# Accumulation points and finding them

1. Feb 15, 2009

### H2Pendragon

This is the definition of accumulation point that my book gives:

A is an accumulation point of $$S \subset \mathbb{R}, \forall \epsilon > 0, S \bigcap B(A;\epsilon)$$ is infinite.

The book I have gives horrible examples on what accumulation points actually are (contradicting itself two out of the three times), but never actually gives instructions on how to find the points.

This is the question I have to solve:

Find the accumulation points of

$$S = \left\{\frac{2}{n} + (1 - \frac{1}{n})cos(\frac{n\pi}{2}) : n \in\mathbb{N}\right\}$$

Can anyone help to actually explain to me, in english, what an accumulation point is? I tried Wikipedia but it's more of this meaningless jargon.

Hopefully, understanding what it is I'm looking for will show me how to answer this question. If not, I could use help there too.

I'd post some relevant work I've done on this problem, but I really have no idea where to start! Why can't analysis books ever actually explain things as if I might actually not understand their initial rambling?

2. Feb 15, 2009

### quasar987

In english, an accumulation point of S is a point A of R such that you can find points of S (different from A) arbitrarily close to A.

Some enlightening examples are:

(1) If S={x} is made of just one point, then it has no accumulation point.
(2) In R, every point is an accumulation point.
(3) The set {1/n: n in N} has 0 as its unique accumulation point.

3. Feb 15, 2009

### H2Pendragon

Why couldn't this book have just said that? That makes complete sense. My book doesn't like making sense. I don't think this was written to be a text book at all, because it certainly doesn't read like it's trying to teach!

Ok so for the question I'm looking for points a,b,c, etc. where I can find infinitely many points arbitrarily close to those points.

Judging from the set itself, it looks like we get S = {2, 1/2, 2/3, 5/4, 2/5, -1/2, 2/7, 9/8, 2/9, -7/10, 2/11...}

Skrew it, I graphed the rest

It looks like it will just go back to looking like a typical cosine graph (which I guess is to be expected since 2/n and 1/n go to 0)

So if we set $$n_{k} = 2k \Rightarrow\left\{\frac{1}{k} + (1 - \frac{1}{2k})cos(k\pi) \right\} \rightarrow (-1,1)$$

and $$n_{k+1} = 2k+1 \Rightarrow\left\{\frac{2}{2k+1} + (1 - \frac{1}{2k+1})cos(\frac{\pi(2k+1)}{2}) \right\} \rightarrow {0}$$

Ok I think I broke my brain. Did any of that make sense? If it does, are the accumulation points for the set -1,0,1?

If it didn't make any sense (because I'm supposing it doesn't at this point, I've had a bit of a head cold all weekend), where should I actually be heading?

Last edited: Feb 15, 2009
4. Feb 15, 2009

### quasar987

That's perfect.

5. Feb 16, 2009

### Office_Shredder

Staff Emeritus
From the definition of the book, if x is in S, then x is an accumulation point of S

6. Feb 16, 2009

### quasar987

How come?

For instance, take S={x}. Then $\forall \epsilon > 0, \ S \cap B(A;\epsilon)=\{x\}$, which is not infinite!

7. Feb 16, 2009

### Office_Shredder

Staff Emeritus
infinite and non-empty look very similar on my laptop screen