$, then the set $\{a_{n_k}\,|\, k\in\mathbb{Z}_+\}$ contains infinitely many elements.
Proof $\quad$ To obtain a contradiction, suppose that $\{a_{n_k}\,|\, k\in\mathbb{Z}_+\}$ contains only finitely many elements. Let $k^*$ be such that $a_{n_k{^*}}$ is the greatest of them. Then, for all $k\in\mathbb{Z}_+$, we have $$a_{n_{k^*}}\geq a_{n_k}>L-\frac{1}{k}.$$ Since this is true for all positive integers $k$, we know that $$a_{n_{k^*}}\geq L.$$ But we also know that $$a_{n_{k^*}}\leq L,$$ so that $$a_{n_{k^*}}=L,$$ a contradiction. $\blacksquare$
Since set $\{a_{n_k}\,|\, k\in\mathbb{Z}_+\}$ contains infinitely many elements, we can construct a legitimate subsequence of $(a_n)$ out of it, perhaps after dropping “duplicate indices,” (i.e., such indices $k',k''$ such that $k'\neq k''$ but $n_{k'}=n_{k''}$). With a slight abuse of notation, I will denote this subsequence also as $(a_{n_k})_{k\in\mathbb{Z}_+}$
Now fix $\varepsilon>0$. Let $K$ be an integer so large that $1/K<\varepsilon$. Suppose that $k>K$. Then, $$\big|\,a_{n_k}-L\,\big|=L-a_{n_k}<\frac{1}{k}<\frac{1}{K}<\varepsilon.$$ By definition of convergence, you can see that $(a_{n_k})_{k\in\mathbb{Z}_+}$ converges to $L$.
As for a counterexample, suppose that
\begin{align*}
a_n=\begin{cases}1&\text{if $n=1$,}\\0&\text{if $n>1$.}\end{cases}
\end{align*}
Then, $L=1$, but all subsequences of $(a_n)$ converge to $0$.
A: Let's prove (a) : If $L\notin S := \{a_n\}$, then since $L$ is the supremum, for any $\epsilon > 0$, $L - \epsilon$ is not an upper bound for $S$. In particular, for $\epsilon = 1/k$, there exist $s \in S$ such that
$$
L - 1/k < s \leq L
$$
Write $a_{n_k}$ for this $s$. Now do you see that $a_{n_k} \to L$?
