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Suppose $X$ is a separable metric space, let $D(X)$ denote the Cantor-Bendixson derivative of $X$, and $D_\alpha(X)$ the $\alpha$-th derivative of $X$.

We denote $\operatorname{Ker}(X)$ the kernel of $X$, which is the least derived subspace of $X$ such that $D_\alpha(\operatorname{Ker}(X))=\operatorname{Ker}(X)$ for all $\alpha$. This implies that the kernel is a perfect space.

We say that a topological space is zero-dimensional if it is Hausdorff and has a basis of clopen sets.

Theorem: Suppose $X$ is a separable metric space, and $\operatorname{Ker}(X)$ is zero-dimensional, then $X$ is zero-dimensional.

The proof we were given is fairly constructive, and quite long and filled with details. However being the lazy person that I am, I was looking for a shorter proof.

My efforts came to this conclusion:

Suppose $\alpha=\gamma+\beta$ for $\gamma,\beta<\alpha$ and that for all $\beta<\alpha$ we have that if $D_\alpha(X)$ is zero-dimensional, then so is $X$, then $D_\alpha(X)$ being zero-dimensional implies $X$ is zero-dimensional.

Proof: Note that $D_\alpha(X) = D_\beta(D_\gamma(X))$, therefore $D_\gamma(X)$ is zero-dimensional, and therefore $X$ is.


This proof, however, does not hold for indecomposable ordinals (e.g. $\omega$, $\omega^\omega$, etc.) which makes it of very little use. I am certain that such argument can be made for indecomposable ordinals (or limit ordinals in general), I just can't find it.

Any help will be most appreciated.

Edit: It seems likely that the requirement that $X$ is a separable and metric space is redundant. In most likelihood it will be true for Hausdorff and (completely?) regular spaces.

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I'm starting to feel weird asking questions and having to come up with the answers as well... I suppose, though, it is better to write my answer than to delete the question. –  Asaf Karagila Jul 7 '11 at 19:55

1 Answer 1

(I believe that I have found a proof, I am not 100% sure about its correctness though.)

Theorem: Suppose $X$ is a metric space, then if $D_\alpha(X)$ is zero-dimensional, so is $X$.

Denote by $D_\alpha$ the $\alpha$-th derivative, and let $I_\alpha=D_\alpha\setminus D_{\alpha+1}$, that is the isolated points of $D_\alpha$.

Now suppose $D_\alpha$ is zero-dimensional, that is Hausdorff and has a basis of clopen sets. Denote such basis by $\mathcal B$.

We generate a clopen basis for $X$ by induction. Denote the topology $\tau$.

Take $B_0$ as the basis for $\tau^0$ formed as follow: $U\in B_0$ if and only if one of the following holds:

  1. $U\in\mathcal B$; for every $x\in U$, $x$ is not a limit point (in $X$) of the sequence from $X\setminus D_\alpha$;
  2. $U = V\cup\{x_n\in X\setminus D_\alpha\mid n<\omega\}$ where $x_n$ is a sequence whose limit is in $V$, for some $V\in\mathcal B$;
  3. $U=\{x\}$ for $x\notin D_\alpha$;
  4. $X\setminus U$ has one of the above properties.

Claim: $U\in B_0$ is clopen in the topology $\tau^0$.
Proof: The proof is trivial, since $X\setminus U\in B_0$.


Suppose $B_\gamma$ for $\gamma<\beta$ was defined and generates a zero-dimensional topology. We define a basis for $\tau^\beta$, $B_\beta$. Take $U\subseteq X$, then $U\in B_\beta$ if and only if one of the following holds:

  1. $U\in \bigcup_{\gamma<\beta} B_\gamma$ and for every $x\in U$, there is no sequence in $I_\beta$ such that $x$ is a limit point of it;
  2. $U = V\cup\{x_n\in \bigcup_{\gamma<\beta} I_\gamma\mid n<\omega\}$ for some $V\in \bigcup_{\gamma<\beta} B_\gamma$ and $\lim x_n\in V$;
  3. $U=\{x\}$ for $x\in I_\beta$;
  4. $X\setminus U$ has one of the above properties.

Claim: $U\in B_\beta$ is clopen in the topology $\tau^\beta$. (trivial as before)


Claim: $\tau=\tau^\alpha$.
Proof: Suppose $x_n\to x$ in $\tau$. If $x_n\in D_\alpha$ then $x\in D_\alpha$ and the convergence is same in $\tau^\alpha$. Suppose $x_n\notin D_\alpha$. If $x\in D_\alpha$ then for every clopen set containing $x$ we adjoined infinitely many $x_n$, so the convergence remains.

Otherwise, since $D_\beta$ are always closed we have that $x\in I_\beta$ such that $x_n\notin D_\beta$, in particular on the $\beta$-th level, $\tau^\beta$ was such that every set containing $x$ was adjoined with infinitely many $x_n$'s.

For the other direction we have that if $x_n\to x$ in $\tau^\alpha$ then once again we deal with the cases - if the sequence in $D_\alpha$ we are done, otherwise we have some $\beta$ in which $x\in I_\beta$. In particular $x_n\to x$ implies every open set with $x$ has infinitely many $x_n$'s, which was taken care of if and only if $x_n$ was a sequence of points which were removed before $D_\beta$ and $\tau$ has converging to $x$. In particular $x_n\to x$ in $\tau$.


We have that $\tau$ has $B_\alpha$ as a basis of clopen sets, as needed. (Separability and second-countability are retained by this process, and I believe the proof is easily generalized to Hausdorff spaces by replacing sequences with nets.)

A nice corollary is that if the kernel of $X$ is zero-dimensional, then $X$ is zero-dimensional.

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Try applying your construction to $\omega + 1$ with the usual topology. Unless I'm missing something, clause (2) of your construction puts every $\{\omega\} \cup A$ with $A \in [\omega]^\omega$ in your basis, and you end up generating the discrete topology. It's not clear that at each stage you actually have a base rather than a subbase. –  Brian M. Scott Jul 10 '11 at 3:18
    
Note that any argument is going to have to use more than 1º countability, by virtue of [R.C. Solomon's example][1] of a 1º countable, scattered (of rank $3$) $T_{3\frac{1}{2}}$ space that is not $0$-dimensional. [1]: blms.oxfordjournals.org/content/8/3/239.extract –  Brian M. Scott Jul 10 '11 at 3:19
    
@Brian: I discussed my teacher about this proof and he remarked a similar problem about clause (2), I have an idea how to correct this and will do so after I take the exam (which is in about one hour from now). As for the second comment I'm not sure what you mean by 1º. Lastly, many thanks for proof reading this!! –  Asaf Karagila Jul 10 '11 at 5:05
    
1º is just an abbreviation for 'first', since my comment was starting to run long. (At that point I was still hoping to combine the comments.) Good luck with the exam! –  Brian M. Scott Jul 10 '11 at 5:32

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