# use diagonalization to prove uncountability

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• Sep 26th 2013, 09:48 AM
emakarov
Re: use diagonalization to prove uncountability
Quote:

Originally Posted by stribor40
When you have numbers...

0, 1, 2, 3, 4, 5, 6, 7, ...
1, 2, 3, 4, 5, 6, 7, 8, ...
2, 3, 4, 5, 6, 7, 8, 9, ...
3, 4, 5, 6, 7, 8, 9, 10 ...

How does chosing something different from diagonal (0,2,4,6..) to something 1,5,7,9...ensuring that is different from everything else in the list (regardless of being repeating or non repeating)

Well, is 1, 5, 7, 9, ... different from the first line? Is it different from the second, third, fourth lines? The answer to why the modified diagonal is different is because we modify it to be different. In the notation of post #13, the \$\displaystyle n\$th element \$\displaystyle d_n\$ of the modified diagonal, by definition, is, e.g., \$\displaystyle x_{n,n}+1\$, and the \$\displaystyle n\$th element of the \$\displaystyle n\$th line is \$\displaystyle x_{n,n}\$. So, the sequence \$\displaystyle d\$ is different from the \$\displaystyle n\$th line at least in the \$\displaystyle n\$th element (maybe in other places as well), and this is true for all \$\displaystyle n\$.
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