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This question already has an answer here:

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?

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marked as duplicate by José Carlos Santos, John Douma, Asaf Karagila elementary-set-theory Jun 18 at 6:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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/… $\endgroup$ – Doug M Jun 17 at 17:28
  • $\begingroup$ In general, when you seem to have disproved a major result be very suspicious about any claim along the lines of "i can argue" (or "it is clear," or etc.). $\endgroup$ – Noah Schweber Jun 17 at 18:15
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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 $[0,1]$.

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  • $\begingroup$ oh thx man, i failed to see this😅 $\endgroup$ – sasasqt Jun 17 at 18:36

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