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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?

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Is the Sierpinski carpet really homeomorphic to $T$? That sounds very unlikely, since $T$ is totally disconnected whereas the Sierpinski carpet contains a subset homeomorphic to $[0,1]$. – Henning Makholm Aug 30 '13 at 20:18
Yes, you are absolutely right. I edited the question. – John Aug 30 '13 at 20:33
The title is not a complete sentence, so does not contain any question. Nor does the body clearly state one. What are you asking? – Marc van Leeuwen Aug 30 '13 at 20:46
up vote 7 down vote accepted

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).

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Very elegant, +1! – John Aug 30 '13 at 20:38

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