numerable diagonalization [closed]

Question

If I understand Cantor demonstrated by using a strategy using the diagonal of a matrix, that there are decimal numbers which cannot be a part of an enumeration.

We can build a Table, the rows corresponding to natural numbers from zero to greater numbers top down:

1 2 3 . . .

Let's see numbers with 0 as unity and decimals. We assume that every number has his immediate successor in the row below. For convenience we can write the numbers in binary (with only zeroes an ones).

We observe the number corresponding to the diagonal and replace each zero by one and each one by zero. The result is a number out of the enumeration.

But on the first glance I feel that it's possible to include this diagonal into an enumeration.

We can enumerate as this.

1. First row

2. First diagonal

3. Second row

4. Second diagonal.

We call first diagonal the diagonal built as this

 x

 yx

 yyx

and so on. The second

 x

 yyx

 yyyyx

The third

 x

 y

 yx

 yy

 yyx

And so on.

We have an enumeration of numbers containing all the numbers build by changing diagonal numbers.

So the demonstration of Cantor find a number out of the count but we can build a new count.

Therefore what is the step to find a non enumerable number in this condition?

Answer 1

You can always add one more, but then there's going to be a different one missing - and Cantor's diagonal construction will tell you what it is.

Any proposed enumeration must be completely defined and fixed - you can't keep patching it. What you're doing is kind of like trying to buy two $20 items with just one $20 bill, by constantly moving it to a differently teller when they ask for payment. You say "See, any time either one asks for twenty, they get a twenty. I've paid for both items." And of course that doesn't work - there has to be a single fixed moment when both tellers are satisfied at that same time, and that can't be done.

Answer 2

The answer is obvious.

We can transform these diagonals in new horizontal lines numerated as I described in the question and upon the new table we can build a new diagonal and so the problematic diagonal will recur anyway.

(But we can everytime expand the countable set and include every new number.)