# Math Help - [SOLVED] Complex limit points

1. ## [SOLVED] Complex limit points

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

2. 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.

3. 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

4. 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.

5. 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.

6. I see it now. Thank you both.

-Dan