How to define the set $\lim_{n\to\infty}\{1/n,2/n,...,n/n\}$ rigorously? 
How to define the set $\lim_{n\to\infty}\{1/n,2/n,...,n/n\}$ rigorously?

For example, when asked to give a dense set in $[0,1]$, we let $S_n=\{1/n,2/n,...,n/n\}$ and take the union of all $S_n$ as answer, but actually $S_n$ alone, take $n$ to infinity, is enough, but is it a well defined set?
p.s. the limit here I used is just in an intuitive sense, and is an abuse of notation
 A: In some contexts, we would say that
$$
\liminf_{n \to \infty} S_n = \bigcup_{n =1}^\infty \bigcap_{m=n}^\infty S_m\\
\limsup_{n \to \infty} S_n = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty S_m
$$
and we say that $\lim_{n \to \infty} S_n$ exists iff these two sets are the same.  When the limit exists, we say
$$
\lim_{n \to \infty} S_n = \liminf_{n \to \infty} S_n = \limsup_{n \to \infty} S_n
$$
more generally, we always have 
$$
\liminf_{n \to \infty} S_n \subseteq \limsup_{n \to \infty} S_n
$$
In your case, we find
$$
\liminf_{n \to \infty} S_n = \{1\}\\
\limsup_{n \to \infty} S_n = \Bbb Q \cap (0,1]
$$
so that $\lim_{n \to \infty} S_n$ fails to exist.
A: Among the very many topologies that we may consider on sets of subsets of a given set, let us consider the one given by the Hausdorff distance on the set $C$ of all the closed subsets of the interval $I = [0,1]$. $C$ will thus become a metric space (even more: since $I$ is compact, then so will be $C$, by the Blaschke selection theorem; this, in particular, guarantees that your sequence $(S_n) _{n \ge 1}$ will have at least one limit point).
If $D$ is the Hausdorff distance and $d$ the usual distance on $I$, let us now show that $D(S_n, I) \to 0$ (and thus that $S_n \to I$).
According to the definition,
$$D(S_n, I) = \max \Big( \sup \limits _{x \in S_n} \inf \limits _{y \in I} d(x,y), \sup \limits _{y \in I} \inf \limits _{x \in S_n} d(x,y) \Big) .$$
Since $S_n \subset I$ one clearly sees that $\inf \limits _{y \in I} d(x,y) = 0$ (it is reached for $y=x$), so our distance becomes just $D(S_n, I) = \sup \limits _{y \in I} \inf \limits _{x \in S_n} d(x,y)$. Now, break $I$ into $n$ pieces, $I = \bigcup \limits _{k=1} ^n I_k$, with $I_k = [\frac {k-1} n, \frac k n]$ (they are not disjoint but we don't need that). Therefore,
$$D(S_n, I) = \max \limits _{k=1} ^n \sup \limits _{y \in I_k} \inf \limits _{x \in S_n} d(x,y) .$$
But when $y \in I_k = [\frac {k-1} n, \frac k n]$, then the point from $S_n$ that is nearest to $y$ is the endpoint of $I_k$ nearest to $y$; in formulae, if $y \in I_k$ then $\inf \limits _{x \in S_n} d(x,y) = \min \Big( d(x, \frac {k-1} n), d(x,\frac k n) \Big)$. But the smallest of these two numbers is clearly smaller than half the length of the interval $I_k$; in formulae, $\inf \limits _{x \in S_n} d(x,y) \le \frac 1 {2n}$. Since this is independent of $y$ and $k$, you immediately get
$$D(S_n, I) \leq \frac 1 {2n}$$
which shows that $D(S_n, I) \to 0$ as claimed.
Of course, nobody forces you to use this topology (you might consider order-related topologies, for instance). In other topologies you may get a different limit (or none at all). You might also consider exploring the concept of directed limit of a family of (sub)sets - a concept that is not based on topology; in this latter case, note that you may view the index set $\Bbb N$ as a directed set in two different ways: one is given by the usual order, the other given by the (partial) order induced by division ($m \preceq n \Leftrightarrow n|m$). I think, though, that the choice of the Hausdorff distance is the most appropriate in the larger context of your question because this was motivated by the study of topological density and the Hausdorff distance is deeply connected to the concept of topological closure.
