# Supremum of a non-decreasing sequence is the limit of the sequence

Suppose $(s_n)$ is a non-decreasing sequence of real numbers. Prove that $\sup\{ s_1,s_2,....\} = \lim s_n$

### Attempt:

Suppose $A = \lim s_n$. We show $A = \sup \{ s_n \}$. Let $\epsilon > 0$ be given and choose some $N$ such that $| s_n - A| < \epsilon$ for all $n > N$. We have $s_n < A + \epsilon$. Hence, $A + \epsilon$ is upper bound for the set $\{s_n\}$. In particular, $\sup \{s_n \} \leq A + \epsilon$ and this implies that $\sup{s_n} \leq A$. Also, we have $A - \epsilon < s_n \leq \sup \{ s_n\}$. Therefore, $A \leq \sup \{s_n\}$. Hence, $A = \sup \{ s_n \}$.

Is this a correct approach? It seems I havent used the fact that the sequence is non-decreasing, so I kind of suspicious about my solution.

• You write, "We have $s_n < A + \epsilon$. Hence, $A+\epsilon$ is upper bound for the set $\{s_n\}$." Keep in mind that this is true only for all $n > N$, so you have only shown that $\sup\{s_n : n>N\} \leq A + \epsilon$. – Théophile Feb 26 '15 at 20:07

## 1 Answer

Your solution is correct. You used that $(s_n)$ is nondecreasing to claim that $A + \epsilon$ is an upper bound for the set $\{s_n\}$. Indeed, since $s_n < A + \epsilon$ for all $n > N$ and $(s_n)$ is nondecreasing, for $k = 1,2,\ldots, N$, $s_k \le s_{N+1} < A + \epsilon$. Therefore, $A + \epsilon$ is an upper bound for $\{s_n\}$.