Infinitely many accumulation points in a bounded sequence? Any bounded sequence of real numbers contains at least one accumulation point. If it doesn't converge it has more than one. In fact, $$a_n \equiv n (mod m)$$ has exactly m limit points.
Question: Can a bounded sequence of real numbers have infinitely many limit points?
 A: 1, 1,1/2, 1,1/2,1/3, 1,1/2,1/3,1/4, ...
A: Yes. Take the sequence that enumerate all rational between $[-1,1]$.
A possible sequence that enumerate rational in $[0,1]$: $$x_{\frac{q(q-1)}{2}+p}=\frac{p}{q}.$$
A: Take an infinite family of bounded sequences in $[0, 1]$ converging to $1/k$ for each $k \in \mathbb{N}$.
Interweave the sequences (dovetail them) to produce an explicit example.
That is, for $a_k \rightarrow 1$, $b_k \rightarrow 1/2$, $c_k \rightarrow 1/3$, $\ldots$:
$a_1, a_2, a_3, a_4 \ldots$
$b_1, b_2, b_3, b_4 \ldots$
$c_1, c_2, c_3, c_4, \ldots$
$\vdots$
becomes:
$a_1, b_1, a_2, c_1, b_2, a_3, \ldots$
where this final sequence has the desired property.
A: The sequence
$$0,1,0,1/2,1,0,1/3,2/3,1, 0,1/4,2/4,3/4,1, \dots$$
has every point in $[0,1]$ as an accumulation point.
A: There are uncountably many ways to do this. One way, which does not  enumerate all of $\Bbb Q$, is to write the natural numbers in their usual decimal form $0,1,...,9,10,11,12,..., 6792,6793,...,$ and assign to each one the reversed decimal fraction, thus generating the sequence of decimal rational numbers $0.0,0.1,...,0.9,0.01,0.11,0.21,..., 0.2976,0.3976,...$. It is easily seen that every real number in the interval $[0\,\pmb,\,1]$ is distant by no more than $10^{-n}$ from one of the numbers in the first $10^n$ terms of this sequence.
