[SOLVED] Complex limit points

**topsquark** · February 16th 2008, 08:38 AM

This is the first of a probably numerous series (pun intended) of questions about complex series. My series work has never been good, and I don't write proofs well.

So here we go with the first:

A point $\alpha$ is said to be a limit point of a set E of points in the complex plane (as opposed to a sequence $z_n$ ) if every neighborhood of $\alpha$ contains infinitely many (distinct) points of E. Show that a sequence $z_n$ can have a limit point $\alpha$ in the sense that, given any $\epsilon$ the inequality $|z_n - \alpha | < \epsilon$ holds for infinitely many points of $z_n$ , without being a limit point of the set of points corresponding to the terms of $z_n$ .

I can't even think of an example that fits this situation, much less writing a proof for it. Any pointers (not full solutions please) would be appreciated.

-Dan

**Plato** · February 16th 2008, 09:04 AM

Dan, are you quite sure that you have quoted this problem correctly?
It clear that $\alpha$ can be a limit point of the set $\left\{ {z_n } \right\}$ without being the limit of the sequence $\left( {z_n } \right)$ . Unless you textbook is making some unusual point, I don’t think that statement is true. (Or maybe I have misread the statement!)

Post Script
I have just thought of a reading of the phrase “given any $\varepsilon >0$ the inequality $\left| {z_n - \alpha } \right| < \varepsilon$ holds for infinitely many points of ${z_n }$ ”. If we read that as meaning infinitely many terms of the sequence then any sequence that is constant from some index on will fit that reading. Here is a trivial example: $\left\{ {z_n :\forall n\left[ {z_n = i} \right]} \right\}$ . The number i is limit of the sequence but not a limit point.

**topsquark** · February 16th 2008, 10:28 AM

Originally Posted by Plato

Dan, are you quite sure that you have quoted this problem correctly?
It clear that $\alpha$ can be a limit point of the set $\left\{ {z_n } \right\}$ without being the limit of the sequence $\left( {z_n } \right)$ . Unless you textbook is making some unusual point, I don’t think that statement is true. (Or maybe I have misread the statement!)

Post Script
I have just thought of a reading of the phrase “given any $\varepsilon >0$ the inequality $\left| {z_n - \alpha } \right| < \varepsilon$ holds for infinitely many points of ${z_n }$ ”. If we read that as meaning infinitely many terms of the sequence then any sequence that is constant from some index on will fit that reading. Here is a trivial example: $\left\{ {z_n :\forall n\left[ {z_n = i} \right]} \right\}$ . The number i is limit of the sequence but not a limit point.

I'm going to have to "chew" on this for a bit. I'll get back to you. Thank you!

-Dan

**ThePerfectHacker** · February 16th 2008, 03:01 PM

This is not true. Consider $z_n = 0$ then there is no limit point for the set $\{z_1,z_2,...\}$ because it is a finite set. And no neighborhood has will have infinitely many point of this set because the set is finite. So you might say what happens if $z_n$ has infinitely many point. But that is still not good enough because consider $z_n = n$ then $\{1,2,3....\}$ and the points are never close enough together so that within a certain small neighrhood they accumulate. However, if $z_n$ is a sequence of distinct point that converges then $\{z_1,z_2,z_3,...\}$ has an accumulation point. In fact, the accumation point (limit point) is the limit of the sequence. Let $z_n \to z$ . Then for any $\epsilon > 0$ we can find $N$ so that if $n\geq N\implies |z_n - z|<\epsilon$ thus $z_N,z_{N+1},z_{N+2},...$ are all within the $\epsilon$ -neighborhood of $z$ and the crucial step here is that all these point are distinct hence there are infinitely many of them.

**Plato** · February 16th 2008, 03:15 PM

It is true if one reads the statement as “each neighborhood contains infinitely many terms of the sequence”.
I suspect that was the intention of the author of the problem.

**topsquark** · February 17th 2008, 05:01 AM

I see it now. Thank you both.

-Dan

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