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Since fat Cantor set is homeomorphic to ternary Cantor set it is also a perfect set, i.e. every point of a fat Cantor set is limit point of itself. But I am not being able to prove it formally. In case of ternary Cantor set I have proved this using the ternary representation of points of the set. But this method cannot be used in case of fat Cantor set as at each step, the length of the open intervals that we are removing is decreasing.

Thanks in advance for any help!

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It is really just a matter of the way the set is constructed. A fat Cantor set $C$ has the form $$C = \bigcap_{n \ge 0} C_n$$ where each $C_n$ has the form $$C_n = \bigcup_{k=1}^{2^n} I_{n,k}$$ and where each $I_{k,n}$ is a closed interval with $\max_k \ell(I_{k,n}) = \ell_n$ for some sequence $\{\ell_n\}$ with $\ell_n \to 0$. Moreover each endpoint of every $I_{n,k}$ is an element of $C$.

Select a point $x \in C$ and let $G$ be a neighborhood of $x$. Since $\ell_n \to 0$ there exist indices $n$ and $k$ with $$x \in I_{n,k} \subset G.$$ Thus $G$ contains at least two points of $C$: the endpoints of $I_{n,k}$. Thus $x$ is not an isolated point of $C$.

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