康托尔集相关集合相等性的证明咨询

Alright, let's work through these two problems about the Cantor set ( K ) carefully.

First, recall that the Cantor set ( K ) includes all real numbers in ([0,1]) with a base-3 (ternary) expansion that has no digit 1 (we allow the standard infinite trailing 2s for finite expansions, e.g., ( \frac{1}{3} = 0.1_3 = 0.0222\ldots_3 \in K )).

Let's verify the definition of ( S ):

• Right-to-left inclusion: Take any ( \frac{k}{3^n} ) where ( k \leq 3^n ) and ( k ) isn't divisible by 3 (so ( \frac{k}{3} \notin \mathbb N )). If ( k ) ends with 2 in base-3, its finite ternary expansion uses only 0s and 2s, so it's directly in ( K ). If ( k ) ends with 1, we can rewrite the fraction as an infinite expansion with trailing 2s (like ( \frac{1}{3} = 0.0222\ldots_3 )), which has no digit 1—so it's still in ( K ). Thus this fraction belongs to ( S ).
• Left-to-right inclusion: Take any ( x \in S \subseteq K ). Since ( x \in K ), it has a ternary expansion without 1s. If ( x ) is a finite ternary fraction ( \frac{k}{3^n} ), ( k ) can't be divisible by 3 (otherwise it would be excluded from the set). Even for infinite expansions that equal finite fractions (like ( 0.0222\ldots_3 = \frac{1}{3} )), they still correspond to ( \frac{k}{3^n} ) where ( k ) isn't divisible by 3. So all elements of ( S ) fit the given form.

This confirms the characterization of ( S ).

We'll prove this by showing mutual inclusion between the two sets.

Step 1: Show ( K \bigcap \left{ \dfrac{a+b}{2} \mid a,b \in K, a \ne b \right} \subseteq S ) #

Let ( x = \frac{a+b}{2} \in K ), where ( a,b \in K ) and ( a \neq b ). Write ( a = \sum_{n=1}^\infty \frac{x_n}{3^n} ), ( b = \sum_{n=1}^\infty \frac{y_n}{3^n} ) with ( x_n,y_n \in {0,2} ). Let ( n_0 ) be the smallest index where ( x_{n_0} \neq y_{n_0} )—without loss of generality, ( x_{n_0}=0 ), ( y_{n_0}=2 ).

Then:
[
x = \sum_{n=1}^{n_0-1} \frac{x_n}{3^n} + \frac{1}{3^{n_0}} + \sum_{n=n_0+1}^\infty \frac{x_n + y_n}{2 \cdot 3^n}
]
Since ( x \in K ), it must have a ternary expansion without digit 1. The term ( \frac{1}{3^{n_0}} ) can only fit this if we replace it with ( \frac{2}{3^{n_0}} - \sum_{n=n_0+1}^\infty \frac{2}{3^n} ). For this to match the expression above, the trailing sum must equal ( \sum_{n=n_0+1}^\infty \frac{2}{3^n} ), which means ( x_n = y_n = 2 ) for all ( n > n_0 ).

Substituting back, we get:
[
x = \sum_{n=1}^{n_0-1} \frac{x_n}{3^n} + \frac{2}{3^{n_0}}
]
This is a finite fraction ( \frac{k}{3^{n_0}} ) where ( k = \sum_{n=1}^{n_0-1} x_n 3^{n_0 -n} + 2 ). Since ( k ) ends with 2 in base-3, it's not divisible by 3, so ( \frac{k}{3} \notin \mathbb N ). Thus ( x \in S ).

Step 2: Show ( S \subseteq K \bigcap \left{ \dfrac{a+b}{2} \mid a,b \in K, a \ne b \right} ) #

Take any ( x = \frac{k}{3^n} \in S ), so ( x \in K ) and ( k ) isn't divisible by 3. Since ( x \in K ), its ternary expansion is ( 0.d_1d_2\ldots d_n ) with ( d_i \in {0,2} ) and ( d_n=2 ) (because ( k ) isn't divisible by 3).

Define:

• ( a = x = \sum_{i=1}^n \frac{d_i}{3^i} )
• ( b = \sum_{i=1}^{n-1} \frac{d_i}{3^i} + \sum_{i=n+1}^\infty \frac{2}{3^i} )

Clearly ( b \in K ) (its ternary expansion has no digit 1) and ( a \neq b ). Now compute the midpoint:
[
\frac{a+b}{2} = \sum_{i=1}^{n-1} \frac{d_i}{3^i} + \frac{d_n + 2}{2 \cdot 3^n} + \sum_{i=n+1}^\infty \frac{2}{3^i}
]
Since ( d_n=2 ), ( \frac{d_n+2}{2} = 2 ), so:
[
\frac{a+b}{2} = \sum_{i=1}^n \frac{d_i}{3^i} = x
]
Thus ( x ) is the midpoint of two distinct elements of ( K ), and since ( x \in K ), it belongs to the left-hand set.

Conclusion #

Both inclusions hold, so the two sets are equal.

内容的提问来源于stack exchange，提问作者MathMan