I'm trying to show that the interval $(0,1)$ is uncountable and I want to verify that my proof is correct
My solution:
Suppose by way of contradiction that $(0, 1)$ is countable. Then we can create a one-to-one correspondence between $\mathbb{N}$ and $(0,1)$.
$1 \rightarrow 0.a_{0,0}$ $a_{0,1}$ $a_{0,2}$ $a_{0,3} \dots$
$2 \rightarrow 0.a_{1,0}$ $a_{1,1}$ $a_{1,2}$ $a_{1,3} \dots$
$3 \rightarrow 0.a_{2,0}$ $a_{2,1}$ $a_{2,2}$ $a_{2,3} \dots$
$4 \rightarrow 0.a_{3,0}$ $a_{3,1}$ $a_{3,2}$ $a_{3,3} \dots$
$\dots$
Create a number of the form $0.b_0 $ $b_1$ $b_2$ $b_3 \dots$ in the following way. Define $b_i = 1$ if $a_{i, i} \neq 1$ and define $b_i = 2$ if $a_{i, i} = 1$.
Note, since decimal representations are not necessarily unique ( i.e $.599999 = .600000$), choose the representation consisting of infinitely many zeros instead of infinitely many nines.
Creating our number $0.b_0 $ $b_1$ $b_2$ $b_3 \dots$ in this way gives us a number that is not in the above list of decimals in one-to-one correspondence with $\mathbb{N}$, but is in $(0,1)$. This is a contradiction. Thus showing that $(0,1)$ is uncountable.
Is there anything I need to add or is there any way that I can make my proof clearer? Thank you.
Best Answer
To me it is fine.
I know a similar method, which is clearer to me. Suppose $(0,1)$ is countable, and create a one-to-one list as you have done. Now, choose the number $0,b_0b_1b_2...$ so that $b_0\neq a_{0,0}$, $b_1\neq a_{1,1}$, and in general, $b_i\neq a_{i,i}$. That number cannot be in the list.