Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
add comment

1 Answer

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).

share|improve this answer
    
Very elegant, +1! –  John Aug 30 '13 at 20:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.