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**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?