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Prove or disprove the following statements of sequences:

  1. There is a bounded sequence ${a_n}$ with three limit points -8, 22 and 23.
  2. There is an unbounded sequence ${a_n}$ with three limit points -8, 22 and 23.
  3. There is a monotonic sequence ${a_n}$ with three limit points -8, 22 and 23.
  4. There is a Cauchy sequence ${a_n}$ with three limit points -8, 22 and 23.

My problem is that I don't really know how to do this...

1) I would say that there is such a bounded sequence,
f.ex. the sequence ${a_n}=(-1)^n$ if $n=2k |\forall k \in \mathbb{N}$then ${a_n}+14$ else
if $n=2k+1$ then ${a_n}*8$ so that there are the lpts (lpt=limit point) -8 and 22 and 23


3)If ${a_n}$ is monotonic: then $a_n+1>a_n$ or $a_n+1<a_n$,
so that $a_n = -8$ and $a_(n+1) = 22$ and $a_(n+2) = 23$
so that it works, and there also is such a sequence with these limit points.

4)It is not possible because every real Cauchy sequence converges and therefore only has one limit a, which is also the only limit point.

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up vote 2 down vote accepted

(1) You’re working much harder than necessary: what’s wrong with $\langle -8,22,23,-8,22,23,\dots\rangle$?

(2) Instead of repeating the block $-8,22,23$ over and over, why not use $-8,22,23,n$, where $n$ is the number of the block?


(3) No, you can’t make the sequence monotonic and keep all three cluster points. If $-8$ is a cluster point, there are arbitrarily large $n$ with $a_n\in(-9,-7)$, and if $22$ is a cluster point, there are also arbitrarily large $n$ with $a_n\in(21,23)$. Show that $\langle a_n:n\in\Bbb N\rangle$ has a subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$ such that $a_{n_k}\in(-9,-7)$ when $n$ is even, and $a_{n_k}\in(21,23)$ when $n$ is odd, and explain why this subsequence shows that $\langle a_{n_k}:k\in\Bbb N\rangle$ cannot be monotone.

(4) You’re right, and your reasoning is correct.

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(1) that 22 is too near to 23 f.ex. an oscillating sequence? – phil Dec 20 '12 at 17:33
For (3) you can ignore $23$: the point is that if the original sequence has at least two cluster points, it must have a subsequence that oscillates between rising and falling. – Brian M. Scott Dec 20 '12 at 17:35
If I would connect each point of a oscillating sequence, would the cluster point become/has to be a sort of a saddle point?(1)so if the 22 is a cluster point there will be arbitrarily large n with $a_n \in (21,23)$ and if there is also 23 as a cluster point there is be arbitrarily large n with $a_n \in (22,24)$, wouldn't there be a sort of interference of both points, so that the middle point between 22 and 23 would be a cluster point? – phil Dec 20 '12 at 18:04
@phil: You keep trying to make it too hard. Pick any two of the cluster points $-$ I chose $-8$ and $22$, but that was an arbitrary choice $-$ pick subsequences converging to each, and show that this means that the original sequence has an oscillating subsequence and therefore cannot be monotone. The third cluster point of the original sequence is completely irrelevant. – Brian M. Scott Dec 20 '12 at 18:09
this comment was for the (1) and (2).For (3): $a_(n_k) = -8$ then $\lim\limits_{n\rightarrow\infty} (n_k) = -8$,$a_(n_{k+1}) = 22$ then $\lim\limits_{n\rightarrow\infty} (n_{k+1}) = 22$, in both cases the subsequence converges, that means that the sequence diverges, and therefore the sequence cannot be monotonic and bounded.So it is impossible to be monotonic.Is that correct? – phil Dec 20 '12 at 18:35

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