Existence of a Maximal Element of the Set of Subsequential Limits of a Bounded Sequence OK, so I've been burned by this all day now and I've given up.
Supposing that we have a bounded sequence, I cannot grasp how the maximal element (as my professor put it) could exist if we have a sequence, say $z_n$, of subsequential limits that converges, satisfies boundedness and also has some $m>0$ for which all $M>m$ allow $z_M$ to be greater than any other subsequential limit besides those forming the convergent sequence I'm talking about whose index is greater than $M$. That is, at some point in this sequence of limits of convergent subsequences, we reach a point where the only subsequential limits greater than the one we are at are precisely those for which the index is greater than the index we are currently at (i.e., $z_m>z_n\iff m>n$ and $z_m$ is a member of the convergent sequence of subsequential limits I'm talking about). Isn't this just the same as saying that there exists no maximal element of $(0,1)$, because for any element $a<1$ one picks, we can find a $1>b>a$?
As an aside, how would this extend to a general metric space? 
I'm essentially at my wit's end here. Could anyone explain this to me? Thanks!
Edit:
The precise formulation from my notes:

Let $(a_n)\subset\mathbb{R}$ be a bounded sequence consider the set $S$ defined to be the set of all possible limits of convergent subsequences of $(a_n)$. Then $\max(S)$ exists.

Clarification:
I believe that the example I provided is a case where the maximum (if (?) the maximum is indeed what my professor meant) cannot exist. Here is an attempt to clarify what I stated originally.
Let $S$ be the set of all possible limits of convergent subsequences of a bounded sequence in $\mathbb{R}$ and suppose that
$$(\forall{m\in\mathbb{N}})(\exists N>0\in\mathbb{R})\ \text{  such that }\ |a_m|\le N$$
Additionally, assume that our set of the limits of subsequences of $(a_n)$ (i.e., the set $S$) has,  itself, a subset of subsequences, $B\subset{S}$, whose limits we order into a sequence, say $(b_n)$, that converges to a number $b\in\mathbb{R}$ where $|b|\le{N}$ with the property that $$(\forall{x,y\in\mathbb{N}})(b_x>{b_y}\iff{x>y})$$ and that for some $m'\in\mathbb{N}$, any subsequence $(c_k)$ of $(a_n)$ converging to a limit $c\ge{b_{m'}}$ implies that $(c_k)\in{B}$.
I contend that such a case permits no maximum because it is somewhat analogous to asking about the maximum member of the open interval $(0,1)$.
Thanks again!
 A: All right, all right, first a solution to the problem. I'll leave a few things for you to prove, but shouldn't be too hard, if you need help let me know and I'll expand later. It will also be a little long, but nothing too hard to follow.

Let $(a_n)\subset\mathbb{R}$ be a bounded sequence consider the set $S$ defined to be the set of all possible limits of convergent subsequences of $(a_n)$. Then $\max(S)$ exists.

Note $$C = \{ (a_{n_k}) : \text{is a subsequence of  } a_n \text{ that converges to some value in } \mathbb{R} \}$$ and
$$A = \limsup a_n = \lim_{n \to \infty} \sup_{k \geq n} a_n$$
To see $A$ exists, let $A_n = \sup \{a_k \colon k \geq n\}$, then the $A_n$ form a decreasing sequence of values in $\mathbb{R}$ wich are bounded below and thus converge to some value when $n \rightarrow \infty$.
Note that if the sequences wasnt bounded $\limsup$ could even be $-\infty$! (and $+\infty$). The good thing about $\limsup$ is that it always defined (on the extended reals of course).
Now, let $(a_k)$ be a sequence from $C$, we have the following properties of limsup: (These are the things I wont prove. )


*

*$\lim a_{n_k} = \limsup a_{n_k}$

*If $A = \limsup a_n$ then given $\varepsilon > 0$, the set $\{a_n > \limsup - \varepsilon\}$ is infinite and the set $\{a_n > \limsup + \varepsilon\}$ is finite.

*If the index set of the sequence $(n_k) \subseteq (n_j)$ for another index, that is the sequence $(a_{n_j})$ has all the elements from $(a_{n_k})$ in the same order plus maybe a few more, then $$\limsup_{n_k \to \infty} a_{n_k} \leq \limsup_{n_j \to \infty} a_{n_j} \leq \limsup_{n \to \infty} a_n$$

*There exists some subsequence wich converges to $\limsup$


With this done its is now trivial that, $C$ has a maximum value and its $\limsup a_n$
Now, with respect to your counter example, indeed, you can have a lot of values in the middle (You wont be always be able to order all the subsequential limits in an ordered sequence, because you can write a sequence wich has subsequences to all values of $\mathbb{R}$, and you cant count $\mathbb{R}$) But you could order them with $\leq$. But because of what I showed, you see that $\limsup$ is greater than everyone (that is all subsequential limits, you can still have finitely many values above $\limsup$ by $2.$) and you have a subsequences that converges to it, so the maximum must be in $C$.
