A sequence in $[0,1]\cap\mathbb Q$ with no convergent subsequence in $[0,1]\cap\mathbb Q$ Recently, a friend of mine asked me about this problem. If $E$ is the set of rationals in the closed interval $I = [0,1]$, then find a sequence in $E$ that does not have a convergent subsequence that converges to a point in $E$.
My immediate solution was $x_n  = \dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}\to e-2$ (this is a convergent sequence itself, so any convergent subsequence must converge to the same limit).  Then, I also figured that for any irrational $s\in I$, we can simply look at the continued fraction of $s$ and obtain a sequence of the rationals with desired property.
But as it turns out, neither of these solutions was easy enough to understand for my friend. In fact, many of the students in his class, (first undergraduate course in real anaylysis) seem to have struggles understanding these constructions.
So I was wondering if I was missing some very obvious and easy examples of such sequences.
 A: Instead of the continued fraction of an irrational number, you could consider its representation as a decimal number and truncate that. This would give a sequence with similar properties.
For a maybe "more basic" example, consider the set $E=[0,1] \setminus \{\frac 12\}$ and the sequence $x_n = \frac 12 + \frac 1n$. Then the sequence is bounded, but has no convergent subsequence in $E$ by the same argument.
A: In fact, virtually every 'common' sequence is such an example!  For any $r\in I\setminus\Bbb{Q}$, and for any increasing (or decreasing) sequence $\{r_n\}$ with $\lim_n r_n=r$, the sequence $\{r_n\}$ can't have a subsequence that converges to any number but $r$.  The proof is a basic exercise using the epsilon-delta definition of the limit: just show that for any other possible limit $L$, all sufficiently large-index members of the sequence will have to be at some distance from $L$ that you can bound from below, so they can't possibly converge to $L$.
(Much more is true — if the sequence converges to $r$ then it can't have any subsequence converging to any other number but $r$ regardless of its increasing nature — but I feel like the increasing property makes it easier for people to 'get a handle' on the sequence and understand why its limit must be the given number.)
