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Let $(a_j)_{j=0}^\infty \subseteq \mathopen]0,1\mathclose[$. We define the mirroring function: $$ r(x) := 1 - x, \qquad (x \in [0,1]) $$ and the scaling function: $$ s_j(x) := \frac12 (1 - a_j) x, \qquad (x \in [0,1],j \in \mathbb{W}) $$ Define $C_{j,0} = [0,1]$ for all $j \in \mathbb{W}$. We define $C_{j,k}$ inductively: $$ C_{j,k+1} := s_{j-k}[C_{j,k}] \cup r\circ s_{j-k}[C_{j,k}] \qquad (k \in [0\ldotp\ldotp j-1]) $$ and $C_j := \bigcap_{k=0}^j C_{j,k}$. Finally, define the Smith-Volterra-Cantor set to be $C :=\bigcap_{j=0}^\infty C_j$.

How do I show that $C$ is perfect? If $a_j = \frac13$ for all $j$, we obtain the usual Cantor set and a proof for its perfectness is given here, but I am trying to prove the general case.

I have already shown that $C$ is compact. (Being a intersection of closed sets and subset of $[0,1]$). Moreover, $C$ is nowhere dense and totally disconnected.

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    Do you know how to show that the middle-thirds Cantor set is homeomorphic to $\{0,1\}^{\Bbb N}$? – 2017-01-04
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    @BrianM.Scott There is a [proof](http://math.stackexchange.com/questions/69905/the-cantor-set-is-homeomorphic-to-infinite-product-of-0-1-with-itself-cy) about that on this site. – 2017-01-04
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    It does leave some details to be filled in, but if you’re comfortable with that, I can use it as a basis for an answer to your question here. – 2017-01-04
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    @BrianM.Scott Using that argument is perfectly fine. Thanks. – 2017-01-04
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    I just realized that whether $K$ is perfect depends on the scaling sequence: I’ve not completely checked the details, but I think that if $a_0=\frac34$, say, and $a_j=4^{-(j+1)}$ for $j>0$, then $K=\{0,1\}$. You need to choose the scaling sequence so that the closed intervals making up $K_j$ eventually do split into two non-empty pieces. – 2017-01-04
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    @BrianM.Scott I think my definition of SVC set is wrong. I am still coming up with a real definition. – 2017-01-04
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    The [usual definition](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set#Other_fat_Cantor_sets) removes the middle $a_j$ from each closed interval of $K_j$. This splits each closed interval in two. Each point of $K$ corresponds to a nest of these closed intervals that can be described by an infinite sequence of zeroes and ones: $0$ for left half of previous interval, $1$ for right half. This gives you the natural homeomorphism with $\{0,1\}^{\Bbb N}$. – 2017-01-04

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