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Aug
16
asked “So That” vs. “Such That”
Aug
15
accepted Limit point versus limit and implications for convergence
Aug
15
comment Limit point versus limit and implications for convergence
So that sequence I have constructed works as a counterexample then. Thanks @DavidMitra
Aug
15
comment Limit point versus limit and implications for convergence
Does a sequence of this form "count"?: $(a_n) = 1$ for $n = 2k, k \in \mathbb N$, $(a_n) = n$ for $n = 2k - 1, k \in \mathbb N$?
Aug
15
asked Limit point versus limit and implications for convergence
Aug
15
accepted Proving that a sequence is Cauchy
Aug
15
comment Proving that a sequence is Cauchy
Thanks! I see the convergent series, which gives me the $N$ I need to pick.
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 $\sup S$