$\forall\ n \in N $ I have $C_n = \bigcup{\left[{k\over3^n}, {k+1\over3^n} \right]}$ (union of intervals) for every $k\in I_n$ where:
$$I_0 = \{0\} \text{ and }I_n = I_{n-1} \cup I_{n-1}', \text{where }I_{n-1}' = \left\{ k+2*3^{n-1} | \forall k \in I_{n-1}\right\}$$
We say that C = $\bigcap C_n, \forall n \in N$.
So C = $C_0 \cap C_1 \cap ... \cap C_n$, right?.
But doesn't this mean that C = $C_n$? I mean the last $n$, to say so... Because when making the intersection, only the smallest intervals remain.
I am asked to: prove that $\frac14 \in C$ and to find out (with demonstration) the cardinal of $C$.
I observed this:
- $I_n$ has $2^n$ elements
- $I_0 = \{0\}, I_1 = \{0,2\}$
- $I_2 = \{0,2,6,8\}, I_3=\{0,2,6,8,18,20,24,26\}, $
- $I_4=\{0,2,6,8,18,20,24,26,54,56,60,62,72,74,78,80\}$ and so on... (I can't find a formula for a general term in $I_n$.)
- $I_n$ could be re-written as $I_0 \cup I_{1}' \cup I_2' \cup ... \cup I_{n-1}'$ but I do not know how much it helps
- $\frac14$ belongs to C for n=1,2,3,4
- I tried to prove that there is a x from $I_n$ so : $\frac{x}{3^n} \le \frac14 \le \frac{x+1}{3^n}$ but could not succeed.
- I also believe that, knowing that any interval $\left[a,b\right]$ has the cardinal equal to the power of continuum ($\aleph$ or $2^{\aleph_0}$), it kind of result that the cardinal of C is the same...
Could anyone help me prove that $\frac14 \in C$ and to find out the cardinal of C and get to an end to this exercise? P.S. : Sorry for my bad english and thanks in advance!
