1. Let $\{A_n\}$ be a sequence of perfect sets in a metric space. Then, is $\bigcap_{n\in\omega} A_n$ perfect?

  2. Recently, i have studied some Cantor-like sets and i got this very natural question.

Let $C_1$ be a closed connected set in a metric space $X$ and $\alpha\in \mathbb{N}^\mathbb{N}$

Define $A(n)$ be a union of disjoint $n$ closed connected subsets of $A$. (With Axiom of choice, it can be well-defined)

Define recursively, $C_{n+1}=C_n (\alpha_n)$.

I believe $\bigcap_{n\in\mathbb{N}}C_n$ is perfect.

Is it true or is there any generalization similar to this? Sorry in advance that i'm not really good at describing things..

  1. For any closed connected metric space $A$, there exists a perfect subset $B$ such that $Int(B)=\emptyset$? Or possibly, "Every perfect set in a metric space contains a perfect subset with empty interior"?
  • 1
    $\begingroup$ I'd recommend breaking these up into three separate questions. $\endgroup$ Commented Dec 4, 2012 at 17:27

2 Answers 2


The intersection of even finitely many perfect sets (closed sets with no isolated points) need not be perfect. Consider the sets $$A_n=\left\{\langle x,y\rangle\in\Bbb R^2:y=nx\right\},$$ for a counterexample. Even if we're given a nested sequence of perfect sets, their intersection need not be perfect--consider the intervals $$\left[0,\frac1{n+1}\right]$$ on the real line. This likewise provides a counterexample to your second question.

In answer to the third question, note that $\emptyset$ is a perfect set. In general, there need be no non-trivial example, as the one-point space is connected, and has $\emptyset$ as its only perfect subset.

  • $\begingroup$ Thank you. The reason why i'm asking second question is SVC,middle-$\alpha$ Cantor sets, Cantor-Fractals and all other Cantor-like sets look very alike but have all different definitions and way they are defined are quite ugly.. So i want to know if there is a simple description of this kind of sets. And yes, i'm looking for a non-trivial example for the third question. $\endgroup$
    – Katlus
    Commented Dec 4, 2012 at 17:43
  1. No. Let $A_n = [ \frac{-1}{n} , \frac{1}{n} ]$ for all $n$. Clearly each $A_n$ is perfect, but their intersection is the imperfect $\{ 0 \}$.

  2. As the answer to (1) shows, you will need to be careful with the details. If at each stage you take $n$ subintervals such that the left-endpoint of the left-most subinterval is the left-endpoint of the original interval, and similarly with the right-endpoint of the right-most subinterval, the answer will be similar to the Cantor set construction.

    Otherwise, you will probably want to make sure that each interval is split into at least two subintervals. Then you might be able to argue as follows: Given $x \in C = \bigcap_n C_n$ and $\epsilon > 0$, if some non-degenerate subinterval of $C$ contains $x$, there is nothing to do. Otherwise there must be an $n$ such that the entire subinterval of $C_n$ which contains $x$ is a subset of $( x - \epsilon , x + \epsilon )$. Starting with a subinterval of this interval in $C_{n+1}$ which does not contain $x$, we can recursively construct a convergent sequence consisting of endpoints of subintervals whose limit belongs to $C$ but is different from $x$. This point must be within $\epsilon$ of $x$. (Details are, as it seems is common with me, left to someone else.)

  3. (I'll have to think about this for a bit to see if there is a non-trivial example.)

  • $\begingroup$ Thank you very much for the answer to the second question! It seems "splitting into at least two subintervals" can cover every cantor-like sets $\endgroup$
    – Katlus
    Commented Dec 4, 2012 at 18:00

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