# Accumulation points and divergence

1. Dec 23, 2017

### fishturtle1

1. The problem statement, all variables and given/known data
Show that the sequence with two distinct accumulation points must diverge. (Hint: look at the proof of divergence for {$(-1)^k$}.

2. Relevant equations
Some definitions and propositions I'm trying to use:
2.2.3: A sequence cannot converge to two different numbers. If {$x_k$} is a sequence in $\mathbb{R}$ which converges to both y and z, then y = z.

2.2.5 A real sequence {$x_k$} diverges to (positive) infinity (denoted $x_k \rightarrow \infty$) if for every $M \epsilon \mathbb{R}$ there exists $K$ such that, for every $k \ge K$, we have $x_k \ge M$.

2.3.9 For any sequence {$x_k$} in $\mathbb{R}$, an accumulation point of the sequence is any point which is the limit of some subsequence {$y_j$} of the sequence {$x_k$}.

The proof/excerpt for {$(-1)^n$} diverges is this:
While every convergent sequence must be bounded, it is not true that every bounded sequence vineries: in other words, there exist bounded sequence which diverge .For example, consider the sequence
$$\lbrace(-1)^n\rbrace = +1, -1, +1, -1, ...$$
which alternates the two values +1 and -1. This is bounded, since the absolute value of al elements is 1. Nonetheless the sequences diverges. To see this consider the possibilities. It can't converge to +1, because every second element satisfies
$$|(-1)^k - 1| = 2$$
so it cannot approximate +1 with accuracy, say $\epsilon = 1$. However, if
$y \neq 1$, we see that every even index term has
$$|(-1)^k - y| = |y - 1| > 0$$
and so we can't approximate y with accuracy, say, half the distance from y to 1:
$\epsilon = |y-1|/2$.
3. The attempt at a solution
Proof: Let {$x_k$} be a sequence and {$y_j$}, {$z_h$} be subsequences of {$x_k$} with distinct accumulation points. Then $\lim_{j\to\infty} y_j = a$ and $\lim_{h \to\infty} z_h = b$ where $a \neq b$.
.....

edit1: one of the things I'm stuck on is that, what if I have the sequence {$x_k$} = (-10, -5, 0, 5, 10, 5, 0, 0, 0, 0, 0). Then I looked at the subsequences {$y_j$} = (-10, -5, 0, 5, 10) and {$z_h$} = (5, 0, 0, 0). Then {$x_k$} converges to 0, {$y_j$} approaches 10 and {$z_h$} approaches 0. So what's the problem?

2. Dec 23, 2017

### Staff: Mentor

The sequence you show in your edit, {xk} is a finite sequence. Not sure if that's what you intended. When we talk about sequences converging, we're usually talking about infinite sequences.
For the problem you're working on, set d equal to the distance between two accumulation points (in absolute value), then take $\epsilon < d$. Can you show that there is no index N so that all of the elements with index >= N are within $\epsilon$ of either of the accumulation points?

BTW, this probably belongs in the Calc & Beyond section, as the concepts here aren't typically precalc topics.

3. Dec 26, 2017

### fishturtle1

Trying to use your suggestion but I'm not sure

Let b and c be distinct accumulation points of the sequence {x_n}. Let d = |b - c|. We'll show that d cannot be the limit of {x_n}. Suppose for any $\epsilon$ > 0 there exists an N such that for all n > N, we have
$|x_n + -|b - c|| < \epsilon$
$||x_n + -|b - c|| \le |x_n| + |-|b-c|| < \epsilon$
$|x_n| + |b-c| < \epsilon$

but i don't see a way to get a contradiction..

i'm also kind of confused, if I understand what you said, we are trying to show $d$ is not the limit of {x_n}, but how does that show that a limit doesn't exist

4. Dec 26, 2017

### Staff: Mentor

d is just the distance between b and c. It is not a limit point.
Assume that b is the limit of the sequence $\{x_n \}$. Let $\epsilon = \frac d 2$. Can you show that no matter how large you let N get, some of the elements of the sequence will be farther away from b than $\frac \epsilon 2$?

5. Dec 26, 2017

### fishturtle1

Ok, so i have the inequality $|x_n - b| < \frac{|c - b|}{2}$ where $b, c$ are fixed real numbers. And i'm trying to show, algebraically, that this inequality isn't true for all n. But i don't think this can be done algebraically, since we can't say anything about $x_i$. So do I need to show this with some theorems/defintions?..

6. Dec 27, 2017

### Staff: Mentor

You have two accumulation points, so there are two convergent subsequences. Call them $\{a_{n_k}\}$ and $\{a_{n_l}\}$, where $n_k$ and $n_l$ are sequences of positive integers. Assume that the subsequence $\{a_{n_k}\}$ converges to b, and that the subsequence $\{a_{n_l}\}$ converges to c. Use the definition of convergence of a sequence to show that for suitably large numbers n, all of the elements of the first subsequence are "close to" b and all of the elements of the second subsequence are "close to" c. Use this idea to show that there are some elements of the overall sequence that are either too far away from b or too far away from c for the sequence to converge.