i think i understand the proof using diagonalization.
butttttt!
consider the following construction step:
for c(olumn) = 1 to countable infinitely
for r(ow) = 1 to 2^c
[r][c]=(r <= 2^c/2 ? 0 : 1)
before each time the value of c get updated:
copy the whole block from [1 - 2^c][1 - c] to [2^c+1 - 2*2^c][1 - c]
c=1:
\ 1 2 3 4 5 6 7 . .
1 0 (this means 0.0
2 1 (this means 0.1
3
4
5
6
7
.
.
after block copied
\ 1 2 3 4 5 6 7 . .
1 0
2 1
3 0
4 1
5
6
7
.
.
c=2:
\ 1 2 3 4 5 6 7 . .
1 0 0 (means 0.00
2 1 0 (means 0.10
3 0 1 (means 0.01
4 1 1 (means 0.11
5
6
7
.
.
etc
i can argue that for arbitrary r, all real numbers in [0..1] with length(:=number of digits after binary point) smaller than log(2,r) are enumerated.
since c is counting to countable infinity, all real numbers will eventually be enumerated.
whats wrong with this?
You are enumerating finite strings of 0 and 1 of arbitrary length. You will miss every real number whose binary expansion doesn't terminate, in particular, all irrational numbers in $[0,1]$.