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To prove that every convergent monotone increasing sequence $(a_n)_{n \in N}$ is bounded above by its limit?

We prove by contradiction. Let $l = \displaystyle \lim_{n \rightarrow \infty} a_n.$ Suppose there exists $n_1$ such that $a_{n_1} > l.$ From the definition of the limit for every $\epsilon > 0$, $\exists n_0 \in N$ such that $|a_n - l| < \epsilon$ $\forall n \geq n_0.$
So take $\epsilon = a_{n_1} - l > 0.$ Then $\exists n_2$ such that $|a_n - l| < a_{n_1} - l$, $\forall n \geq n_2.$

So we have $a_n < a_{n_1}, \forall n \geq n_2.$ But i don't know if $n_2 \geq n_1$ or not. If $n_2 \geq n_1$ we have proved it.

Any help will be appreciated:)

Roba
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    Did you check monotone convergence theorem? – beingmathematician May 12 '25 at 08:09
  • it says that a monotone increasing sequence is convergent iff it is bounded. So i prove the other direction that a convergent monotone increasing sequence is bounded by its limit. i can't find a proof for it. – Roba May 12 '25 at 08:11
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    Can you check this https://math.stackexchange.com/a/1398686/872549 – beingmathematician May 12 '25 at 08:17
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    You can definitely take $n_2 >n_1$ since you can replace $n_2$ by any larger integer. – Kavi Rama Murthy May 12 '25 at 08:36
  • I think i can take cases . In case $n_2 \geq n_1$ we are done. In case $n_1 \geq n_2$ we still have a contradiction :) – Roba May 12 '25 at 08:36
  • I do not intend for this to be taken in a negative way, but I am quite surprised by this question given the other questions asked by this same user about functional analysis and Banach space theory. – Dean Miller May 12 '25 at 08:59
  • @DeanMiller Sorry, i suddenly stopped thinking but i am the same person. I just want to make sure of all details and prove every small information on my way. – Roba May 12 '25 at 09:10
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    @Roba That is all good. Don't worry about it. – Dean Miller May 12 '25 at 09:14

2 Answers2

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Here is the other approach mentioned in the comments. Let $(a_{n})_{n\in\mathbb{N}}$ be an increasing sequence in $\mathbb{R}$ that is also convergent. Since convergent sequences are bounded [see here], we have that $(a_{n})_{n\in\mathbb{N}}$ is an increasing sequence in $\mathbb{R}$ that is bounded above. This implies [see here] that the sequence $(a_{n})_{n\in\mathbb{N}}$ converges to $\sup\{a_{k} : k\in\mathbb{N}\}$. Clearly we have $a_{n} \leq \sup\{a_{k} : k\in\mathbb{N}\}$ for each $n\in\mathbb{N}$ and the result follows.

Dean Miller
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Your first steps are correct, but you need to use the other hypotesis: the sequence is monotone increasing.

Let $l$ be the limit of $(a_n)$ and suppose that $n_0$ is such that $a_{n_0} > l$. Let $\epsilon = a_{n_0} - l $.

Because the sequence is monotone increasing, $a_n>a_{n_0}$ for all $n > n_0$. Therefore, $a_{n} - l > \epsilon$ for all $n>n_{0}$, which violates the assumption that $l$ is the limit.

XavierO
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