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Suppose $(s_n)$ is a sequence such that $\lim_{n\to \infty} s_n = 7$ and $s_n<7$ for all $n\in \Bbb N$. Let $S=\{s_n\mid n\in\Bbb N\}$, i.e., let $S$ be the set of all values that appear in the sequence $(s_n)$. Prove that $\sup S = 7$.

Can anyone help me out here?

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  • $\begingroup$ A good place to start would be the definitions of $\sup$ and limits. Can you intuitively see why $\sup S = 7$? $\endgroup$ – nukeguy Feb 12 '15 at 22:13
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You are given that $s_n < 7$ for all $n\in \Bbb N$, which implies that $7$ is an upper bound for $S$. On the other hand, since $\lim s_n = 7$, for every $\epsilon > 0$, there exists an $N$ such that $s_N > 7 - \epsilon$. This means that for every $\epsilon > 0$, $7 - \epsilon$ is not an upper bound for $S$. Therefore, $7$ is the least upper bound of $S$, i.e., $\sup S = 7$.

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Hint

  • $s^* = \sup S $ if and only if $s^* \geq \tilde{s}$ for every $\tilde{s}\in S$ and for every $\epsilon >0$ there exist $s \in S$ such that $|s^*-s|<\epsilon$.

  • $\lim_{n\to \infty}s_n = s_*$ if and only if for every $\epsilon > 0$ there exists $N \in \Bbb N$ such that $|s_n-s_*|<\epsilon$ for every $n > N$.

Can you merge these definitions (also using $s_n<7$ for every $n$)?

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  • $\begingroup$ This isn't quite correct - the definition of $\sup S$ also requires $s^*$ to be an upper bound of $S$. $\endgroup$ – Jason Feb 13 '15 at 2:36
  • $\begingroup$ @Jason You are right (and I edited). Anyway, the main idea stays the same. $\endgroup$ – Surb Feb 13 '15 at 9:51

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