I'm trying to understand this concept of Analytic subset of Cantor Space that is, which subsets of the Cantor Space are Analytic subsets and how to determine that.

Here is what I've known this far (from Wikipedia):

Polish Space: A topological space that is separable and completely metrizable i.e. a separable space which can be induced by a complete metric.

Analytic Subsets: A subset of a Polish space is analytic if it is continuous image of a Polish space. These subsets are closed under countable union and intersection, continuous image, *inverse image. *(This is not getting conceptually clear to me - subsets of Polish space that are continuous images of Other Polish spaces? Too complicated !)

Now the Cantor space is denoted by $C=\{0,1\}^{\mathbb N}.$ Now, to find out the analytic subsets of $C$, I need to know first whether $C$ itself is Polish or not, right? If the Cantor space itself is separable and be induced by a complete metric. But are those true? I know an uncountable separable completely metrizable space has the Cantor Space as a subspace. But if $C$ is not $Polish$ itself, then how will that given definition of analytic subsets work here? (According to Wikipedia, only Polish spaces can have Analytic subsets; correct me if I'm wrong.)

A subset of $C$ here consists of a family of infinite binary sequences. How do I decide when this subset is Analytic?

It is because of Some further results on ideal convergence in topological spaces by Pratulananda Das where Analytic P ideals of $\mathbb N$ are used but I do not know how those things look. I realize that this should have been my first post concerning this paper. Anyways, kindly help me to understand this concept. Hope I've conveyed my query properly. Thanks.

  • $\begingroup$ Re: I do not know how analytic P-ideals look like. In Das' paper you find a result of Solecki, which says that analytic P-ideals are precisely the ideals of the form $$\mathcal I=\operatorname{Exh}(\phi)=\{A\subseteq\omega; \|A\|_\phi = 0\}$$ where $$\|A\|_\phi=\limsup_{n\to\infty} \phi(A\setminus n)=\lim\limits_{n\to\infty} \phi(A\setminus n).$$ and $\phi$ a lower semicontinuous submeasure. Probably you could learn more about this type of ideals from the relevant chapters of the I. Farah. Analytic Quotients and V. Kanovei: Borel Equivalence Relations. $\endgroup$ – Martin Sleziak Oct 28 '16 at 4:14
  • $\begingroup$ Or at least by skimming through those chapters, you should see several examples of various classes of analytic ideals. (I should add that this is not a topic I am too familiar with, but I've stumbled upon it a few times.) Also, every Borel set is analytic, so if you can show that some ideal is, for example, $F_\sigma$ of $F_{\sigma\delta}$, then you know that it is analytic. $\endgroup$ – Martin Sleziak Oct 28 '16 at 4:17

The Cantor space is indeed a Polish space; consider the metric given by $d(f, g)=0$ if $f=g$ and $d(f, g)=2^{-n}$ if $n$ is the least natural number such that $f(n)\not=g(n)$. Alternatively, view the Cantor space as a subset of $\mathbb{R}$ (the Cantor set) and use the induced metric. It's a good exercise to show that each of these metrics is compatible with the topology, and is complete.

Meanwhile, you state that you find the definition of an analytic set to be too complicated. This is, unfortunately, the nature of the beast: analytic sets are quite complicated objects! We may improve things a little by the following: an analytic subset of a Polish space $X$ is a continuous image of the Baire space $\mathbb{N}^\mathbb{N}$ (this is also a Polish space; in fact, the first metric I described above for the Cantor space works here). This is simply because every Polish space is the continuous image of Baire space (this is a good exercise). However, this doesn't make things much simpler. Analytic sets are pretty complicated objects: every Borel set is analytic, for example.

The following is a good exercise for getting a handle on how analytic sets behave:

Consider $\mathbb{R}$ with the usual topology; this is a Polish space. Show that $[0, 1], (0, 1), \mathbb{R}\setminus\mathbb{Q}$, and $\mathbb{Q}$ are all analytic subsets of $\mathbb{R}$.

Where things get really fun is beyond the Borel. We can view an element of Cantor space $f$ as coding a binary relation $R_f$ on $\mathbb{N}$: we define $R_f$ by saying $R_f(a, b)$ holds iff $f(\langle a, b\rangle)=1$, where "$\langle\cdot,\cdot\rangle$" is some bijection from $\mathbb{N}^2$ to $\mathbb{N}$ (e.g. the Cantor pairing function). Now, for $\mathfrak{P}$ some property of relations on $\mathbb{N}$, let $S_\mathfrak{P}$ be the set of $f\in 2^\mathbb{N}$ such that $R_f$ has property $\mathfrak{P}$ (e.g. if $\mathfrak{P}$ is "holds for infinitely many pairs," then $S_\mathfrak{P}$ is the set of all $f\in 2^\mathbb{N}$ which have infinitely many $1$s). Then if we let $\mathfrak{Q}$ be the property, "Is not a well-ordering of $\mathbb{N}$," it turns out that $S_\mathfrak{Q}$ is analytic but not Borel. Indeed, $S_\mathfrak{Q}$ is in some sense the most complicated analytic set! See Kechris' book on descriptive set theory for more details; he does a good job of describing the basic properties of analytic sets.

  • $\begingroup$ what are $f$ and $g\ ?$ Seems like they are functions from the Cantor's space to somewhere. To where exactly? $\endgroup$ – user118494 Oct 28 '16 at 4:23
  • $\begingroup$ @user118494 No, $f$ and $g$ are elements of the Cantor space. But elements of the Cantor space are functions from $\mathbb{N}$ to $\{0, 1\}$! $\endgroup$ – Noah Schweber Oct 28 '16 at 4:24
  • $\begingroup$ Since you wrote $f(n)\neq g(n)$ so I thought they must be functions! $\endgroup$ – user118494 Oct 28 '16 at 4:26
  • 1
    $\begingroup$ Oh! Is it that thing $\{0,1\}^{\mathbb N}\ ?$ This is the function that identifies the Cantor Space with $P(\mathbb N)$. $\endgroup$ – user118494 Oct 28 '16 at 4:31
  • 1
    $\begingroup$ Yes, in general "$A^B$" denotes the set of functions from $B$ to $A$. You can think of the Cantor space as either $\{0, 1\}^\mathbb{N}$ or as $\mathcal{P}(\mathbb{N})$; the two interpretations are related via the notion of the characteristic function of a set. $\endgroup$ – Noah Schweber Oct 28 '16 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.