Re: Implementable Set Theory and Consistency of ZFC

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

Jesse F. Hughes wrote:

I don't follow your so-called answer at all.
Here's my statement of ~Infinity:
Every set has a finite number of members. In other words, every set
can be put in one-to-one correspondence with some set {0,1,...,n}.

I have as much an intuitive understanding of ~Infinity as you
have. But how does that relate to the negation of a complicated
axiom like: forall x there exists a set that contains x and the
successor of x, where this 'successor' is defined as ( x union {x}
)? Starting with {} = 0.

But OK. Every set can be put in one-to-one correspondence with some set
.. huh ? _What_ did you say?

Every set can be put into one-to-one correspondence with a set of the
form {0,1,2,...,n}. In other words, every set is equinumerous with a
natural number.


Proof:
Each set n has at most log_2(n)+1 members (except 0, which has 0
members
of course).

Seems to me that this proof is very similar to the proof of (5) found
in your paper.

Sure. Except that I think this proof is wrong. Each set can be put in
one-to-one correspondence with some natural coding c < 2^n - 1. Where
n is the number of bits employed. Such a set has at most n = log_2(c+1)
members. Where the right hand side is rounded to the integer above it.
To be precise, in Delphi Pascal terms: Trunc(ln(k+1)/ln(2)). The empty
set has n = log_2(0+1) <= 0 members here, while in your "proof" it has
log_2(0)+1 (<= minus infinity) members. Wow!

I explicitly stated that 0 was an exception to my claim.

According to my construction (where I was rounding *down* the log, not
up), we have:

0 has at most 0 members
1 has at most 1 member
2 has at most 2 members
3 has at most 2 members
4 has at most 3 members

and so on. If I haven't made a stupid mistake. But if so, it matters
little.


So what's wrong with concluding that (1)-(4) also
entail ~Infinity?

The above & the axiom of Infinity is more complicated than my
intuitive understanding of it.

Fine. By ~Infinity, I mean "Each set is equinumerous with a set of
the form {0,1,...,n}." Is it your belief that (1)-(4) prove *that*
statement?

--
Sam Vimes could parallel process. Most husbands can. They learn to
follow their own line of thought while /at the same time/ listening to
what their wives say.... At any time they could be challenged and
must be ready to quote the last sentence in full. -- Terry Pratchett
.

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