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Sierpinskis Gasket - Linear Algebra

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Where
    :- ∆0 = ∆ is the original triangle ABC.
    :- DEF are the midpoints of AB, BC, AC respectively.
    :- f1, f2, f3 map the triangular region ABC to the triangular region ADF, DBE and FEC respectively.
    :- ∆n+1 = f1(∆n) ∪ f2(∆n) ∪ f3(∆n) for n≥0.

    (these are just definitions of Sierpinskis gasket)

    1. We need to first sketch ∆1 and ∆2 and prove that ∆n+1 ⊆ ∆n for all n ≥ 0.

    We then define S = intersection of ∆n from n=1 to ∞.

    2. We then need to prove that S is non-empty and that S = f1(S) ∪ f2(S) ∪ f3(S).


    2. Relevant equations

    Given above.

    3. The attempt at a solution

    I attempted 1. by using induction but did not get very far and for 2. I have proved that S ⊇ f1(S) ∪ f2(S) ∪ f3(S) but I am struggling to prove S ⊆ f1(S) ∪ f2(S) ∪ f3(S)


    Thanks

    Tom
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    So you basically at the n+1 step map the n-th triangle to three of its interior triangles, and leaving the middle triangle intact. Doesn't this by itself prove the inclusion, even proper inclusion.

    For the intersection, take [tex]x\in S[/tex] thus for every [tex]n \geq 1[/tex] x is in delta(n), thus [tex]x\in f_1(\Delta n)\cup f_2(\Delta(n)\cup f_3(\Delta(n))[/tex] for every n>0, thus you can keep iterating it ad infinitum, thus [tex]x\in f_1(S)\cup f_2(S)\cup f_3(S) [/tex].
     
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