Taking a bizarre limit Consider the set of integers, $\Bbb{Z}$. Now consider the sequence of sets which we get as we divide each of the integers by $2, 3, 4, \ldots$.
Obviously, as we increase the divisor, the elements of the resulting sets will get closer and closer.
Question: In the limit as $\text{divisor}\to\infty$, what will the "limiting" set be? 
(I don't think it could be $\Bbb{R}$.)
 A: The typical way to define limits of sets is via
$$\liminf_{n\to\infty} A_n = \bigcup_{n\geq 1} \bigcap_{k \geq n} A_k \\ \limsup_{n\to\infty} A_n = \bigcap_{n\geq 1} \bigcup_{k\geq n} A_k$$
Using these and $A_n = f_n(\mathbb{Z})$ where $f_n(x) = x/n,$ we have
$$\liminf_{n\to\infty} A_n = \mathbb{Z} \\ \limsup_{n\to\infty} A_n = \mathbb{Q} $$
In particular, the limit doesn't exist.
A: Let $A_n=\{x/n:x\in\mathbb{Z}\}$ with $n$ any integer greater than $1$ then it is easy to see that $\mathbb{Z}\subset A_n\subset \mathbb{Q}$.  We claim that
$$\limsup_n A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)=\mathbb{Q}$$ and $$\liminf_n A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)=\mathbb{Z}.$$
See set-theoretic limit for more details about  the limit of a sequence of sets.
A: 1) Let $S_n = \{ \frac{z}{n} \ | \ z \in \mathbb{Z} \}$ and $p_i$ be $i$-th prime integer.
2) It has no limit! Because since $(n,n+1)=1$ we have $S_n \cap S_{n+1} = \mathbb{Z}$, so always new set miss any non integer rationals included in previous one and get some new ones.
3) But $\limsup$, exists. If you consider $a_n=\prod_{i=1}^n p_i^n$, then set sequence $S_{a_n}$ is an strict increasing sequence, with respect to inclusion order, that for every $m$ there is a $k$ that $m | a_k$, so $S_m \subseteq S_{a_k}$, Therefore it tends to $\mathbb{Q}$.
4) Also $\liminf$, exists. As we see $S_n \cap S_{n+1} = \mathbb{Z}$, so it tends to $\mathbb{Z}$.
A: The three answers thus far assume by limits of the sets you mean the common value of the set-theoretic $\liminf$ and $\limsup$ (where convergence means they agree). This is a highly reasonable assumption, given that you did not specify a meaning for the limit of sets yourself.
However, I want to point out that there are other possibilities for defining a limit of sets. For example, given a sequence of sets $(S_n)$ with $\forall n, S_n \subseteq X$ for some topological space $X$, you could define
$$\lim_n S_n = \{x\in X\mid \exists (s_n) \subset X, s_n \to x \wedge \forall n, s_n \in S_n\}$$
By this definition with $X = \Bbb R$, the limit of your sets is indeed $\Bbb R$.
A: If you are physicist or applied mathematician, if you define $S_n = \{ z_i/n, z_i=i, i = 1..\infty \in \mathbb{Z}\}$, then you can just say that $\lim_{n\rightarrow\infty} S = \{ \lim S_n \}$, which will yield you $\lim S = \{0, 0, 0, 0,...\}$. This will happen because $\lim$ is just working on every fixed element, pretty similar to partial differential.
