# Accumulation points and divergence

fishturtle1

## Homework Statement

Show that the sequence with two distinct accumulation points must diverge. (Hint: look at the proof of divergence for {##(-1)^k##}.

## Homework 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##.

## 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?

Mentor

## Homework Statement

Show that the sequence with two distinct accumulation points must diverge. (Hint: look at the proof of divergence for {##(-1)^k##}.

## Homework 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##.

## 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?
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.

fishturtle1
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.
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

Mentor
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}.
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##?
fishturtle1 said:
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

fishturtle1
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##?