Sequences with Infinite Number of Accumulation Points 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. 
 A: 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?
A: 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.
A: 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.
