Is it easier to make perfect sets using intersection [closed]

A perfect set $A$ is one in which every point is a limit point. So it has to be closed. Does this mean that if we want to generate perfect sets inductively it is usually best to just intersect collections of other sets?

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closed as not a real question by Jonas Meyer, Asaf Karagila, joriki, t.b., Mariano Suárez-Alvarez♦Aug 24 '11 at 15:00

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Well, $\mathbb{Q}$ is perfect but it isn't closed in $\mathbb{R}$. – t.b. Aug 23 '11 at 8:05
I cannot see what this is asking... – Mariano Suárez-Alvarez Aug 23 '11 at 8:13
It seems that the question rests on the assumption that a set in which every point is a limit point has to be closed, which, as Theo pointed out, is incorrect. The question doesn't make much sense without that assumption, so, in its own words, "it has to be closed". – joriki Aug 23 '11 at 9:09

Intersecting arbitrary collection of sets will get you nowhere, even if they are decreasing and closed. Instead here is a nice way of procuring a very special perfect set from a topological space.

Suppose $X$ is a topological space (for sake of convenience $T_1$), for $x\in X$ if $\{x\}$ is open, we say that $x$ is an isolated point.

Let $I(X)=\{x\in X\mid \{x\}\text{ is open}\}$, then we define $D(X)=X\setminus I(X)$. We say that $D(X)$ is the Cantor-Bendixson derivative of $X$.

Now define a sequence:

• $X_0 = X$
• $X_{\alpha+1}=D(X_\alpha)$
• $\displaystyle X_\delta=\bigcap_{\alpha<\delta} X_\alpha$, for $\delta$ a limit ordinal.

If there exists some ordinal $\alpha$ such that $X_\alpha=D(X_\alpha)$ then we say that $\alpha$ is the Cantor-Bendixson rank of $X$, and $X_\alpha$ is the kernel of $X$, denoted by $\ker(X)$.

A few observations:

1. If $\ker(X)$ exists then it is perfect, since $D(\ker(X))=\ker(X)$ implies $I(\ker(X))=\varnothing$.
2. If $X$ is perfect then $I(X)=\varnothing$ and therefore $D(X)=X$, and $X=\ker(X)$.
3. If $X$ is a set then for some ordinal $X_\alpha=X_{\alpha+1}$.
4. If $Y\subseteq X$ is perfect in the induced topology from $X$, then $Y\subseteq\ker(X)$.
5. If $\{x_i\mid i\in I\}$ is a convergent net in $D(X)$ then its limit is in $D(X)$, therefore $D(X)$ is always closed. In particular $\ker(X)$ is closed.

An interesting theorem is that if $X$ is a separable metric space then its rank is always a countable ordinal.

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Can the rank be uncountable in a non-separable metric space? – Ricky Demer Aug 23 '11 at 9:08
@Ricky, that is a good question. I believe so. Let me try and construct an example (granted Theo or Brian M. Scott, who probably have one at hand won't get here before that :-)) – Asaf Karagila Aug 23 '11 at 9:11
@Ricky: Here is an almost trivial observation, take $X_\alpha$ to be a metric space with rank $\alpha$, and take $X=\bigcup X_\alpha$ (wlog they are disjoint, of course) and define the distance between points from different parts to be $42$. It is a metric space, and it has rank of at least $\alpha$ for any countable $\alpha$. Therefore the rank is at least (and in fact equal to) $\omega_1$. – Asaf Karagila Aug 23 '11 at 10:10
@Ricky: A couple of references I know off-hand are: Stone's "Kernel constructions and Borel sets" [Trans. Amer. Math. Soc. 107 (1963), pp. 58-70] ams.org/journals/tran/1963-107-01 and Chapter 1 of Sierpinski's "General Topology" amazon.com/dp/0486411486 Also, there are essentially three different ways to obtain what Asaf called the kernel of $X$: Cantor's method of iterating the derived set function (1883), Lindelof's method of using condensation points (1905), and Hausdorff's method of forming the union of all subsets of $X$ that are dense in themselves (1914). – Dave L. Renfro Aug 23 '11 at 14:36
@Ricky: The three methods I mentioned for obtaining the kernel of $X$ are outlined at groups.google.com/group/sci.math/msg/1b7439a55bf25eb9 For generalizations and applications of the method of iterating the – Dave L. Renfro Aug 23 '11 at 14:40