# Sierpinskis Gasket - Linear Algebra

1. Mar 10, 2012

### Upsidealien

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. Mar 10, 2012

### MathematicalPhysicist

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 $$x\in S$$ thus for every $$n \geq 1$$ x is in delta(n), thus $$x\in f_1(\Delta n)\cup f_2(\Delta(n)\cup f_3(\Delta(n))$$ for every n>0, thus you can keep iterating it ad infinitum, thus $$x\in f_1(S)\cup f_2(S)\cup f_3(S)$$.