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I am having trouble on understanding a proof from this website https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/02%3A_Sequences/2.05%3A_Limit_Superior_and_Limit_Inferior.

Suppose $\limsup a_n = L$. $(\forall \epsilon \in \mathbb{R}^+) (\exists N \in \mathbb{N}) (\forall n \in \mathbb{N}) (n \geq N \implies | \limsup a_n - L| < \epsilon$. Let $ \epsilon = 1 $, there exists an $N_1 \in \mathbb{N}$ s.t. $1 - \epsilon < \limsup a_{N_1} < 1 + \epsilon$. I am confused on why we can say there exists $n_1 \in \mathbb{N}$ s.t. $1 - \epsilon < a_{n_1}< 1 + \epsilon$. I know that the set $\{a_k : k \geq N_1 \}$ has to be bounded above since $\sup a_{N_1}$ is an upper bound $\forall k \geq N_1$ and $\sup a_{N_1} < L + 1$. Hence, $\sup a_{N_1}$ exists, but what confuses me is that how do we know that $\sup a_{N_1} \in \{a_k : k \geq N_1 \}$.

After that I think I understand the rest of the proof by using the archmidean's property $(\forall \epsilon > 0) (\exists N \in \mathbb{N}) (1/\epsilon < N)$ with $1 - 1/k < a_{n_k} < 1 + 1/k$ and setting k>N.

3j iwiojr3
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We don't know that sup $a_{N_1}\in\{a_k\,:\,k\geq N_1\}$. If, for instance, you take the sequence $a_n=1-(1/n)$, then the sup (1) is never an element of the sequence. Or am I misunderstanding what you meant?

yona
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  • the thing I am confused about is "Let $ \epsilon = 1 $, there exists an $N_1 \in \mathbb{N}$ s.t. $1 - \epsilon < \limsup a_{N_1} < 1 + \epsilon$. I am confused on why we can say there exists $n_1 \in \mathbb{N}$ s.t. $1 - \epsilon < a_{n_1}< 1 + \epsilon$". I am using what the author said. Is this technically wrong since we don't know $\limsup a_{N_1} \in {a_k : k \geq N_1}$ – 3j iwiojr3 Jul 16 '22 at 04:18
  • Here's a different post about this question on math stack https://math.stackexchange.com/questions/581128/prove-that-subsequence-converges-to-limsup – 3j iwiojr3 Jul 16 '22 at 04:18
  • Because there's a chance that $\sup a_{N_1} > a_k \forall k \geq N_1$ – 3j iwiojr3 Jul 16 '22 at 04:21
  • It should be $L-1 < \lim\sup a_{N_1}< L+1$. We don't necessarily have $\lim\sup a_{N_1}\in{a_k,:,k\geq N_1}$. – yona Jul 16 '22 at 04:46
  • As for why such an $N_1$ exists, recall that we are taking a limit, so $\sup_{k\geq n} a_k$ converges to something (it's a monotone decreasing sequence). – yona Jul 16 '22 at 04:50
  • I agree and understand what you're saying, but I don't get how to move on to show that a subsequence of $a_n$ converges. Note, to clarify the title we want to show that a subsequence of $a_n$ converges. – 3j iwiojr3 Jul 16 '22 at 04:59
  • By the definition of supremum, for every $\varepsilon>0$, where $\sup_{k\geq N_1} a_k=s_{N_1}$, there exists a $j\geq N_1$ such that $|s_{N_1}-a_j|<\varepsilon/2$ (create a subsequence out of the $a_j$ while decreasing $\varepsilon$). Since $s_n\to L$ (by definition of $\lim\sup$), there exists an N_2 such that $k\geq N_2$ implies $|s_k-L|<\varepsilon/2$. Use the triangle inequality bound $|a_j-L|<\varepsilon.$ – yona Jul 16 '22 at 05:24
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    Oh! Thank you so much I get the proof now – 3j iwiojr3 Jul 16 '22 at 05:38