I have a problem that asks me investigate these things about the set B.
$B=\{\frac{{(-1)}^nn}{n+1}:n=1,2,3…\}$
(a) Find the limit points of B.
(b) Is B a closed set?
(c) Is B an open set?
(d) Does B contain any isolated points?
(e) Find the closure
I am new to this and so I would like someone to go over my work and see if i made any mistakes.
(a) This a set of a sequence that we will call $b_n$ it is obvious that the two subsequences $b_{2n}\to 1$ and $b_{2n+1}\to -1$. Thus we have at least two limit points $1,-1$.
My first intuition was to say since every subsequence of $b_n$ has to converge to $1,-1$ or diverge, then those are all the limit points. However, that does not work since the subsequence only considers the terms in the order of the original sequence. Thus i came up with this:
Let us assume that there is another limit point, $a\not=1,-1$. WLOG let us say $0\leq a$ then all we care about is the set $\{b_{2n}=\frac{2n}{2n+1}:n=1,2,3…\}$ as the other part of the sequence is negative so they wont be in the very small neighborhood of $a$ when looking at the definition of a limit point. Now let us pick such an $N$ such that $|b_{2n}-1|<\frac{1-a}{2}$. Now pick the neighborhood $V_\frac{1-a}{4}(a)$ clearly it can only have a finite amount of elements at most it contains the set $\{b_2,b_4,…b_{2N}\}$ consider the set $\{|a-b_2|,|a-b_4|….|a-b_{2N}\}$ it is finite aswell and so it has a minimum say x. Now we consider the neighborhood $V_{\min\{\frac{x}{2},\frac{1-a}{4}\}}(a)$ it does not have any elements of $b_{2n}$ and so $a$ cannot be a limit point.
This was a very long proof, but is it correct? Is there a more succinct proof of this?
(b) No it does not contain $1,-1$
(c) This I can answer with $(d)$ since B has isolated points, it cannot be open. Since every element of an open set is a limit point.
(d) We found out in (a) that $1,-1$ are the only limit points. Thus every elements of $B$ must be an isolated point
(e) $B\cup\{1,-1\}$
Best Answer
There is an easier way to prove there can't be any other limit points. If $a_n\to a$ then $|a_n|\to |a|$. But note that in your exercise the sequence of absolute values converges to $1$, hence this is true for every subsequence as well.
Edit: you brought a good point. Well, let's call the original sequence $a_n$ and let's take any sequence $b_k$ which contains different elements from $B$. All the elements of $b_n$ are elements in $a_n$ and hence there is a subsequence $a_{n_k}$ and a permutation $\sigma:\mathbb{N}\to\mathbb{N}$ such that $b_k=a_{\sigma(n_k)}$. Now let $\epsilon>0$. Since $|a_{n_k}|\to 1$ we know there is $k_0\in\mathbb{N}$ such that $||a_{n_k}|-1|<\epsilon$ for all $k>k_0$. Alright, so in that case the elements in the sequence $b_k$ which might satisfy $||b_k|-1|\geq\epsilon$ are the elements $b_{\sigma^{-1}(n_1)},...,b_{\sigma^{-1}(n_{k_0})}$, and this is a finite number of elements. From here we conclude $|b_k|\to 1$.