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I'm having some difficulty solving this problem :

Find two sequences $ a_n, b_n $ such that their accumulation points are given by the set $$ K = \{ - \infty, + \infty, 2, 4, 6, \text{...} \} \; \text{and} \; \Bbb{Z} \cup \{ - \infty, + \infty \} \; \text{respectively}. $$

Both sequences have an infinite number of accumulation points and this is where I struggle.

If the set $ K $ had a finite number of accumulation points, say $ K = \{ - \infty, 2 \} $, I can easily construct the sequence $ a_n $ :

$$ a_n = \begin{cases} -n, & \text{if $ n\pmod 2 = 0 $} \\ 2 + \frac{1}{n}, & \text{if $ n \pmod 2 = 1 $} \end{cases} $$

but I don't know how to expand this idea for an infinite number of accumulation points.

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  • $\begingroup$ All the answers handle the case of countably many accumulation points. However, we can do better - for any nonempty $C\subseteq \mathbb{R}$ closed, we can find a sequence whose accumulation points are precisely the points in $C$! Note that if $C$ is "big enough" - e.g., take $C=[0, 1]$ - then the idea of tackling each point in $C$ individually won't work, since $C$ is uncountable. This is a good exercise. $\endgroup$ – Noah Schweber Oct 26 '15 at 23:46
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One way to handle the first problem is similarly to Omnomnomnom’s suggestion for the second: arrange to repeat the desired finite limit points infinitely often. That automatically takes care of $+\infty$, and we can fold in a sequence converging to $-\infty$. This can be done in many ways; after a little tinkering with the negative subsequence I chose this sequence as being easier than most to describe.

$$\color{purple}-2,\color{green}{2,-4},\color{orange}{2,4,-6},\color{blue}{2,4,6,-8},2,4,6,8,-10,\ldots$$

The $n$-th block has $n$ terms: $2(1),2(2),\ldots,2(n-1),-2n$. The last term of the $n$-block is the $(1+2+\ldots+n)$-th term of the sequence, which makes it the $\frac{n(n+1)}2$-th term of the sequence. In other words, if $T_n=1+2+\ldots+n=\frac{n(n+1)}2$, the $n$-th block consists of the terms $a_{T_{n-1}+1},a_{T_{n-1}+2},\ldots,a_{T_n}$. If $k=T_{n-1}+\ell$, then $a_k=2\ell$ unless $\ell=n$, in which case $a_k=-2n$.

If you want to express this in more traditionally formulaic notation, you can write

$$a_k=\begin{cases} 2\left(k-\frac{n(n-1)}2\right)=2k-n(n-1),&\text{if }\frac{n(n-1)}2<k<\frac{n(n+1)}2\\\\ -2n,&\text{if }k=\frac{n(n+1)}2\;. \end{cases}$$

However, it’s not really necessary to do so: as long as you have a definition that clearly makes it possible unambiguously to calculate $a_k$ from $k$, you’ve defined a sequence.

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  • $\begingroup$ This is very helpful, thanks. $\endgroup$ – Von Kar Oct 25 '15 at 10:08
  • $\begingroup$ @VonKar: You're welcome. $\endgroup$ – Brian M. Scott Oct 25 '15 at 10:30
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Consider this sequence: $$ 0,\\ 0,-1,1,\\ 0,-1,1,-2,2,\\ 0,-1,1,-2,2,-3,3,\\ \vdots \vdots $$ What are its accumulation points?

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  • $\begingroup$ It's accumulation points is the set $ \Bbb{Z} \cup \{ - \infty, + \infty \} $. Is there a way I can rewrite this sequence in a more mathematical way? $\endgroup$ – Von Kar Oct 24 '15 at 14:13
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One general way to create a sequence with infinitely many accumulation points is as follows:

  • For each accumulation point, write down a sequence that converges to that accumulation point. Let's say your accumulation points a{re $p_0, p_1, p_2, \ldots\}$ then you would have a sequence $(a_{00},a_{01},a_{02},\ldots)$ converging to $p_0$, a sequence $(a_{10},a_{11},a_{12},\ldots)$ converging to $p_1$ and so on.

  • Now write a table of your sequences, and read it by antidiagonals. That way you get the sequence $$(a_{00}, a_{01}, a_{10}, a_{02}, a_{11}, a_{20}, a_{03}, a_{12}, a_{21}, a_{30}, \ldots)$$ which has all $p_n$ as accumulation points.

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  • $\begingroup$ This generalisation is very helpful, thanks. $\endgroup$ – Von Kar Oct 25 '15 at 13:28

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