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$\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!

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  • $\begingroup$ Is $C$ supposed to be the Cantor set? Because the definition is the intersection of the union of closed intervals. In either case, the definition is not clear because you have two variables $n$ and $k$ there. $\endgroup$ – Asaf Karagila Oct 25 '14 at 16:28
  • $\begingroup$ For every n from N. I edited right now :D $\endgroup$ – nightwing96 Oct 25 '14 at 16:33
  • $\begingroup$ Sorry, still not clear enough. I'm honestly thinking there should be an intersection there in addition to the union. $\endgroup$ – Asaf Karagila Oct 25 '14 at 16:36
  • $\begingroup$ Re-re-edited :D That's the full-original exercise, but I had modified it because I thaught C = Cn. $\endgroup$ – nightwing96 Oct 25 '14 at 16:45
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    $\begingroup$ @MarianFX see here $\endgroup$ – John Oct 25 '14 at 16:57

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