Re: Peano's Axioms are Inconsistent

RussellE wrote:
On Aug 7, 8:57 pm, Joshua Cranmer <Pidgeo...@xxxxxxxxxxxxxxx> wrote:
What the end result means is that, although every cell is written to,
there exists no time at which all cells have been written to--you need
to actually pass infinity to be able to do that.

We agree there is no step when the
set of blank positions is empty.

Then you say BNTM writes to every position.
You can't have it both ways.
You are saying there is a step when the set
of blank positions is empty.

I formally showed that this is not the case. You are imagining an implication which you have not justified.

The shortest explanation is that infinite sets can and will act very differently from finite ones. Infinite sets can even act differently from each other--the set of real numbers has different properties than the set of integers.

If you look at the size of the set of natural numbers, it is very plainly aleph-null. You can only visit one number per step, which means you need at least as many steps as you have elements in the set. But aleph-null is not a natural number--the set of natural numbers does not contain its size. (You can prove this by noting that if a certain function f: A -> B maps A onto B, |A| >= |B|. I'm not sure what this is called, but I think it's an extension of the pigeonhole principle).

There is no natural number that can represent the size of the set of natural numbers. We use the natural numbers to represent each step, so there is no "step" which has an empty set of blank positions.

In short: if you accept that aleph-null is a natural number, then yes, Peano's Axioms (with this addendum) is inconsistent. But no one includes aleph-null in the set of natural numbers, so Peano's Axioms are safe.

You have to prove there exists a step for every
position. And the only way you can do that
is by proving the set of blank positions is empty.

The set of blank positions is empty if and only if an element of X has no successor. Peano's Axioms definitively set that every element has a successor.

By proving BNTM writes to every position,
you are proving there are no more positions
to write to.

Only once BNTM has written to every position, which it will never have time to finish.

One of the elements of X has no successor.

Which one?

--
Beware of bugs in the above code; I have only proved it correct, not tried it. -- Donald E. Knuth
.

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