# Is the Mandelbrot set weakly self-similar?

A subset $$F$$ of an Euclidean space $$E$$ will be called weakly self-similar if for all $$x \in F$$ there is $$\epsilon_x>0$$ such that for all positive $$\epsilon \le \epsilon_x$$ there are $$y \in F$$, $$\epsilon' \ge \epsilon_x$$ and a similarity $$s$$ of $$E$$ such that $$s(F \cap B(y,\epsilon')) = F \cap B(x,\epsilon)$$ with $$B(z,r)$$ the ball of center $$z$$ and radius $$r$$.

Question: Is the Mandelbrot set weakly self-similar?

If so: Is there $$\alpha>0$$ such that for all $$x \in F$$, we can take $$\epsilon_x>\alpha ?$$

• @ClaudeChaunier: in the definition written, weakly self-similar does not main self-similar with a weak similarity (sorry for the confusion). If $\epsilon < \epsilon_x$ then $\epsilon < \epsilon'$, and so $s$ is not an isometry. – Sebastien Palcoux Nov 11 '18 at 19:16
• Sorry, there is no confusion in your definition, I should have seen that the part around $y$ is required to be bigger. Thank you to have pointed to it. – Claude Chaunier Nov 12 '18 at 6:16