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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complex-analysis
analysis
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0What is $C$ here? And is $s$ a typo for $S$? – 2012-11-29
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0@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 edit – 2012-11-29
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0I 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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0@BrianM.Scott They should both be S' and C is the complex numbers I just didn't know how to code that yet – 2012-11-29
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0It’s `\Bbb C` or `\mathbb C`. – 2012-11-29
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0Ok i'll do that now – 2012-11-29
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0@BrianM.Scott Perfect that should be how I want it to look now – 2012-11-29