What does this mean? Your complete imprecision is making it impossible to answer the questions, despite wanting to help, because they don't make sense.
What set? Define it clearly. Don't talk about infinite strings of zeros followed by stuff, because in the context of decimal or binary expansions that doesn't make sense. Strings have a start, then they go on one place at a time.,
> I am flipping all the bits along a diagonal and they
> are all zeros before flipping.
If they are all zero before flipping then 0.11111... isn't there. You've stated that the n^th number has 0 in the n^th place. That means 0.11111... is not the n^th number for any n.
If 0.99999... = 1, then using my argument of flipping the bits around the decimal, wouldn't infinity be in the set?
>> If they are all zero before flipping then 0.11111... isn't there.
This helps. Since the first set has infinitely many zeros on the left, flipping around the decimal means there would have to be infinitely many zeros somewhere on the right. 0.1111.... however has infinitely many ones before these supposed infinitely many zeroes, which cannot be for the same reason that a finite string with one cannot be after infinitely many zeroes at the right of a decimal.
This means that the first set does not have 111111...
So if I keep incrementing binary numbers, I'll never reach 11111... within the limits of enumerability.
I need to think and read more. The above seems to imply that the simple infinity is not in the enumerable set. But I may be confused again.
>> You've stated that the n^th number has 0 in the n^th place. That means 0.11111... is not the n^th number for any n.
I have been confused about this. Somehow 0.11111... is not in the set, in spite of the mathematical induction proof I supply. I need to think more. The proof must be incomplete (or imprecise as you say).
What set? Define it clearly. Don't talk about infinite strings of zeros followed by stuff, because in the context of decimal or binary expansions that doesn't make sense. Strings have a start, then they go on one place at a time.,
If they are all zero before flipping then 0.11111... isn't there. You've stated that the n^th number has 0 in the n^th place. That means 0.11111... is not the n^th number for any n.If 0.99999... = 1, then using my argument of flipping the bits around the decimal, wouldn't infinity be in the set?