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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.

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    What is $C$ here? And is $s$ a typo for $S$?2012-11-29
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    @BrianM.Scott You were too quick for me I had to check a previous question to edit my original question because i'd for gotten how to do bits of the coding it should now be how I wanted it as I have re-editted your edit2012-11-29
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    I saw; that’s fine. But we still need to know about $C$. Is $C$ really $\Bbb C$, the complex numbers?2012-11-29
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    @BrianM.Scott They should both be S' and C is the complex numbers I just didn't know how to code that yet2012-11-29
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    It’s `\Bbb C` or `\mathbb C`.2012-11-29
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    Ok i'll do that now2012-11-29
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    @BrianM.Scott Perfect that should be how I want it to look now2012-11-29

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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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    I get the general jist of what is happening but I don't understand where the $2^-n$ and $2^-i$ is coming from. Would I be able to show that $l_i$ is less than $a_{k}_{i}^n$ and therefore less than z2012-11-29
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    or should I use proof by induction using the last few terms somehow?2012-11-29
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    @Adam: I could have used $\frac1n$ and $\frac1i$ instead of $2^{-n}$ and $2^{-i}$; I just needed sequences converging to $0$. $\ell_i$ is a positive integer serving as a subscript; it makes no sense to compare it with the compled number $a_{k_i^n}$. Fix $\epsilon>0$; there is an $m\in\Bbb Z^+$ such that $|z-z_n|<\epsilon/2$ for all $n\ge m$. There’s also an $m'\in\Bbb Z^+$ such that $|a_{k_i^n}-z_n|<\epsilon/2$ for all $i\ge m'$. Can you show that $|a_{\ell_i}-z|<\epsilon$ for all $i\ge\max\{m,m'\}$?2012-11-30
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    would I use the triangle inequality?2012-11-30
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    @Adam: Yes, you will.2012-11-30