Question

Find the limit points and isolated points of: A = { (n+1)/2n + (-1)^n/2}. Justify your answer

Answer 1

\[A = \left\{ \frac{ n+1 }{ 2n }+\frac{ (-1)^{n} }{ 2 } \right\}\]
The limit points are 0 and 1, right?
As for the isolated points....aren't there infinitely many? Well, maybe not infinitely many, but an incredibly large amount? An isolated point is any point in a set that is not a limit point, correct?

n =1 -> 1/2
n = 2 -> 5/4
n = 3 -> 1/6
n = 4 -> 9/8 ...
I'm just going to get an infinite amount of points as n goes to infinity, right? So how do I state my isolated points? The interval (0, 5/4) or something?

Answer 2

\(n\) belongs to the natural numbers, right? No negative integers?

Answer 3

Your limit points are correct. And your intuition on the number of isolated points is also correct. There is an infinite number of them. I don't remember if the interval notation notation applies in this case... something tells me no, since this is a set of discrete numbers.

Instead, I would denote the set of isolated points as
\[A\backslash\{0,1\}\]
(i.e. the set \(A\) excluding 0 and 1). This is because every point in the set is an isolated point except for 0 and 1.

Answer 4

Right, that makes sense, I realized that didnt seem right afterwards. And yes, n belongs to the natural numbers. Well, glad I understand well enough to state what the limit and isolated points are, but how would I go about proving that? I have to somehow show:
\[N'_{ \epsilon } (0) \cap A \neq \emptyset \]
\[N'_{\epsilon}(1) \cap A \neq \emptyset\]

Answer 5

Er....and I have to show those are the only limit points. Then if I can do that, A \ {0,1} are isolated points if I can show they belong to A. Or something like that.

Answer 6

For limit points, if I'm remembering correctly, you would let \(x\) be some (real?) number that's not one of the alleged limit points. So you'd analyze the cases where (1) \(x\in(-\infty,0)\), (2) \(x\in(0,1)\), and (3) \(x\in(1,\infty)\).

Then using the definition of the set you should be able to establish that any points in those intervals can't be limit points because they fail to satisfy the definition. Or something like that... my memory of real analysis/topology is fuzzy at best.

Answer 7

The first case is trivial because \(A\) contains only positive numbers.

Answer 8

I have that first case, lol. Haven't gotten much beyond that, though. But yes, everything sounds good so far, your memory seems to match up with anything I've learned or think I've learned, lol.

Answer 9

Maybe a more appropriate way is to assume \(x\) is a limit point, then any neighborhood \((x-\delta,x+\delta)\) will contain a point of \(A\) other than \(x\).

If we suppose \(x=0\), then we need to find some point \(x+\delta>x\) for some \(\delta\), which we can always do. This means \(x=0\) is certainly a limit point, but the reasoning here is a bit hand-wavy in my opinion because I've assume the thing we want to show... Maybe you can put your own spin on this?

Answer 10

Right. Well, I know that you can always say there exists a positive integer n such that 0 < 1/n < epsilon. But I don't think that says anything about set A. I was hoping to find a way to say (n+1)/2n + (-1)^n/2 < epsilon, but not sure how I could do that. One way of thinking was to say n = 2k+1 and that for all k, we have (k+2)/(2k+2) -1/2. That certainly approaches 0 as k -> infinity, so I thought maybe letting epsilon equal that. But even then, not sure what I could add to guarantee we have another point within A. Just some of my train of thought.

Answer 11

And I dont think we can say "for some" with these limit points. I have this:

" A point p element of R is said to be a limit point of A if for every epsilon > 0, the deleted neighborhood of p intersect A is non-empty." The for every part seems to make this more difficult than I'd like.

Answer 12

Well I have to admit it's far too late for me to be tackling something I haven't dealt with in over a year... Sorry I couldn't help! I'm sure there are a few other people around here that are more able to help out than me.

Answer 13

I appreciate you coming by either way, you've always been a great help. Have a good night ^_^