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 Aug14 comment Counterexample Check for Sum of Limit Points of Subsequences Wondering @BrianM.Scott Aug14 accepted Counterexample Check for Sum of Limit Points of Subsequences Aug14 comment Counterexample Check for Sum of Limit Points of Subsequences @Rolando posted what looked like a proof of this conjecture, and then deleted it after you posted your answer. It looked something like this: Let $\epsilon >0$ be given. Since $a_n \to c$ and $b_n \to d$, there are $N_1, N_2$ such that for $l \ge N_1$, $k \ge N_2$ $|a_l - c| \le \epsilon/2$ and $|b_k - d| \le \epsilon/2$. Let $m = \max\{N_1, N_2\}$ and therefore we have for $n \ge m$ we have $|a_n - c + b_n - d| \le \epsilon$. Is the problem with his proof that he is not considering subsequences, and that he interpreted the limit points as limits? Aug14 asked Counterexample Check for Sum of Limit Points of Subsequences Aug14 accepted Convergent Subsequence in $\mathbb R$ Aug14 comment Convergent Subsequence in $\mathbb R$ Never mind, of course you are right, because the denominator $\to \infty$ so the sequence $\to 0$, sorry I lost perspective. Thanks again! Aug14 comment Convergent Subsequence in $\mathbb R$ Is it fair to say that since those subsequences are the only ones I have to consider because they are the only two convergent subsequences? Other than ones like $\langle a_{4k} \rangle$ which are contained within the one considered? Aug14 asked Convergent Subsequence in $\mathbb R$ Aug14 comment Proving that $\mu$ is $\sup S$ @JayeshBadwaik then why the $\Longleftrightarrow$? Aug14 comment Proving that $\mu$ is $\sup S$ Thank you. This is clear to me now @PeterTamaroff Aug14 accepted Proving that $\mu$ is $\sup S$ Aug14 revised Proving that $\mu$ is $\sup S$ redefined question Aug14 comment Proving that $\mu$ is $\sup S$ Ah! So Since we assumed that there is no $x \in [\mu -\epsilon, \mu]$ and arrived at $\mu \ne \sup S$ that means that $\mu$ is the supremum? Aug14 comment Proving that $\mu$ is $\sup S$ @BiditAcharya why does $\lambda \notin S \Longrightarrow \mu \ne \sup S$ Aug14 revised Proving that $\mu$ is $\sup S$ asked again Aug14 revised Proving that $\mu$ is $\sup S$ asked again Aug14 comment Proving that $\mu$ is $\sup S$ Ok I found my mistake there thank you @ArthurFischer Aug14 revised Proving that $\mu$ is $\sup S$ fixed mistake in forward direction Aug14 asked Proving that $\mu$ is $\sup S$ Aug14 comment Appropriate Notation: $\equiv$ versus $:=$ @AnonymousCoward see here: tex.stackexchange.com/questions/4216/how-to-typeset-correctly