Is there a cantor set in $\mathbb{R}^{n}-\{0\}$ which intersects every ray from the origin? Let $T$ denote the Cantor ternary set on the real line with the usual topology. Any topological space $C$ homeomorphic to the Cantor ternary set is called a Cantor set.
The case $n=1$ is clear. For $n\ge 2$, is there a cantor set in $\mathbf{\mathbb{R}^{n}-\{0\}}$ which intersect every ray from the origin?
 A: For $n=2$, consider the function $f$ defined as: For $x\in T$, write $x$ as a ternary fraction with only the digits 0 and 2. Replace every 2 with an 1; $f(x)$ is then the value of the resulting digit sequence interpreted as a binary fraction.
Clearly $f$ is surjective $T\to [0,1]$. (It is the restriction of the Cantor function to $T$)
Prove that the graph of $f$ -- that is, $\{(x,f(x))\mid x\in T\}$ -- is homeomorphic to $T$. Then distort this graph so it wraps around the origin.

Generalizing this to higher $n$ should just be a matter of splitting $f$ into $n-1$ functions that extract disjoint sequences of digit positions from $x$. 
Alternatively, compose $f$ with a space-filling curve. (This option is probably easier to prove).
A: The same idea works with no change: choose any surjective continuous map $p$ from the Cantor set $C$ to the unit sphere in $\mathbf{R}^n$, any injective continuous map $i$ from $C$ to the set of positive reals, and consider $j(x)=i(x)p(x)$ and $C'=j(C)$. Then $C'$ is a Cantor set intersecting every ray.
(How to construct $p$? a general result is that every nonempty metrizable compact set $X$ admits a continuous surjection from a Cantor set. One can be more explicit here of course since the sphere is a finite union of closed balls and each closed ball is homeomorphic to $[0,1]^{n-1}$.)
