The plane is a disjoint union of sequences converging to $0$? Can $\mathbb{R^2}$ be written as a disjoint union (necessarily uncountable) of  sequences converging to $0$? 
 A: In any normed space $\mathbb{X}$, let $$f(x)=\frac{x}{1+\|x\|}.$$ Obviously, $f$ is injective, we have $$f^{-1}(x)=\frac{x}{1-\|x\|}$$ if $\|x\|<1.$ Let $$s(x)=(x, f(x), ff(x),fff(x),\ldots).$$ Since we can trace back any member of such a sequence back to the first element (just apply $f^{-1}$ as long as possible, i.e. as long as $\|x\|<1$), they are disjoint for different starting values. Let $\mathbb{X}_0=\{x\in\mathbb{X}:\|x\|\ge1\},$ and $\mathbb{X}_{n+1}=f(\mathbb{X}_n)$. Then, 
$$\mathbb{X}=\bigcup_{x\in\mathbb{X}_0\cup\{0\}}s(x).$$
That's easy to see, because $\mathbb{X}_n=\{x\in\mathbb{X}:\frac1n>\|x\|\ge\frac1{n+1}\}$ for $n\ge1$, so they are disjoint, and any $x\neq0$ is in some $\mathbb{X}_n$.
A: HINT: Let's show how to write $(0,1)$ as a disjoint union of sequences converging to $0$. Every number with continued fraction $[0, a_1,a_2, \ldots, ]$ belongs to the sequence $([0,n, a_2, \ldots])_{n\ge 1}$, and this sequence has limit $0$. 
$\bf{Added:}$ 
Easier to look at it as follows: Do this for every ray in $\mathbb{R}^2$, thus reduce to $(0,\infty)$. Consider the map $x \mapsto \frac{1}{x}$ from $(0,\infty)$ to itself. Now reduce to partitioning $(0,\infty)$ into sequences that have limit $\infty$. This is simple, the sequences are $(n+a)_{n\ge 0}$ for $a\in (0,1)$. 
Note that this is essentially the same solution as the above, and as the solution of @Professor Vector:
