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Aug
15
asked Proving that a sequence is Cauchy
Aug
14
comment Counterexample Check for Sum of Limit Points of Subsequences
Ah ok. Thank you all for your help, I am self-studying introductory real analysis and slowly getting the hang of the material.
Aug
14
comment Counterexample Check for Sum of Limit Points of Subsequences
Wondering @BrianM.Scott
Aug
14
accepted Counterexample Check for Sum of Limit Points of Subsequences
Aug
14
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?
Aug
14
asked Counterexample Check for Sum of Limit Points of Subsequences
Aug
14
accepted Convergent Subsequence in $\mathbb R$
Aug
14
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!
Aug
14
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?
Aug
14
asked Convergent Subsequence in $\mathbb R$
Aug
14
comment Proving that $\mu$ is $\sup S$
@JayeshBadwaik then why the $\Longleftrightarrow$?
Aug
14
comment Proving that $\mu$ is $\sup S$
Thank you. This is clear to me now @PeterTamaroff
Aug
14
accepted Proving that $\mu$ is $\sup S$
Aug
14
revised Proving that $\mu$ is $\sup S$
redefined question
Aug
14
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?
Aug
14
comment Proving that $\mu$ is $\sup S$
@BiditAcharya why does $\lambda \notin S \Longrightarrow \mu \ne \sup S$
Aug
14
revised Proving that $\mu$ is $\sup S$
asked again
Aug
14
revised Proving that $\mu$ is $\sup S$
asked again
Aug
14
comment Proving that $\mu$ is $\sup S$
Ok I found my mistake there thank you @ArthurFischer
Aug
14
revised Proving that $\mu$ is $\sup S$
fixed mistake in forward direction