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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!

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  • $\begingroup$ I think you're talking about the limit superior, but with the wording you used is extremely dificult to understand what you want to say. $\endgroup$ – Aram Nov 4 '14 at 19:00
  • $\begingroup$ @Aram I believe you're right, from what I could tell upon looking to see if my question is a duplicate. I will edit my post to include the precise formulation from my notes in class. My professor is brilliant, but occasionally his style of lecture mimics that of a child in a candy shop: one moment we will be talking about Ramsey Theory, and the next Real analysis. So it is possible that I hastily scribbled down the formulation in a less-than-precise manner! Or, are you referring to my wording? I will attempt to clear up both. $\endgroup$ – Nobody Nov 4 '14 at 19:23
  • $\begingroup$ max(S), always exists when the sequence is bounded and I repeat is called the limit superior. What I'm still not getting from the edit is, $B$ is a set that has the limits of all subsequential limits right? When you mean $(c_k) \in B$ you mean that the limit of the sequence $c_k$ is in $B$?. I dont see how that last part is true, because you could pick some subset of $S$ of increasing values wich converged to $b$, and then just take away the limit of $c_k$ and that would still be true. Are you trying to construct $B$ or show its existence? $\endgroup$ – Aram Nov 4 '14 at 22:14
  • $\begingroup$ @Aram sorry, let $B$ be the set of subsequences, and let $(b_n)$ be the sequence of their limits. So suppose that such a $B$ and $(b_n)$ exist. I think that is a counter example to the claim. $\endgroup$ – Nobody Nov 4 '14 at 23:52
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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. )

  1. $\lim a_{n_k} = \limsup a_{n_k}$
  2. 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.
  3. 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 $$\limsup_{n_k \to \infty} a_{n_k} \leq \limsup_{n_j \to \infty} a_{n_j} \leq \limsup_{n \to \infty} a_n$$
  4. 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 $C$.

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  • $\begingroup$ So I agree that we can always find $A$. I believe 1. should follow because $a_{n_{k}}$ is a convergent sequence, hence, $\limsup\limits_{n_k\rightarrow\infty} a_{n_{k}}=\liminf\limits_{n_k\rightarrow\infty} a_{n_{k}}$. 2. works because is the latter set were not finite, then we could construct a convergent subsequence to some point ${a_n>\limsup+\epsilon}$ 3. I'd need to think about, but since all elements of a subsequence are elements of $(a_n)$ it couldn't possibly be the case that a subsequence had a supremum greater than that of the original sequence Continued below... $\endgroup$ – Nobody Nov 5 '14 at 1:14
  • $\begingroup$ @nihil 3. comes from the property that if $A \subseteq B$ then $sup A \leq sup B$. $\endgroup$ – Aram Nov 5 '14 at 1:17
  • $\begingroup$ because that would be the new supremum of $(a_n)$ and I suppose that same argument would then work down the line as well. I'm beginning to suspect that I may well have overdone this. As I said, supposing that $\limsup_{n_k \to \infty} a_{n_k}>\limsup_{n \to \infty} a_{n}$, then there are points in the sequence $a_{n_{k}}$ for which as $n_k \to \infty$ are always greater than the points in $a_n$ as $n\to\infty$, but since every point in the former is in the latter, it must be that $\endgroup$ – Nobody Nov 5 '14 at 1:25
  • $\begingroup$ I think I might just be a little too tired to be doing this today! Thank you for your help, I'll sleep on it a little and let you know. I appear to have accidentally pressed return without enter above, but I think I'm intuiting the same thing as you remarked. In fact, I think my 'counterexample' is rather confused thinking right now... $\endgroup$ – Nobody Nov 5 '14 at 1:44
  • $\begingroup$ No I'm still confused hahaha... So what if we take a sequence with no accumulation points in the interval $(0,\frac{1}{2}]$ and combine it with a sequence $(a_n)$ where $$a_{m}=\begin{cases} \left(\left[\sum\limits _{i=1}^{m}\frac{1}{i}\right]\mod1\right)+\frac{1}{2}, & \left(\left[\sum\limits _{i=1}^{m}\frac{1}{i}\right]\mod1\right)<\frac{1}{2}\\ \left(\left[\sum\limits _{i=1}^{m}\frac{1}{i}\right]\mod1\right), & \left(\left[\sum\limits _{i=1}^{m}\frac{1}{i}\right]\mod1\right)\ge\frac{1}{2} \end{cases}$$? What would be the maximum here? I'm not seeing it in this case. $\endgroup$ – Nobody Nov 5 '14 at 2:10

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