Let $\alpha$ be a fixed real number s.t. $0<\alpha<1$. In stage one of the construction, remove the centrally situated open interval in $[0,1]$ of length $\alpha$. In stage 2, remove two central intervals each of relative length $\alpha$, one in each of the remaining intervals after stage 1, and so on.

Let $C_\alpha$ denote the set which remains after applying the above procedure indefinitely.

So that's the set up. I am now trying to prove that the complement of $C_\alpha$ in $[0,1]$ is the union of open intervals of total length equal to 1. I'm having trouble understanding how there can be an open interval at all in the complement, because to me it seems like any open interval would have to contain a element of the cantor set. But then it's common knowledge that cantor sets are nowhere dense, and therefore every open interval on $[0,1]$ must be contained in it's complement?

If anyone could shed some light on this for me I'd really appreciate it!

I believe I have proven that $m_*(C_\alpha)$=0.

To see this, I will prove that after the k'th stage, the remaining set has total length $(1-\alpha)^k$.

The base case is trvial, assume by induction that this holds for the case $n=k$.

To get to the kth step, we removed $2^{k-1}$ intervals, and so at this step we have $2^k$ intervals total. I want to calculate $\alpha_k$.

$((1-\alpha)^k/2^k)$/$(\alpha_k)$ = $1/\alpha$

Because on step k the total length is $(1-\alpha)^k$, and this total length is divided amongst $2^k$ intervals, and so this equation makes sense because $\alpha_k$ is defind as being relative to $\alpha$.

So $\alpha_k=\alpha(1-\alpha)^k/(2^k)$

The length at the k+1 stage will be the length of the kth stage minus the total length of the $2^k$ segments we remove, i.e.


The length of this interval will go to zero as $k \to \infty$

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    $\begingroup$ I think you're trying to prove a false statement. Lebesgue measure of a Fat Cantor set is not necessarily zero. In fact $\lambda(C_\alpha)=\frac{3(1-\alpha)}{3-2\alpha}$ $\endgroup$ Commented Jul 7, 2018 at 14:41
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    $\begingroup$ Moreover, not every open interval $I\subset [0,1]$ is contained in $C_\alpha^C$ (e.g. $I=(0,1)$) $\endgroup$ Commented Jul 7, 2018 at 14:46
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    $\begingroup$ @francescop21: I believe you're thinking of a different construction where you remove an interval of absolute length $\alpha^n$ at stage $n$. I think Michael has correctly shown that his construction, where relative length $\alpha$ is removed, results in a set of measure zero. (An easier way to see the formula: if you remove a fraction $\alpha$ of the previous set, then what's left is a fraction $1-\alpha$ of it. So after $n$ iterations it is a fraction $(1-\alpha)^n$ of its original length.) $\endgroup$ Commented Jul 7, 2018 at 15:23
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    $\begingroup$ @francescop21: Indeed, Michael's construction is not what people usually call a fat Cantor set. $\endgroup$ Commented Jul 7, 2018 at 15:24
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    $\begingroup$ A hint for the original question: the complement of $C_\alpha$ consists of the union of all the open intervals you removed! And their total length is one minus the length of $C_\alpha$, which you just showed equals zero. The complement contains lots of open intervals, though it doesn't contain every open interval. "...it seems like any open interval would have to contain a element of the cantor set": I don't understand why you would think that. $\endgroup$ Commented Jul 7, 2018 at 15:29

1 Answer 1


Let $C_0=[0,1.$ For $n\geq 0$ we have $C_n=\cup F_n$ where $F_n$ is a finite set of pairwise-disjoint closed intervals, each of positive length. And the measure $m(C_n)$ is $\sum_{f\in F_n}m(f).$

For $n\geq 1,$ at stage $n$ we remove an open interval $I_f$ from each $f\in F_{n-1},$ with $m(I_f)=\alpha \cdot m(f).$ So the measure $$m( C_n)=\sum_{f\in F_{n-1}}(1-\alpha)m(f)=(1-\alpha)m(C_{n-1}).$$ So by induction on $n$ we have $m(C_n)=(1-\alpha)^n\cdot m(C_0)=(1-\alpha)^n. $

Therefore $m(\cap_{n\geq 0}C_n)\leq \inf_{n\geq 0}m(C_n)=0.$

For $n\geq 0$ let $L(n)$ be the length of the longest member of $F_n.$ Each $f\in F_n$ has a central open piece $I_f$ of length $\alpha \cdot m(f)$ removed at stage $n+1,$ resulting in $2$ members $f',f''$ of $F_{n+1},$ each of length $\frac {1-\alpha}{2}m(f).$ Therefore by induction on $n$ we have $$L(n)= ((1-\alpha)/2)^{-n}L(0)<2^{-n}L(0)=2^{-n}.$$ So if $K$ is an interval in $[0,1]$ of length $m(K)>0,$ then for any $n$ large enough that $2^{-n}<K$ we have $K\not \subset C_n,$ and a fortiori we have $K \not \subset \cap_{n\geq 0}C_n.$

The open intervals removed at stage $n\geq 1$ are disjoint from each other and are disjoint from the open intervals removed at any previous stage. So the family $I$ of all open intervals that are removed is a pair-wise disjoint family, so $$\sum_{i\in I}m(i)=m(\cup_{i\in I}\;i)=m(\cup I)=m([0,1]\backslash \cap_{n\geq 0}C_n)=1.$$

BTW some students erroneously suppose that $C=\cap_{n\geq 0}C_n$ consists entirely of the end-points of the members of $I$. But all limit-points of all convergent sequences of those end-points also belong to $C$.

  • $\begingroup$ Wow, great write up! Thanks so much! Your closing statement really peaked my interest, that all limit-points of all convergent seqeunces of those end points also belong to $C$. I was aware that more than just the end up points of the removed intervals end up in $C$, and that the points in $C$ can be completely understand as ternary expansions without any 1's, but the characterization you provided is new to me. Any chance you could tell me more about that? I'm slow at understanding this stuff but you are a really great explainer. $\endgroup$ Commented Jul 8, 2018 at 16:15
  • $\begingroup$ Suppose $I_f$ always contains the mid-point of $f $ but $m(I_f) $ is arbitrary ( but not $0$ ), provided that $f$ \ $I_f$ consists of 2 closed intervals, each of positive length. Then $C$ is closed nowhere-dense but $m(\cup I)=\sum_{i\in I}m(i)$ can be any member of $(0,1,$ so $m(C)$ can also be any member of $(0,1)$... Such $C$ is called a "fat Cantor set". $\endgroup$ Commented Jul 9, 2018 at 2:49
  • $\begingroup$ (1)..By a construction analogous to Cantor's we can show that if $X$ is a non-empty complete metric space with no isolated points then $X$ has a subset $D$ whose cardinal is the cardinal of the set of all binary sequences, which is the cardinal of the reals.(With a little more work, we can make $D$ homeomorphic to the Cantor set).... So $C,$ in your Q, or the fat Cantor set in my other comment, has the cardinal of $\Bbb R$.... (2). A countable union of fat Cantor sets can have measure $1$ but its complement is still dense in $[0,1]$ by the Baire Category Theorem. $\endgroup$ Commented Jul 9, 2018 at 2:59

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