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The cantor set $C$ is defined as

$$ A_0 := [0, 1], \quad A_{n+1} := \frac{1}{3}(A_n \, \cup \, (2+A_n)), \quad C:= \bigcap_{n\in\mathbb{N}_0}A_n \text{.} $$

I need to show that $C$ is perfect (every point in $C$ is a limit point). I strongly believe that my proof idea is right but I am stuck at the point where I need to show that each $A_n$ consists of $2^n$ closed intervals and that the boundary points of these intervals are in $C$.

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    Can you show that each $A_n$ consists of $2^n$ closed intervals (each of length $3^{-n}$) and are stuck at showing the boundary points are in $C$, or are you stuck already on the first part? – 2017-01-23
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    First part seems possible. – 2017-01-23
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    Then try to show that the $2^{n+1}$ boundary points of the intervals in $A_n$ are also boundary points of the intervals in $A_{n+1}$. – 2017-01-23
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    If you know that the numbers in $C$ are precisely those in $[0,1]$ with ternary expansions having no $1$'s, then you get a nice simple proof. – 2017-01-23
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    @zhw. I don't know that. Is there an easy way to show that? – 2017-01-23

2 Answers 2

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Let $x \in C$, and let $S$ be an open interval containing $x$. Let $I_n$ be the closed interval in $A_n$ (of which there are $2^n$) that contains $x$. Choose $n$ large enough so that $I_n \subset S$. Let $x_n$ be an endpoint of $I_n$ such that $x_n \neq x$.

Now, by construction, $x_n \in C$ (you should verify this). Hence, $x$ is a limit point: an arbitrary open interval $S$ containing $x$ contains $x_n \in C$.

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The ternary expansion of $x \in [0,1)$ is found by this construction: divide the interval into three equal parts, that is, $$ [0,1) = [0,1/3) \cup [1/3 \cup 2/3) \cup [2/3 \cup 1). $$ To find the ternary expansion of $x$, see in which of the three subintervals $x$ lies in. If $x \in [0,1/3)$, then the first digit in the expansion is $0$; if $x \in [1/3,2/3)$, then the first digit in the expansion is $1$; if $x \in [2/3,1)$, then the first digit in the expansion is $2$. To find the next digit, further divide the subinterval in which $x$ lies into three equal parts and repeat the same process.

By the way the Cantor set is contructed, the digit $1$ can never appear in a ternary expansion of $x \in C$ because the middle one-third is removed successively from each sub-interval. Of course, this is unless $x = 2/3$, in which case, its ternary expansion is $0.1$. However, $2/3$ can also be represented as $0.022222\dots {}= 0.0\bar{2}$ and we'll make an exception and choose this as the ternary expansion of $2/3$. Similarly, for consistency we can choose the ternary expansion of $1$ to be $0.22222\dots {}=0.\bar{2}$ instead of $1.00000\dots$. In general, if $x$ has a ternary expansions ending in an infinite string of twos, then we will choose that over a representation ending in an infinite string of zeros.

Conversely, any real number in $[0,1]$ whose ternary expansion consists only of $0$'s and $2$'s is in the Cantor set by the same reasoning: the numbers that are removed in the construction of the Cantor set are precisely those that have a $1$ somewhere in their ternary expansion.

Now, to show that the Cantor set is perfect, it suffices to show that every point in $C$ is a limit point of $C$. So, let $x \in C$. Then, $x$ has a ternary expansion of the form $0.x_1 x_2 x_3 \dots$ where each $x_n$ is either $0$ or $2$. The sequence of numbers $y_k = 0.x_1x_2\dots x_k$, formed by truncating the ternary expansion at the $k$th position, clearly converges to $x$. Moreover, $y_k \in C$ for each $k$ because its ternary expansion consists only of $0$'s and $2$'s. Hence, $x$ is a limit point of $C$, and so $C$ is perfect.