OK, so you're not talking about the usual diagonalisation. I'll try to follow what you've said and respond as I go.
> In this first set, I do not have a decimal point anywhere.
OK, fine. But then you talked about flipping them around to come after the decimal point. When you do that you have only those strings that only have a finite number of 1s in them.
> As far as representing it as a string, I may still start
> from the right and work towards the left.
Yes you can, but that's not what people do when talking about Cantor and diagonalisation, so it's now completely unclear what you're talking about.
However, you start with finite strings of 0s and 1s, basically the non-negative whole numbers, represented as binary strings. Note that these are all finite, and the non-zero parts are still finite, even if you prepend an infinite number of 0s.
> Stating the above for the decimal case, let n=1 be the
> place right after the decimal, n=2 to the right of it, and
> so on.
See, now you're talking about stuff after the decimal point. I'll continue ...
> Now for place n=1, the both zero and one are covered
> (first two cases below).
But for diagonalisation that's irrelevant. We only ask what is the first digit of the first number.
> For n=2 place, again both zero and one are covered for
> all possible combinations above for n=1 place
Again, irrelevant. We only ask what is the 2nd digit of the 2nd number.
So here if we construct the diagonal of this sequence as you've listed it we have 0.00000... Let me highlight the diagonal for you from this quoted section:
However, you start with finite strings of 0s and 1s, basically the non-negative whole numbers, represented as binary strings. Note that these are all finite, and the non-zero parts are still finite, even if you prepend an infinite number of 0s.
See, now you're talking about stuff after the decimal point. I'll continue ... But for diagonalisation that's irrelevant. We only ask what is the first digit of the first number. Again, irrelevant. We only ask what is the 2nd digit of the 2nd number. So here if we construct the diagonal of this sequence as you've listed it we have 0.00000... Let me highlight the diagonal for you from this quoted section: If we now flip this we get 0.11111....Now observe that all of the strings you have contain only a finite number of 1s. That means that 0.11111... is not in your sequence.
You haven't made an induction argument. For each place, both possibilities are eventually covered. but for each sequence that you give, it is eventually all zeros. No, it doesn't. Infinity does not fit in that set, and 0.1111... is not in your defined set of numbers.