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I am struggling to internalize how do we construct sets with CB ranks $\omega$ and $\omega+1$. I can imagine one of them but then I don't understand the other. Here's what I mean. I know how to construct a set with rank n for some natural. For rank 1 take $\{0\}$. Then add the sequence $1/n$ to get rank 2, then add sequences to all previous isolated points and so on recursively. What will I get if I extend this for every natural? My understanding is that I will get the set with rank $\omega$? As the empty set should come out in the end. If that is so then how would I construct one with $\omega + 1$? I need some point to survive in order to do it?

If my understanding is wrong and actually the result of $\omega$ derivations is not empty but rather $\{0\}$ then that will be the set with $\omega + 1$ rank. But then how do I get the one with just $\omega$ rank?

I'm missing some intuition and if you could please answer from a set theory perspective rather than a topology one that will be greatly appreciated.

Chris Eagle
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toniuyt
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    This MSE answer should be helpful in getting fairly straightforward examples for what you want, and for much larger ordinals (but not especially large; starts getting descriptively difficult much past ${\omega}^{\omega}).$ – Dave L. Renfro Oct 09 '24 at 01:01
  • To get to $\omega + 1$ you can place copies of the set of rank $\omega$ that you already constructed in the intervals $(\frac{1}{n+1}, \frac{1}{n})$ and add $0$. – Jonathan Schilhan Oct 09 '24 at 09:58
  • That's what I'm thinking as well but since the set with rank ω still boils down to the empty set I don't understand how the numbers of the form 1/n will survive after infinite steps? – toniuyt Oct 09 '24 at 12:25
  • @user9855939 Yes, the sets of rank $\omega$ will boil down to the empty set after $\omega$ many steps. But then you are still left over with $0$. So you need one extra step to remove $0$. – Jonathan Schilhan Oct 09 '24 at 13:42
  • But isn't the derivative of 0 part of the w steps we take? If it's not then aren't the total steps w + 1? Taking the w derivative is like taking the intersection of all steps right? So we will always get empty set in the end? I struggle to realize how the intersection will contain anything if in the end it all boils down to empty set. – toniuyt Oct 09 '24 at 14:37
  • @user9855939 $0$ is not removed in any finite step because at each finite step there is something left in each of the copies of the rank $\omega$ set, so $0$ stays non-isolated. – Jonathan Schilhan Oct 09 '24 at 18:11
  • Ok so then isn't this w + 1 steps? If 0 stays non-isolated after w steps then after 1 more we get to the empty set. Why do we need the construction in your first comment for w+1 if we have 0 left out? – toniuyt Oct 09 '24 at 21:43
  • @user9855939 Yes, I gave an example of a set of CB rank $\omega +1$. Isn't this what you asked for? I don't understand the rest of your comment. – Jonathan Schilhan Oct 09 '24 at 23:24
  • Let us continue this discussion in chat. – toniuyt Oct 10 '24 at 08:38

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Here is a general recursive procedure for constructing a set $A_\alpha$ of CB rank $\alpha$ for any countable ordinal $\alpha$:

First we may simply let $A_0 = \emptyset$ and $A_1 = \{0\}$, as you also note.

Then suppose that we already have constructed $A_\beta$ for each $\beta< \alpha$.

If $\alpha$ is of the form $\beta +1$ (i.e. a successor ordinal) we let $A_\alpha$ be the set that is obtained by placing a homeomorphic copy of $A_{\beta}$ in the interval $(\frac{1}{n+1}, \frac{1}{n})$ for each $n$ (e.g. by stretching and translating) and by adding the element $0$.

Otherwise, if $\alpha$ is a limit ordinal let $(\beta_n)$ be a cofinal sequence in $\alpha$. Then we let $A_\alpha$ be the set that is obtained by placing a homeomorphic copy of $A_{\beta_n}$ in the interval $(\frac{1}{n+1}, \frac{1}{n})$ for each $n$.

Inductively you can prove now that $A_\alpha$ has rank exactly $\alpha$

Jonathan Schilhan
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