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Prove if $\{s_n\}$ is monotonically increasing. Then $\{s_n\}$ converges if and only if it is bounded.


My Proof

Since $\{s_n\}$ is monotonically increasing, $s_n \leq s_{n+1}$.

Part 1: We first assume $s_n \to p$, and show that $\{s_n\}$ is bounded.

Since $s_n \to p$ and $s_n \leq s_{n+1}$, $p$ is an upper bound of $\{s_n\}$, as $p \geq s_n$ for all $n$. To show that it is the least upper bound, take $\epsilon > 0$, then $p - \epsilon < p$ is not an upper bound of $\{s_n\}$ and $p = \sup \{s_n\}$

Now we also have $s_0$ as a lower bound of $\{s_n\}$, as $s_0 \leq s_n$ for every $n$. To show that $s_0$ is the greatest lower bound, we fix $\epsilon > 0$, then $s_0 + \epsilon > s_0$ is not a lower bound of $\{s_n\}$. Therefore $s_0 = \inf \{s_0\}$. And thus $\{s_n\}$ is bounded.

Part 2: Conversely we assume $\{s_n\}$ is bounded and use it to show $s_n \to p$.

$p = \sup \{s_n\}$ and $s_0 = \inf\{s_n\}$ by the same arguments made above. And since $s_n \leq s_{n+1}$ and $p = \sup \{s_n\}$, it follows that $s_n \to p$.

Thus a monotonically increasing $\{s_n\}$ converges if and only if it is bounded. $\square$


Is my proof correct, and if so how rigorous is it? I'm specifically looking for comments on my level of rigor, I don't want my proofs to be hand-wavy or sketchy at all. Also any comments on my proof-writing is greatly appreciated.

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  • $\begingroup$ In the first part 1, you have not really shown that from $s_n \to p,$ you have $p$ as an upper bound of ${s_n}.$ You'd need to rule out that any particular $s_n$ was greater than $p.$ [Of course that is easy, just saying you didn't spell out an argument for it.] $\endgroup$ – coffeemath Nov 6 '16 at 17:47
  • $\begingroup$ @coffeemath, if I added the following, "If we had $s_n > p$ for any $n$, then $s_n \not \to p$ as $s_n \leq s_{n+1}$, and $s_n$ would converge to some $p' > p$, contradicting our assumption. Thus no $s_n$ is greater than $p$ and we have $p \geq s_n$ for all $n$, making $p$ and upper bound of $s_n$.", would that then be rigorous? $\endgroup$ – Proven Nov 6 '16 at 17:59
  • $\begingroup$ Proven: The approach of your last comment would assume the result was true for the subsequence $s_{n+1},s_{n+2},...$ Maybe try using the definition of $s_n \to p$ (along with triangle inequality) to conclude $s_n$ is bounded. $\endgroup$ – coffeemath Nov 6 '16 at 18:03

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