Smith-Volterra-Cantor set is perfect

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.

• Do you know how to show that the middle-thirds Cantor set is homeomorphic to $\{0,1\}^{\Bbb N}$? – Brian M. Scott Jan 4 '17 at 6:35
• @BrianM.Scott There is a proof about that on this site. – Henricus V. Jan 4 '17 at 6:50
• 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. – Brian M. Scott Jan 4 '17 at 6:54
• @BrianM.Scott Using that argument is perfectly fine. Thanks. – Henricus V. Jan 4 '17 at 6:55
• 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. – Brian M. Scott Jan 4 '17 at 7:17