# Thread: Sequence and accumulation point

1. ## Sequence and accumulation point

Suppose $\displaystyle \{a_n\}_{n\in\mathbb{N}}\to A$ and $\displaystyle \{a_n: \ n\in\mathbb{N}\}$ is an infinite set. Show that A is an accumulation point of $\displaystyle \{a_n: \ n\in\mathbb{N}\}$.

Let $\displaystyle A=\text{sup} \ \{a_n\}$ and Q be a neighborhood of A. There is an $\displaystyle \epsilon>0$ such that $\displaystyle (A-\epsilon, A+\epsilon)\subset Q$.
Since A = sup and if c is an upper bound, $\displaystyle A\leq c$
Let $\displaystyle a_i\in\{a_n\}, \ i=1,2,....$
Assume there are finitely many $\displaystyle a_i\in Q$.
Now, let $\displaystyle A'=\text{max}\{a_i\}$
Since x' is the max, x' is an upper bound of $\displaystyle \{a_n\}$. Therefore, $\displaystyle A'<A$, but $\displaystyle A = \text{sup} \ \{a_n\}$ which is a contradiction. Thus, A is an accumulation point.

Correct?

2. ## Re: Sequence and accumulation point

Originally Posted by dwsmith
Suppose $\displaystyle \{a_n\}_{n\in\mathbb{N}}\to A$ and $\displaystyle \{a_n: \ n\in\mathbb{N}\}$ is an infinite set. Show that A is an accumulation point of $\displaystyle \{a_n: \ n\in\mathbb{N}\}$.

Let $\displaystyle A=\text{sup} \ \{a_n\}$ and Q be a neighborhood of A. There is an $\displaystyle \epsilon>0$ such that $\displaystyle (A-\epsilon, A+\epsilon)\subset Q$.
Since A = sup and if c is an upper bound, $\displaystyle A\leq c$
Let $\displaystyle a_i\in\{a_n\}, \ i=1,2,....$
Assume there are finitely many $\displaystyle a_i\in Q$.
Now, let $\displaystyle A'=\text{max}\{a_i\}$
Since x' is the max, x' is an upper bound of $\displaystyle \{a_n\}$. Therefore, $\displaystyle A'<A$, but $\displaystyle A = \text{sup} \ \{a_n\}$ which is a contradiction. Thus, A is an accumulation point.

Correct?
No, that's not correct. You don't get to just assume that $\displaystyle A=\sup\{a_n\}$. For example $\displaystyle \langle (-1)^n/n\rangle$ is a sequence converging to $\displaystyle 0$, which is neither an upper nor lower bound of the sequence.

Recall that by definition $\displaystyle \langle a_n\rangle\to A$ iff for each $\displaystyle \epsilon>0$ there is $\displaystyle N\in\mathbb{N}$ such that if $\displaystyle n\geq N$ then $\displaystyle a_n\in(A\pm\epsilon)$. So just notice that there are infinitely many elements in $\displaystyle \{a_n:n\geq N\}\subseteq(A\pm\epsilon)$, and the proof is complete.