$\mathbb Z$ and $\mathbb R\setminus\{0\}$ as sets of accumulation points of sequences

Question

Accumulation point(s) was one of the topics in our math analysis lecture today. I'm not familiar with the term of a topological space so I would like to find the solution to a problem task using the tools we were introduced with in the lectures. We are supposed to find,if possible, a sequence whose accumulation points are all the integers, in other words: $$\exists? \;(a_n)_n\;\;\;\text{such that}\;\;\mathbb Z\;\;\text{is the set of all its accumulation points}?$$ I was thinking of the floor & ceiling functions, but the problem is that $$|\mathbb Z|<|\mathbb R|$$ $$\text{if}\;f(x)=\lfloor x\rfloor\;\text{or} \;g(x)=\lceil{x}\rceil$$ $$\nexists\;f^{-1}(x)\;\wedge\;\nexists\;g^{-1}(x)$$ I need an image of an injective function. We proved the Weierstrass theorem and used Cantor's theorem several times, but this sequence, if exists, is divergent. Another question our professor proposed is what an accumulation point can be anyway? And a similar question:

$$\exists? \;(a_n)_n\;\;\;\text{such that}\;\;\mathbb R\setminus\{0\}\;\;\text{is the set of all its accumulation points}?$$ I had sets of linear combinations of exponential functions in my mind, but I don't know how to express that without mixing limits and accumulation points.

Answer 1

Consider $$0,1,0,-1,0,1,2,1,0,-1,-2,-1,0,1,2,3,2,1,0,-1,-2,-3,-2,-1,0, \ldots$$ and verify that the set of accumulation points of this sequence is $\mathbb{Z}$.

It is not possible that the set of accumulation points of a sequence $(a_n)_n$ is $\mathbb{R}\setminus \{0\}$. Indeed, it that were true, we could inductively construct a subsequence $(a_{p(n)})_n$ of $(a_n)_n$ such that $$a_{p(n)} \in \left\langle \frac1{2n}, \frac{3}{2n}\right\rangle$$ because $\frac1n$ is a limit point. Then $a_{p(n)} \to 0$. Indeed, for every $n \in \mathbb{N}$ we have that $a_{p(k)} \in \left\langle -\frac1{n}, \frac1{n}\right\rangle$ for all $k \ge \frac32 n$.

Therefore $0$ is a limit point of $(a_n)_n$ as well.

Answer 2

For the first question: yes, there does exist such a sequence. For example, let $\delta \in (0,1)$, and consider the sequence $(1 + \delta,0+\delta,-1+\delta,0+\delta,2+\delta^2,1+\delta^2,0+\delta^2,-2+\delta^2,-1+\delta^2,0+\delta^2,3+\delta^3,2+\delta^3,1+\delta^3,0+\delta^3,-3+\delta^3,-2+\delta^3,-1+\delta^3,0+\delta^3,\ldots)$.

For each integer $k$, there's a subsequence whose $n$th term is $k + \delta^n$, which therefore converges to $k$, so $k$ is an accumulation point. It is reasonably clear that there are no further accumulation points.

For the second question: no such sequence exists. Suppose that such a sequence $(a_n)_n$ did exist. But then, in particular, there is, for each natural $k$, a natural sequence $(b(k)_n)_n$ such that the subsequence $(a_{b(k)_n})_n$ converges to $\frac{1}{k}$. Without loss of generality (taking further subsequences if necessary), this subsequence has $|a_{b(k)_n} - \frac{1}{k}| < \frac{1}{n}$ for all $n$, so that in particular $|a_{b(k)_n}| < \frac{1}{k} + \frac{1}{n}$. Define a new integer sequence by $c_n := b(n)_n$. Then $(a_{c_n}) \to 0$, since $|a_{c_n}| < \frac{2}{n}$. Thus, $0$ is also an accumulation point of this sequence.