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I have to prove the following: We know that any Cantor set $C$ is uncountable and has zero measure: is it true than any set $L$ which have these properties is equal to a Cantor set?

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To typeset mathematics at this site you can use LaTeX (as you did), but you need to enclose it between dollar signs. $...$ – Martin Sleziak Jul 20 '12 at 7:55
The Cantor set can be proved to be uncountable from the fact that it is a perfect set. – Andrew Salmon Jul 20 '12 at 7:56
What equality are you asking about? One can add or subtract any countable number of points from the Cantor set without changing the uncountable and zero measure properties, but it will not be the same set. You could do the Cantor construction removing the middle $\frac 34$ and again get an uncountable set of zero measure. Stefan Geschke is using homeomorphism-is that what you meant? – Ross Millikan Jul 20 '12 at 16:57
I khnow that i wanted a set independent with cantor – kaveh Jul 21 '12 at 8:18
up vote 6 down vote accepted

For the first question: If $L$ is uncountable, has measure zero, and is also compact, then you obtain a homeomorphic copy of the Cantor set after removing countably many points:

Remove from $L$ all open intervals with rational endpoints that have a countable intersection with $L$. For each of these countably many intervals we have removed only countably many points. Hence the remaining set is still uncountable. It is also closed since we removed open sets. Hence it is still compact. Finally, the remaining set has no isolated points.

Since the remaining set is of measure zero, it does not contain an interval. Hence it is zero-dimensional. Now all compact zero-dimensional subsets of the real line that don't have isolated points are homeomorphic to the Cantor set. (This is non-trivial to show.)

So, not all uncountable measure zero sets look like the Cantor set, but when they are compact, they are a Cantor set with countably many additional points.

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first If $L$ is uncountable, has measure zero, and is also compact, then why you obtain a homeomorphic copy of the Cantor set after removing countably many point?if if is a theory please introduce me refrences – kaveh Jul 20 '12 at 8:31
The first part is due to Cantor and Bendixson. A set is called perfect if it is closed and has no isolated points. Cantor and Bendixson showed that every closed set is the disjoint union of a countable and a perfect set. This perfect set is unique. If the closed set you start with is uncountable, then the perfect set is nonempty. – Stefan Geschke Jul 20 '12 at 16:58
That every (nonempty) compact, zero-dimensional subset of the real line without isolated points is homeomorphic to the standard Cantor set is apparently due to Brouwer. See this link: – Stefan Geschke Jul 20 '12 at 17:05
whats part ? (The first part is due to Cantor and Bendixson) – kaveh Jul 21 '12 at 17:05

Yes, the argument that you sketched does prove that the middle-thirds Cantor set has Lebesgue measure $0$.

There are a number of ways to show that $C$ is uncountable, and in fact has cardinality $\mathfrak{c}$. One way it to notice that the points of the Cantor set are precisely the numbers in $[0,1]$ that have ternary (base-three) expansions consisting entirely of $0$s and $2$s. When you remove $\left(\frac13,\frac23\right)$, for instance, you remove every number whose only ternary expansion starts $0.1\dots\;$. When you remove $\left(\frac19,\frac29\right)$, you remove the numbers whose ternary expansions start $0.01\dots\,$, and when you remove $\left(\frac79,\frac89\right)$, you remove those whose ternary expansions start $0.21\dots\;$. Thus, after these three intervals have been removed, every number that’s left has a ternary expansion that begins $0.00,0.02,0.20$, or $0.22$, and in general stage $n$ of the construction removes those numbers whose ternary expansions have a $1$ in the $n$-th place.

If $x\in C$, think of the ternary expansion of $x$ as an infinite sequence of $0$s and $2$s, $\langle x_1,x_2,x_3,\dots\rangle$. If you want to be a bit more formal about it, $$x=\sum_{n\ge 1}\frac{x_n}{3^n}\;,$$ where each $x_n$ is either $0$ or $2$. For $n\in\Bbb Z^+$ let $$\hat x_n=\begin{cases}0,&\text{if }x_n=0\\1,&\text{if }x_n=2\end{cases}\;.$$ In other words, we’re leaving the $0$s alone and replacing each $2$ with a $1$. Now interpret the sequence $\langle\hat x_1,\hat x_2,\hat x_3,\dots\rangle$ as the binary (base-two) expansion of a number $f(x)$:

$$f(x)=\sum_{n\ge 1}\frac{\hat x_n}{2^n}\;.$$

Show that the map $f:C\to[0,1]$ is surjective.

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For the first question, you can finish the argument by noting that Lebesgue measure is monotonic, i.e. that if $A\subseteq B$ where $A$ and $B$ are measurable then $m(B)\geq m(A)$. Since you have sets containing the Cantor set of arbitrarily small measure, the Cantor set must have measure $0$.

For the second, try proving that a number $x\in [0,1]$ is in the Cantor set iff its base $3$ expansion has no ones, say by showing that it is in $C_1$ iff its first digit after the decimal place is not $1$, etc. Using this, you can easily constructed a bijection between the Cantor set and the set $[0,1]$ by replacing every two in the base $3$ expansion with a one and interpreting the result as a binary expansion (this is the restriction of the Cantor function to the Cantor set).

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