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Reading this, I wanted to do the classic demonstration again by myself but there are points that bother me.

Let $$C_0=[0,1], C_1=[0,\frac13]\cup[\frac23,1]...$$We have the classical definition of the Cantor set $$\mathbb K_3:=\bigcap_{n=0}^{+\infty }C_n$$ and we know that $$\mathbb K_3=\{x\in [0,1]:x=\sum_{n=1}^{+\infty}\frac{a_n}{3^n} (\forall i , a_i\in \{0,2\})\}$$ I want to prove that $\mathbb K_3$ is the attractor $\mathcal A$ of $\{f_1,f_2\}$, where $$f_1:[0,1]\to[0,1], x\mapsto \frac13 x \text{ and }f_2:x\mapsto \frac13(x+2)$$

I know that

  • $(\mathbb R,d)$ is a metric space, with $d(x,y):=|x-y|$. $\mathbb R$ is complete by construction;
  • $\forall x,y \in \mathbb R, |f_2(x)-f_2(y)|\leq \frac13|x-y|$(the same for $f_1$)
  • $\mathcal A$ is the unique compact of $\mathbb R$ such that $\mathcal A=f_1(\mathcal A)\cup f_2(\mathcal A) \color{red}{(*)}$;
  • $\color{blue}{\text{I don't remember how to prove that K3 is compact}} $;
  • Using $\color{red}{(*)}$, it is then easy to prove that $$\boxed{\mathbb K_3=\mathcal A(\{f_1,f_2\})}.\square$$ I'd love to hear from you about this demonstration and help me complete it. Thanks in advance.
Stéphane Jaouen
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    The standard proof that an iterated function system has a compact attractor is given in Hutchinson's 1981 paper. The basic idea is to regard the IFS as a contractive map on the space of subsets of the ambient space, with distance measured via the Hausdorff distance. – Xander Henderson Mar 28 '24 at 19:31
  • I can probably answer your question, but, honestly, I am a bit confused about what your question actually is. I think that you are asking about a single step, i.e. "How does one show that an IFS has a compact attractor?" If that is the case, I would suggest that you edit the question to highlight that and make it a bit more obvious. Otherwise, I would suggest that you edit the question to focus on whatever your real concern is. – Xander Henderson Mar 28 '24 at 19:43
  • No, showing that an IFS has a compact attractor is way beyond what I expect. I made it clear in the post that "I don't remember how to prove that K3 is compac". And I also wonder if all the arguments are there for a complete demonstration. That's all. – Stéphane Jaouen Mar 28 '24 at 19:49
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    The standard argument is the more general argument: subject to some minor constraints (to ensure contractivity), the attractor of an IFS is compact (and unique, for that matter). Alternatively, show that the Cantor set is closed and bounded. Bounded should be easy. For closed, something something intersection of closed sets something something. – Xander Henderson Mar 28 '24 at 19:59

1 Answers1

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Yes, the Cantor set is compact by the Heine–Borel theorem, as it's a closed and bounded subset of $[0, 1] \subset \mathbb{R}$ under the standard topology. Boundedness is obvious, and the fact that it's closed is most easily seen by observing that its complement is a union of open intervals (the "middle thirds").

You can read more about its topological properties here.

Sammy Black
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