Is it easier to make perfect sets using intersection

Question

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?

Answer 1

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.