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Consider the map from the set of infinite binary codes $$\{0,1\}^\mathbb{N}$$often written $$2^\mathbb{N}$$ to the Cantor set defined above: $f:2^\mathbb{N}\to C$ defined as for a sequence: $$a=(a_i)_{i=0}^{\infty}$$

$$f(a)=\sum_{i=0}^{\infty}\frac{2a_i}{3^{i+1}}$$

Then f is a homeomorphism of metric spaces, meaning that all the properties above hold for the space $2^\mathbb{N}$. Where 2 denotes the 2-element set $\{0,1\}$ with the discrete topology.

Where $$C_n=\frac{C_{n-1}}{3} \cup \left( \frac{2}{3} + \frac{C_{n-1}}{3}\right) \mbox{where} C_0 = [0,1]. $$

I am working with the Cantor set and want to work with Cantor spaces, but I am have a lot of trouble understanding how to define a Cantor space. I know that a topological space is a Cantor space if and only if it is non-empty, perfect, compact, totally disconnected, and metrizable. I am wondering how this mapping works. Could anyone help me grasp this whole concept of Cantor space and maybe a few mappings?

I was reading the examples portion of http://en.wikipedia.org/wiki/Cantor_space and that is where I am really basing everything off of.

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  • $\begingroup$ Before $f$ can be a homeomorphism, $2^{\mathbb N}$ needs to be a topological space. But you don't seem to have defined a topology/metric for it. $\endgroup$ – Henning Makholm Nov 21 '14 at 22:55
  • $\begingroup$ @HenningMakholm I am just working with something I found online, I'm trying to understand how this works so I don't really know how to define the topological space. $\endgroup$ – H5159 Nov 21 '14 at 22:59
  • $\begingroup$ @HenningMakholm I think I understand now. It is with respect to the discrete topology. I have fixed it. $\endgroup$ – H5159 Nov 21 '14 at 23:13
  • $\begingroup$ The discrete topology is not what you want. You need the product topology on $2^\mathbb{N}$ where $2$ is given the discrete topology as the wikipedia page says. This topology is metrizable by taking $d(x, y)$ to be $0$ if $x = y$ and otherwise to be $1/n$ where $n$ is the smallest index such that $x_n \not= y_n$. $\endgroup$ – Rob Arthan Nov 21 '14 at 23:58
  • $\begingroup$ @RobArthan I am still confused. I know how a space can be called a Cantor space (if it satisfies the properties I listed). In my paper, I want to talk about this Cantor space I have mentioned. Then I want to proceed to prove that it is compact, perfect, totally disconnected, and metrizable. But I'm having trouble starting to grasp my mind around what I'm truly working with. $\endgroup$ – H5159 Nov 22 '14 at 0:07

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