It seems obvious that a limit point of $S'$ should be a member of $S'$ but I have no idea how to even begin with a proof of this.
Let $S=(a_n)_{n=1}^{\infty}$ be a sequence in $\mathbb C$ and $S'$ the set of limits of $S$. Prove that every limit point of $S'$ is a member of $S'$
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0@BrianM.Scott Perfect that should be how I want it to look now – 2012-11-29
1 Answers
Suppose that $z$ is a limit point of $S'$; then there is a sequence $\langle z_n:n\in\Bbb Z^+\rangle$ in $S'$ converging to $z$; without loss of generality you may assume that $|z_n-z|<2^{-n}$ for each $n\in\Bbb Z^+$. Each $z_n$ is a limit point of $\langle a_k:k\in\Bbb N\rangle$, so there is a subsequence $\langle a_{k_i^n}:i\in\Bbb Z^+\rangle$ converging to $z_n$. Without loss of generality you may assume that $|a_{k_i^n}-z_n|<2^{-i}$ for each $i\in\Bbb Z^+$.
Now let $\ell_1=k_1^1$. Given $\ell_i$ for some $i\in\Bbb Z^+$, let $\ell_{i+1}=k_j^{i+1}$, where $j$ is the least integer $\ge i+1$ such that $k_j^{i+1}>\ell_i$. Can you show that $\langle a_{\ell_i}:i\in\Bbb Z^+\rangle$ converges to $z$?
Note: The construction of $\ell_{i+1}$ isn’t as complicated as it may seem. The basic idea is to let $\ell_i=k_i^i$ for each $i\in\Bbb Z^+$, but that might conceivably not result in a strictly increasing sequence of integers, so that $\langle a_{\ell_i}:i\in\Bbb Z^+\rangle$ might not actually be a subsequence of $\langle a_k:k\in\Bbb Z^+\rangle$. The extra complication is simply to ensure that $\langle\ell_i:i\in\Bbb Z^+\rangle$ is strictly increasing.
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0@Adam: Yes, you will. – 2012-11-30