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When we prove that the Cantor set has no isolated point, we can use argument about ternary expansion. I feel we cannot use any similar argument to prove the result of fat Cantor set where the ratio of each interval we remove at each step decreases. Can anyone give me any hint or proof about this fact of fat Cantor set? Thanks in advance for any help!

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The fat Cantor set $C$ that you mentioned is of the form $\bigcap_{n\in\mathbb{Z}^+}C_n$, where:

  • $C_0=[0,1]$;
  • each $C_n$ is an union of finitely many closed intervals ($2^n$ interval, in fact), and you get $C_{n-1}$ from $C_n$ by removing from each of these intervals the middle open interval whose length is $a^n$, where $a\in\left(0,\frac13\right)$ is a fixed number.

Note that the endpoints of each of these intervals belong to $C$. On the other hand, each of the $2^n$ closed intervals whose disjoint union is $C_n$ has length equal to$$2^{-n}\left(1-3a\left(1-\left(\frac23\right)^n\right)\right)\tag1$$and, clearly,$$\lim_{n\to\infty}2^{-n}\left(1-3a\left(1-\left(\frac23\right)^n\right)\right)=0.$$Now, given $x\in C$ and $\varepsilon>0$, take $n\in\mathbb{Z}^+$ such that $(1)<\varepsilon$. Then $x\in C_n$. Let $x'$ be an endpoint of the interval to which $x$ belongs. Then$$|x-x'|\leqslant2^{-n}\left(1-3a\left(1-\left(\frac23\right)^n\right)\right)<\varepsilon.$$This proves that $x$ is not an isolated point of $C$.

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  • $\begingroup$ Thanks! I think the trick here is to use endpoint. In your definition of fat cantor set, you take away middle open interval of length $a^n$ at each step where $a$ is a fixed number. In Folland's book, you take away middle open $\alpha_n$th interval, which means the ratio of the length of the interval you remove over each interval is $\alpha_n$ and $\alpha_n$ decreases rapidly fast (satisfies some conditions). I think your argument would work for that general situation as well. $\endgroup$ – Yuxi Han Aug 10 '18 at 20:27
  • $\begingroup$ @YuxiHan Yes, it does. $\endgroup$ – José Carlos Santos Aug 10 '18 at 20:39
  • $\begingroup$ The fat $C$ is equal to $[0,1]\backslash \cup_{n\in \Bbb N}U_n $ where each $U_n$ is a non-empty open interval of $\Bbb R,$ with $U_n\subset (0,1)$, such that $\overline {U_{n+1}}$ is disjoint from the closure of $\cup_{j\leq n} U_j.$... So $\overline {U_m}\cap \overline {U_n}=\emptyset$ when $m\ne n.$... Now if $p\in C \cap (0,1)$ and $p$ is an isolated point of $C$ then $p=\sup U_m=\inf U_n$ for some $m\ne n,$ implying $p\in \overline {U_m}\cap \overline {U_n},$ which is impossible. It remains to show that $0$ and $1$ are not isolated points of $C$. $\endgroup$ – DanielWainfleet Aug 10 '18 at 21:38

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