Sierpinski Triangle as Finite Union of Dendrites Can the Sierpinski Triangle be written as a finite union of dendrites?  If so, can it also be verified what the minimal number is (assuming you can't do it with just two)?
This is a small piece from a more sophisticated thread, but the particular example can possibly be solved by 'geometric cleverness' so I wanted to make it a separate question.  The main thread is:
Finite Unions of Dendrites
 A: No, it can't be written as a finite union of dendrites.  Actually, it can't be written as a countable union of dendrites.  Since every neighborhood in the Sierpinski Triangle $S$ contains a simple closed curve, a dendrite has empty interior in $S$, so by the Baire Theorem you can't cover $S$ with countably many dendrites.  This is also true of dendroids, $\lambda$-dendroids etc. by the same argument; any class of continua which are acyclic.
This Baire trick was too clever for me, a friend hit me up with this one.
A: I have probably horribly misunderstood the question but here goes:
if you consider a triangle fractal (https://i.stack.imgur.com/shOK1.gif) with depth 1 then there are 3 dendrites, depth two there are 6 dendrites, depth three there are 15 dendrites.
so the Sierpinski triangle is this triangle fractal with depth n=inf. The formula for the number of dendrites for a triangle fractal with depth n is:
$$ 3+3^1+3^2+3^3+...+3^{n-1}=-3(1-3^n)/2 $$
the limit as n->inf for the nth term diverges to infinity, so the Sierpinski triangle can't be written from a finite amount of dendrites.
