Slight edit:
Let us try this:
New diagonal:
page 3. Let x=|N| [:=aleph-0]
the bit in square brackets is my addition, and it is a defintion. It is unclear what you mean here, as |N| is not a number, so the previous arguments don't apply.
'axiom of infinity induction'
please explain what this is
we know what the axiom of infinity is, we know what induction is, but what is the 'axiom of infinity induction'
why are you justified in writing 'length 2^aleph-0'
by construction the cardinality of the columns is aleph-0
You appear to claim that since the 'finite' diagrams (my term) are of width n and length 2^n that it is permissible to write aleph-0 and 2^aleph-0. This is not true. As you prove later. Look at how the finite diagrams are constructed, now rigorously define the limit of this process, the best you can come up with is that there is a natural inclusion of the diagram for x=1 into x=2 into x=3 and so on.
So the limit will be something akin to the nested subset of this. Simple exercise, show that everything so constructed is countable.
Edit: THIS IS THE MOST IMPORTANT POINT, please deal with this if nothing else in the list of criticisms.
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In your case the construction shows clearly that if z is the element of the power set corresponding to position n in the list, then there are only finitnely many non-zero entries in the row. z was an abritrary element of the power set. This is a contradiction. I'm not sure how more clear we can make that.
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Anyway let us ASSUME that the diagram for the infinite case does list all the elements of the power set, then you prove this assumption is wrong. That is the end of the argument. You've just shown aleph-0 is not equal to 2^aleph-0.
So in no way is it justified to CONCLUDE that there are 2^aleph-0 rows.
PAage 4.
problems with cantor's argument.
1. is false. cantor's argument is defined with any ordered subset of the power set.
2. is not relevant.
3. the enumerated rows in your diagrma clearly only have finitely many elements being non-zero so that is incorrect.
4. irrelevant, as at no point in cantor's argument do we invoke the notion of complete as you define it.
5. ditto. Demonstrate that the notion of complete as you've now defined it is in anyway relevant ot the classical cantor argument, which we will reproduce at the bottom of this post.
you present a proof on page 5 of something. you claim on page 6 t prove the opposite. where is the mathematical error on page 5's proof?
page 7 'out of the scope of cantor's definition' please explain what that means, cos you seem to imply it is to do with your inability to accept that the word all has some mathematical meaning. Why is N not the set of all natural numbers. All means in maths that there is no exception.
for all, there exists, do you know what these quantifiers are?
it is your *opinion* that all is not acceptable for infinite setes, there is no basis in mathematics for that statement as far as i can tell. or indeed any other mathematician that I've met.
**Proof of cantor's argument.
Let N be the set of natural numbers, P(N) its power set, |N|:= aleph-0, we show there is no bijection from N to P(N), ie that 2^aleph-0 is not aleph-0, if you wish.
Suppose f is an *injection* f:N-->P(N)
then list the elements of the image by their preimage
f(1), f(2),f(3)...
define a set by S by n is in S iff n is not in f(n)
S is an element of P(N), by construction S is none of the f(i), so we conclude that no injection from N to P(N) can be a bijection, hence the cardinality of N is strictly less than that of P(N)**
no mention of the word all, no proof by contradiction, sets defined simply, so why is that wrong.
for what it's worth I don't like reading your pdfs because downloading takes up bandwidth, and down loading a file from a source I don't trust (in the computing sense) isn't something I encourage people to do.
In fact, you can ignore all my criticisms of your paper, and just tell me where my maths is wrong in my proof between the **'s