# there are uncountable infinity real numbers in interval [0,1], where is my pitfalls? [duplicate]

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?

## marked as duplicate by José Carlos Santos, John Douma, Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 18 at 6:04

• You have made the argument that the cardinatlity of the power set (set of all subsets) of natural numbers $|\mathcal P(\mathbb N)|$ is uncountable. Which it is. math.stackexchange.com/questions/77656/… – Doug M Jun 17 at 17:28
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 $$[0,1]$$.