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So S is a complex sequence (an from n=1 to infinity) has limit points which form a set E of limit points. How do I prove that every limit point of E are also members of the set E. I think epsilons will need to be used but I'm not sure.


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up vote 2 down vote accepted

Let $z$ be a limit point of $E$, and take any $\varepsilon>0$. There is some $x\in E$ with $\lvert x-z\rvert<\varepsilon/2$. And since $x\in E$, there are infinitely many members of $S$ within an $\varepsilon/2$-ball around $x$. They will all be within an $\varepsilon$-ball around $z$, and you're done.

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I don't usually bother with this $\varepsilon/2$-stuff, preferring instead to end the story with lots of points from $S$ within a $2\varepsilon$-ball around $z$. It works just as good, but may be confusing to beginners. – Harald Hanche-Olsen Nov 27 '12 at 22:45
Yes. Really. Not every proof in mathematics needs to be pages long and full of intricate detail. – Harald Hanche-Olsen Nov 27 '12 at 22:46
A sketch is always nice, but the triangle inequality is all you really need (to show that points within the $\varepsilon/2$-ball around $x$ are in an $\varepsilon$-ball around $z$). – Harald Hanche-Olsen Nov 27 '12 at 22:47
Take any member of $S$ inside the $\varepsilon$-ball around $x$, call it $s$. Then $\lvert s-z\rvert=\lvert s-x+x-z\rvert\le\lvert s-x\rvert+\lvert x-z\rvert<\varepsilon/2+\varepsilon/2=\varepsilon$. – Harald Hanche-Olsen Nov 28 '12 at 9:00

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