Building a very odd sequence I have to build a sequence $(x_n)_{n \in \mathbb{N}}$ of $\mathbb{R}^2$ such that, for any $x\in \mathbb{R}^2,$ there is a subsequence of $x_n$ which converges to $x$.
I have the intuition that we should use decimal expansions but I cannot find an explicit formula for $x_n$. Thanks for any help.
 A: $\mathbb{Q}^2$ is dense in $\mathbb{R}^2$ and countable. Pick any enumeration of those.
A: Write the prime factorisation of $n=n=2^a\cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 17^g \cdot ....$
Define
$$x_{n}= \left\{
\begin{array}{lc}
((-1)^a\frac{b}{c}, (-1)^d \frac{e}{f}) &\mbox{ if } c\neq 0 , f \neq 0 \\
\mbox{ anything you want } & \mbox{ if } c=0 \mbox{ or }  f=0
\end{array}
\right.$$
Show taht this sequence has the desired prorerty.
A: Let me expand on what's already here.  As Radost points out, $\Bbb Q^2$ is countable and dense in $\Bbb R^2$.  Because it's countable, we know we can enumerate (i.e., write as a sequence) $\Bbb Q^2$.  In fact, there are many constructive ways to do so.  Pick your favorite; call the sequence you've selected $S$.  (N.S. has suggested one.)  Then $S$ includes all pairs of points with rational cooordinates.
$S$ works.  How dow know that?  Choose $(x, y) \in \Bbb R^2.$  Then because $\Bbb Q^2$ is dense in $\Bbb R^2$, there is some sequence of points in $\Bbb Q^2$ that converges to $(x, y).$  (In fact, of course, there are lots of such sequences.)  The elements of that sequence must appear in $S$ in some order.  Choose an increasing (in $S$) subsequence of the rational sequence you chose.  That subsequence of $S$ converges to $(x, y)$.
