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Is possible for a sequence (made of real numbers) to be bounded, but not having a maximum value, or a limit for this maximum value?

I'm trying to figure if there is something "in between" an open and a closed interval. For example, [a,b] has the same maximum limit than [a,b), so b is the maximum of [a,b] and the limit of [a,b).

By "bounded", I mean the supremum (or infimum for "a"), not just any large value which is simply out of the set.

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    No. Every subset of $\mathbb{R}$ that is bounded above has a supremum. – 2017-02-14
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    @ NeedForHelp but that supremum may be excluded from the interval and not be a limit from the values on the interval? – 2017-02-14
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    If the supremum is not reached then you can approach it as close as you want from elements in the set. It *is* the limit of a sequence of points in the set. – 2017-02-14
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    @NeedForHelp What if the supremum is the smallest number guaranteed not to be in the sequence, but is fundamentally impossible to know what the maximum limit is? (I don't know if that is possible) – 2017-02-14

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Lemma: Let $E$ be a subset of $\Bbb{R}$ and suppose that $E$ is bounded above. Then there exists a sequence $(a_n)$ of elements in $E$ that converges to $\alpha := \sup E$.

Proof: For every $n$, there exists $a_n \in E$ such that $$ \alpha-\frac{1}{n} \leq a_n \leq \alpha \tag{$\star$} $$ Indeed, there must exists an $a_n \in E$ such that $\alpha-\frac{1}{n} \leq a_n$ (otherwise $\alpha-\frac{1}{n}$ would be an upper bound for $E$) and since $\alpha$ is an upper bound for $E$, it must also be the case that $a_n \leq \alpha$.

In view of $(\star)$, the squeeze theorem shows that $(a_n)$ converges to $\alpha$. $\blacksquare$

Now, for your question, let $(a_n)$ be your bounded sequence. This means that the subset $A := \{a_n : n \in \mathbb{N}\}$ consisting of the elements of $(a_n)$ is bounded in $\mathbb{R}$. Applying the above lemma with $E = A$, we see that if $A$ has no maximum then at least its supremum is the limit of a subsequence of $(a_n)$.

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    Ok, but you answer depends on the assumption that the sequence has an upper limit. What if there exists a sequence which does not haves an upper limit (yet is bounded). The interval may be defined as all the points in between any elements of that sequence. – 2017-02-14
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    A supremum for a subsequence does not means that a larger subsequence does not crosses that supremum. – 2017-02-14
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    Note that I asked that the sequence has no limit for his maximum value, (not a limit). That means that there is no limit for $max(a_i)$ as $n \to \infty$ – 2017-02-14
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    @zugatevej **1st comment:** Please define what you mean by "upper limit of a sequence". Also what interval are you talking about?! **2nd comment:** In my answer $\alpha$ is the supremum for the whole sequence $(a_n)$... **3rd comment:** Please define the terms "maximum value of a sequence" and "limit for the maximum value of a sequence". What is $\max(a_i)$ ?! What is the limit of that (which is independent of $n$ !) when $n \to \infty$ ?! – 2017-02-14
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    1st: well, a sequence may oscillate, so the upper limit would be the limit of the maximum value it takes on any finite subset, or $\lim_{n \to \infty} max(a_i)$ for $0\leq i \leq n$ – 2017-02-14
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    About the interval, that sequence may not fill the line (for example, a fractal), so to simplify, the interval may be defined to be continuous, (because I only are interest on his extremes). The interval are all the points between any values in the sequence. If there is an i,j such that $a_i \leq p \leq a_j$, then p is in the interval, and the interval is the union of all possible p, for any i,j – 2017-02-14
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    Or the sequence may be fundamentally chaotic, so is not possible to probe that it has a maximum value without calculating all the infinite values. (Not being practicable to calculate the limit does not probes that the limit exists or not exists). But I don't know if there are chaotic sequences for which a limit does not exist but are bounded. – 2017-02-14
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    @zugatevej No matter how "chaotic" $(a_n)$ is, if bounded, then the associated sequence $b_n := \displaystyle\max_{0 \leq i \leq n}a_i$ will be increasing and bounded above, hence convergent. The value to which $(b_n)$ converges is precisely $\alpha := \sup\{a_n : n \in \Bbb{N}\}$. In general there is no reason why $\alpha$ should be of a known standard form. If you can't get such a standard form (like $a/b$, $e^{a}$, $\log a$, etc.) then you can just define a notation for $\alpha$ (which is what is done when defining $e$ as the limit of $\displaystyle\sum_{k=0}^n\frac{1}{k!}$ for example). – 2017-02-14
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    If $a_n=2-\frac{2}{n}$, then is false that for every $n$, there exists $a_n \in E$ such that $\alpha-\frac{1}{n} \leq a_n$ – 2017-02-15
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    @zugatevej No, it is **not false**. In that case $\alpha = 2$ and given any $n$ just take $a_n := 2-\frac{2}{2n}$. (The letter $n$ is overused so maybe it would be clearer to ask if for every $m$, there exists $a_m$ such that $\alpha-\frac{1}{m} \leq a_m$. As I said, the particular choice $a_m := 2-\frac{2}{2m} = 2 - \frac{1}{m}$ works and gives equality.) – 2017-02-15
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    I do not understand that argument. Maybe you want to rewritten the answer? Maybe you mean that for every n, there exists an m such than $a_m$ .... something – 2017-02-15
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    @zugatevej I say that given your sequence $(a_n)$ you can find a term as close as you want to $\sup\{a_n:n\in\Bbb{N}\}$. If $a_n=2-2/n$ then $\sup$ is $2$ Now if I want to find an element of that sequence that is at most $1/m$ apart from $2$, I need to solve the inequality $2-1/m \leq 2-2/n$. This is equivalent to $n\geq2m$. So any element $a_n$ of the sequence with $n\geq2m$ works. I chose $n_0=2m$ and defined $a_m = a_{n_0}$. We choose index $m$ to say that $a_m$ is at most $1/m$ away from the $\sup$. Write it $a^m$ if you don't want to confuse $a_m$ with the $m$-th term of the sequence. – 2017-02-15