what is wrong with this enumeration procedure for the cantor set?

Question

Consider the following procedure for constructing the cantor set $C$. Whenever I say "$x$ is in $C$", read "$[x,x]$ is in $C$".

Step 0. Take $I_0 = $ {$[0,1]$}, and let $n=0$

Step 1. For each interval $[a,b]$ in $I_n$, add $a$ and $b$ to $C$

Step 2. For each interval $[a,b]$ in $I_n$, add to $I_{n+1}$ the following intervals:

• $[a,a+(b-a)/3]$
• $[b-(b-a)/3, b]$

Step 3. Increment $n=n+1$, and Goto step 1

According to my understanding this should enumerate all elements of the cantor set, but then this is impossible because a diagonalization argument shows that the cantor set is uncountable, so what am I doing wrong?

Answer 1

Step 1 you have to put just b, or a. Depending what step you are. Look in that way you conting 0 a lot of times. But in that way, the number of element is not uncontable...

Answer 2

I'm going to answer two of your questions at the same time:

1. Why does your enumeration procedure fall to the diagonalization argument?

2. Why is $\frac14$ in the Cantor set despite it not being the endpoint of any of the intervals used to construct the Cantor set?

$\frac14 = 0.\overline{02}_3$, and this very number can serve as an witness of the diagonalization argument, namely it is in the Cantor set (which can only consist of real numbers in $[0,1]$ with a ternary representation that does not have any '$1$'s, and it avoids all the elements in the sequence you asked for. Why so? Because it has infinitely many '$0$'s and '$2$'s, whereas each of the endpoints of the intervals, when expressed as ternary without any '$1$'s, must either end with "$\overline0$" or with "$\overline2$".

For a much simpler example of a set constructed as an intersection of closed intervals that includes a real number that is not an endpoint of any of those intervals, and moreover excludes the endpoints of all of those intervals:

$\{0\} = \bigcap_{k=1}^\infty [-\frac1k,\frac1k]$.