# graph of the Cantor set in Mathematica

The interval $$[0,1]$$ of the real line, which we call $$C_0$$, and divide it into three equal subintervals. In this way we obtain the following intervals $$[0,1/3]$$,$$[1/3,2/3]$$,$$[2/3,1]$$, and we get rid of the interval $$[1/3,2/3]$$. Then, we obtain the set $$C_1=[0,1/3]\cup[2/3,1]$$. Now repeat the above process on each of the intervals $$C_1$$, then $$C_2=[0,1/9]\cup[2/9,3/9]\cup[6/9,7/9]\cup[8/9,1]$$ Proceeding in the same way we obtain $$C_3,C_4,C_5,C_6,\dots$$ Thus we define the Cantor set as the intersection of all $$C_i$$ constructed before. Now the goal is to obtain the graph of the Cantor set in the different steps

You can also use new-in-version-10 NumberLinePlot. The last example in NumberLinePlot >> Applications slightly modified:

NumberLinePlot[NestList[# /. {a_, b_} :>
Sequence[{a,2 a/3 + b/3}, { a/3 + 2 b/3, b}] &,
Interval[{0, 1}], 5], {x, 0, 1},AspectRatio->1/4,
PlotStyle -> Directive[CapForm["Butt"],Thick]]/. Point[x_]:>Sequence[]


It may be an available answer:

In=

n = 5;
f = #/3 &; g = #/3 + 2/3 &;
Fission[list_] := Flatten@Through[{f, g}@list]
intervals = NestList[Fission, {Interval[{0, 1}]}, n];
segments[n_] :=
intervals[[n]] /. Interval[{a_, b_}] :> Line[{{a, 0.1 n}, {b, 0.1 n}}]
Graphics[{Black, Array[segments, n + 1]}]


Out=

If you are in 11.1,CantorMesh will make your life easer:

NumberLinePlot[Interval @@@ (MeshPrimitives[#, 1] &@*CantorMesh /@ Range[4] /.
Line -> Flatten), PlotRange -> {0, 1}, AspectRatio -> 1,
PlotStyle -> Directive[PointSize[0], Black]]