Condensation point of Cantor set 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?
 A: 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 $\mathbb R^n$
A: 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.
