Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I proved that every uncountable subset of $\mathbb R^n$ contains a condensation point. However, I can't see easily that the Cantor set contains a condensation point. The above proof certainly shows the existence of a condensation point of the Cantor set, but what's a more direct proof?

share|improve this question
1  
"The above proof certainly shows..." --> What proof? –  BCLC Jul 4 at 3:54
    
You can consider the sequence $\left( \dfrac{1}{3^n} \right)_{n\in\Bbb N}$ and $0$. –  Zircht Jul 4 at 3:55
1  
@Zircht: Are you thinking of showing that $0$ is a limit point? For condensation points, uncountability has to be used. –  Jonas Meyer Jul 4 at 4:10
    
@JonasMeyer Oh, sorry, I was thinking about accumulation points. –  Zircht Jul 4 at 4:49

2 Answers 2

For each $n$, the subset of the Cantor set contained in the interval $[0,1/3^n]$ is homeomorphic to the entire Cantor set, and in particular it is uncountable. Therefore $0$ is a condensation point.

Every point in the Cantor set is a condensation point. The same is true of all perfect subsets of $\mathbb R^n$

share|improve this answer
    
Excellent answer, as simple as possible and no simpler. –  jwg Jul 4 at 7:16

Let $S_1(x) = \frac{x}{3}$ and $S_2(x) = \frac{x}{3} + \frac{2}{3}$. Let $\mathcal{I_n} = \{(i_1,\ldots,i_n) : 1 \leq i_j \leq 2\}$ be n-tuples of 1s and 2s. The middle thirds Cantor set is the unique nonempty compact set such that $S_1(C) \cup S_2(C) = C$. In fact, for any $n \ge 1$, $C = \bigcup_{(i_1,\ldots, i_n) \in \mathcal{I}_n} S_{i_1} \circ \cdots \circ S_{i_n}(C)$. Since $S_1$ and $S_2$ are similarity maps, each $S_{i_1} \circ \cdots \circ S_{i_n}(C)$ is a scaled copy of $C$. Also notice that for any $(i_1,\ldots, i_n) \in \mathcal{I}_n$, $d(S_{i_1} \circ \cdots \circ S_{i_n}(C)) = 3^{-n}$, where $d$ means diameter. That is, $d(A) = \sup\{|x-y| : x,y \in A\}$ for any set $A$.

Let $c \in C$ and $\epsilon > 0$. Choose $N$ large enough so that $3^{-N} < \epsilon$. Since $c \in C$, there is some sequence $(i_1,\ldots,i_N) \in \mathcal{I}_N$ such that $c \in S_{i_1} \circ \cdots \circ S_{i_N}(C)$. Since $d(S_{i_1} \circ \cdots \circ S_{i_N}(C)) = 3^{-N} < \epsilon$, $S_{i_1} \circ \cdots \circ S_{i_N}(C) \subseteq (c-\epsilon,c+\epsilon)\cap C$. That is, $(c-\epsilon,c+\epsilon)\cap C$ contains a copy of the Cantor set, which is uncountable. Therefore $c$ must be a condensation point of $C$.

As Jonas Meyer pointed out, every point in the Cantor set must be a condensation point.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.