Re: An uncountable countable set

Tony Orlow wrote:
....11111 binary (all bit positions finite)


Mike Kelly wrote:
That isn't a natural number, Tony.


Tony Orlow wrote:
Are you sure? Pay close attention.

For any finite bit position n, it and all predecessors can only sum to a
finite bit string value of 2^(n+1)-1. Since there are only finite bit
positions in the string, it can never achieve any infinite value at any
position in the unending string of bits. Therefore the value must be finite.

Breathtaking. You have the annoying habit of ignoring the
completely obvious when it suits you. If ...111 is a binary number
with an infinite number of digits, why is it a finite value?
Shouldn't an infinite value have an infinite number of digits?


Furthermore, since any such number does have a predecessor and
successor, in this case ....1110 and ...0000, respectively, it fits in
the successorship model. The only concept this breaks is that 0 is now a
successor as well, creating an infinite ring of successorship. Other
than that, it works as a natural, ...

If ...111 is a finite value, then there must be another finite value
that is larger than it. You claim that this is 0, but that gives you
a contradiction: 0 = ...111+1 > ...111 > 0, or 0 > 0.
(We can now use this to prove that any finite number is larger than
itself: k = 0+k = (...111+1)+k > ...111+k > 0+k = k, so k > k.)

This should be a rather big clue that something's wrong with your
concept, and that it does not, in fact, "work as a natural".


and in fact, this is the way signed
integers work in your very computer.

Computers only deal with fixed-length finite numbers,
modulo 2^B for some word width B. That does not provide a
very useful model for the entire set of naturals.


So, while ...111 may not be considered a standard natural, I see no
reason why it should not be considered, say, an extended natural.

Which presumably requires some extended Peano axioms in order
to exist?

.

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