Bolzano-Weierstrass bounded Let $a_n$ be a bounded sequence in $R$ that has three subsequences with 1,2 and 3 limit points respectively. Show that there is an $n_0$ in $N$ such that $a_n \leq 4$ for all $n\geq n_0$
Now i know that there is an theorem Bolzano-Weierstrass which says that if $a_n$ is bounded above by $M$ then there exists a sub-sequence $a_{n_k}$ That has a $L$ in the interval [-$M$,$M$]. Now should I just state the theorem or do something specific to proof this?
 A: Suppose there  does not exist such an $n_0$. Then, we claim there is a subsequence $a_{n_k}$ such that $a_{n_k} > 4$ for all $k$. This is because if there are only finitely many $a_n$ having this property, let $N$ be the maximal $n$ for which it happens, and $n \geq N$ will imply $a_n \leq 4$, a contradiction.
Now, $a_{n_k}$ is bounded, say by $M$, where $M > 4$. Then by Bolzano Weierstrass, there must be a limit point of this subsequence in $[4,M]$. But then, since this is again a subsequence of $a_n$, $a_n$ has a limit point in $[4,M]$, a contradiction since $3$ was said to be the maximal limit point of $a_n$. This contradiction means that $a_{n_k}$ couldn't have existed in the first place.
Note : we did not use the fact that $1,2$ are also limit points of $a_n$.
EDIT : I have a direct proof, but I want you to fill in the details.
1) Let $b_k = \sup_{n \geq k} a_n$. Then, $b_n$ is a bounded, increasing sequence.
2) $b_n$ must converge, but in addition, converges to $3$, the largest limit point of $a_n$.
3) By definition, taking $\epsilon = \frac 12$, there exists $K$ such that $|b_k - 3| \leq \frac 12$ for $k \geq K$, since $b_k$ converges to $3$. In particular, by triangle inequality, $|b_k| \leq 3 + \frac 12 \leq 3.5$.
Because $b_K = \sup_{k \geq K} a_n$, by definition of supremum, for all $m \geq K$, $a_m \leq b_K$. However, since $b_K \leq 3.5$ , we have that for $m \geq K$, $a_m \leq 3.5$.

With a similar technique, the following extension can be obtained :

Given a sequence $x_n$ with limit points $a_1,...,a_n$, and any $\delta_1,...,\delta_n > 0$, there exists $n_0$ such that if $n > n_0$ then $x_n \in \cup_{i=1}^n(a_i-\delta_i,a_i+\delta_i)$. In words, eventually every $x_i$ is close enough to at least one of the $a_i$.

