Suppose we were to list all points in the $3/7$th Middle Cantor set where we first remove $3/7$th of $[0,1]$:
$$[0,1/7]\cup[2/7,3/7]\cup[4/7,5/7]\cup[6/7,1]$$
then $3/7$th of remaining intervals, $3/7$th that still remain, $3/7$th of those, etc.
If I'm correct, we can list all points in the set as:
$$\left\{\left(\sum_{q=1}^{w}\frac{j}{7^q}\right)+\frac{h}{7^{w+1}}:j\in\left\{0,2,4,6\right\},h\in\left\{0,1,2,3,4,5,6,7\right\},w\in\mathbb{N}\right\}$$
I tried recreating this in Mathematica:
Table[Sum[Subscript[j, 3][[q, w]]/7^q, {q, 1, w}] +
Subscript[h, 3]/7^(w + 1), {w, 1, 2}, {Subscript[j,
3], {0, 2, 4, 6}}]
But I realized they treated $j$ (or in this case j3
) as a fixed number with q
. As q
changes j3
remains the same which is not what I intended. This gives an unusual graph.
Subscript[B, 3] =
Select[Sort[
DeleteDuplicates[
Flatten[Table[
Sum[Subscript[j, 3]/7^q, {q, 1, w}] +
Subscript[h, 3]/7^(w + 1), {w, 1, 2}, {Subscript[j,
3], {0, 2, 4, 6}}, {Subscript[h,
3], {0, 1, 2, 3, 4, 5, 6, 7}}]]]], Between[#, {0, 1}] &]
Length[Subscript[B, 3]]
ListPlot[Table[{Subscript[B, 3][[x]], 0}, {x, 1,
Length[Subscript[B, 3]]}]]
Which I doubt is correct. How do we correct my method?
cantor = {a_, b_} :> {{a, a + (b - a)*2/7}, {a + (b - a)*5/7, b}};
$\endgroup$