Limit and supremum conceptual question Is it true that for a monotone increasing sequence, the limit of the sequence must be its supremum, but the supremum of the sequence might not be its limit? Else what is the relationship between supremum and limit. Are they the same? How to prove the relationship?
 A: For a monotone increasing sequence, if the limit exists, the limit is the supremum and the supremum is the limit. This is not the case in general. Take for example the sequence: $a_1 = 10$, $a_n=\frac{1}{n}$ for $n> 1$. Then $\lim_{n\to\infty} a_n = 0$, but $\sup_n a_n = 10$. In the rational numbers, denoted by $\mathbb{Q}$, there are monotonically increasing sequences that do not have a limit in $\mathbb{Q}$ and therefore, do not have a least upper bound, either. Take for example, a monotonically increasing sequence of rationals that converges to $\sqrt{2}$, (In the reals). Then, (in the reals), this sequence has a least upper bound, namely $\sqrt{2}$. But in the rational numbers, for any rational $r$ bigger than $\sqrt{2}$, there is always a smaller rational in between $r$ and $\sqrt{2}$. So, for the aforementioned sequence, there is no least upper bound and no limit, (in $\mathbb{Q}$).
A: There are two types of monotone increasing sequences: bounded and unbounded.
If $s_n$ is an unbounded monotone increasing sequence, there is no supremum (or least upper bound) because there is no upper bound in the real numbers.

The second case is that $s_n$ is monotone increasing and bounded. By the Axiom of Completeness, since $s_n$ is bounded we know that there is a supremum of $s_n$ denoated $\sup(s_n)$ such that for all $n$, $s_n \leq \sup(s_n)$. 
Now if our sequence is increasing for each successive term in the sequence, but is always less than the real number $\sup(s_n)$, it seems like the supremum is a reasonable candidate for the limit. 
To prove this, suppose we are given any $\epsilon > 0$. By the definition of the supremum, there exists a natural number $N$ such that $\sup(s_n) - \epsilon < s_N$. Then, since $s_n$ is increasing, for all $n \geq N$, $\sup(s_n) -\epsilon < s_N \leq s_n$
This is saying that given any other other potential upper bound $\sup(s_n)- \epsilon$, we can always find a number $N$ such that every term in the sequence $s_n$ indexed beyond $N$ is greater than the potential upper bound.
However, note that by the definition of supremum $\sup(s_n) - \epsilon < s_n \leq \sup(s_n) < \sup(s_n) + \epsilon$. If we subtract $\sup(s_n)$ from both sides, we get $-\epsilon < s_n - \sup(s_n) < \epsilon$, which by the definition of the absolute value is equivalent to $|s_n - \sup(s_n)| < \epsilon$.
Finally, we realize that this satisfies the definition of the limit: that given any $\epsilon > 0$, we know that there exists a natural number $N$ such that for all $n \geq N$, $|s_n - \sup(s_n)| < \epsilon$.
Therefore, a bounded monotone increasing sequence converges to the supremum.

REFERENCES: http://www.proofwiki.org/wiki/Monotone_Convergence_Theorem_(Real_Analysis)
A: Suppose that$\ $ $lim\ a_n = L$, and that $sup$ {$a_n : n \in \mathbb{N}$} = S.$\ $ Since the sequence is monotone increasing,$\ $ $a_n < L \ \ \forall n$.
If $L < S$,$\ $  $\frac{L+S}{2}$ is an upper bound for {$a_n$} less than the supremum, which is absurd.
If $S < L$, $\ $setting $\ $$\varepsilon = \frac{L-S}{2}$, there exists $n_0$ such that $\ $$|a_n - L| < \frac{L-S}{2}\ \forall n > n_0$, and so you have elements of {$a_n$} greater than S, which is also absurd.
So, the only option left is $\ $S = L.
A: For any monotone increasing sequence in $\mathbb{R}^{\#}$(extended real numbers)  the limit is in $\mathbb{R}^{\#}$ and $\lim_n x_n = \sup_{n\in \mathbb{N}} x_n$ 
Proof: Let  $x_1\le x_2\le \ldots\le x_n\ldots$ in  $\mathbb{R}^{\#}$. Let $x= \sup\{x_n: n\in \mathbb{N}\}$
1) If $x= -\infty$, then $x_n = -\infty$ for all $n$ and so given $L\in \mathbb{R}$ clearly we have $x_n < L$. Thus $\lim x_n = -\infty$.
2) If $x= \infty$. Given $L$, there is some $n_0$ such that $L<x_{n_0}$. Then $n\ge n_0$ clearly implies $L<x_{n_0}\le x_n$, i.e., $\lim x_n = \infty$
3) If $x\in \mathbb{R}$. Given $\varepsilon>0$, there is a $n_0$ such that $x-\varepsilon<x_{n_0}$ and so  for $n\ge n_0$ we have $x-\varepsilon<x_{n_0}\le x_n\le x$ ,i.e., $|x_n-x|<\varepsilon$ for all $n\ge n_0$.
