If the following axiom is added to ZF, would the resulting theory be
equivalent to ZFC?
Axiom:
for all x , for all y ((x is dedekindian and y subset of x)
implies y is dedekindian)
in words: every subset of a dedekindian set is dedekindian.

On May 25, 4:56 pm, Zaljo... at (no spam) gmail.com wrote:

If the following axiom is added to ZF, would the resulting theory be
equivalent to ZFC?
Axiom:
for all x , for all y ((x is dedekindian and y subset of x)
implies y is dedekindian)
in words: every subset of a dedekindian set is dedekindian.

But what does Zuhair mean by "dedekindian"? Does he mean
"Dedekind finite" or "Dedekind infinite"?

Clearly he can't mean the latter, since every D-infinite set
trivially has a D-finite subset, namely 0.

So he can only mean the former. So the axiom asserts that
every subset of a D-finite subset is D-finite. But, as it turns
out, this is already a theorem of ZF!

Theorem:
Every superset of a D-infinite set is D-infinite.

Proof:
Let y be a D-infinite set. Then, by definition of D-infinite,
there exists a bijection f between y and a proper subset of
y, so there exists z such that zey and z is not in the image
of the function f.

Now let x be a superset of y. We now construct the function
F as follows -- for each element w of x, let:

F(w) = f(w), if wey
          w otherwise

So we consider the functions F restricted to y, as well as
F restricted to x\y. The former is simply f, which we already
know is an injection. The latter is simply the identity
function on x\y, so it's an injection whose image is disjoint
from the image of f. Thus F is an injection, and thus a
bijection between its domain and image. But the image of
F can't be all of x, because zex (since zey and ycx) and
z can't be in the image of F -- since it's not in the image of
F restricted to y (or f, by construction) and it's not in the
image of F restricted to x\y (since that's x\y, which is
disjoint from y and zey).

So F is a bijection between x and and proper subset of x,
therefore x is D-infinite. QED

Thus the axiom is already a theorem of ZF and therefore
is clearly not equivalent to AC.

On May 26, 1:14 am, Butch Malahide <fred.gal... at (no spam) gmail.com> wrote:
On May 25, 10:55 pm, Zaljo... at (no spam) gmail.com wrote:

On May 25, 8:00 pm, lwal... at (no spam) lausd.net wrote:

On May 25, 4:56 pm, Zaljo... at (no spam) gmail.com wrote:

If the following axiom is added to ZF, would the resulting theory be
equivalent to ZFC?
Axiom:
for all x , for all y ((x is dedekindian and y subset of x)
implies y is dedekindian)
in words: every subset of a dedekindian set is dedekindian.

But what does Zuhair mean by "dedekindian"? Does he mean
"Dedekind finite" or "Dedekind infinite"?

Oh yes I should have said that.

x is dedekindian iff  (x is dedekindian infinite or x is standard
finite)

In that case, your axiom "every subset of a dedekindian set is
dedekindian" is equivalent to "every set is dedekindian". This is much
weaker than the axiom of choice.

Is it equivalent to axiom of dependant choice?

On May 27, 2:51 pm, Zaljo... at (no spam) gmail.com wrote:





On May 26, 1:14 am, Butch Malahide <fred.gal... at (no spam) gmail.com> wrote:
On May 25, 10:55 pm, Zaljo... at (no spam) gmail.com wrote:

On May 25, 8:00 pm, lwal... at (no spam) lausd.net wrote:

On May 25, 4:56 pm, Zaljo... at (no spam) gmail.com wrote:

If the following axiom is added to ZF, would the resulting theory be
equivalent to ZFC?
Axiom:
for all x , for all y ((x is dedekindian and y subset of x)
implies y is dedekindian)
in words: every subset of a dedekindian set is dedekindian.

But what does Zuhair mean by "dedekindian"? Does he mean
"Dedekind finite" or "Dedekind infinite"?

Oh yes I should have said that.

x is dedekindian iff  (x is dedekindian infinite or x is standard
finite)

In that case, your axiom "every subset of a dedekindian set is
dedekindian" is equivalent to "every set is dedekindian". This is much
weaker than the axiom of choice.

Is it equivalent to axiom of dependant choice?

No.- Hide quoted text -

- Show quoted text -

On May 28, 12:38 am, lwal... at (no spam) lausd.net wrote:

On May 27, 9:01 pm, Zaljo... at (no spam) gmail.com wrote:

On May 27, 8:55 pm, Butch Malahide <fred.gal... at (no spam) gmail.com> wrote:
On May 27, 2:51 pm, Zaljo... at (no spam) gmail.com wrote:
Is it equivalent to axiom of dependant choice?
No.
Is it weaker or stronger?

I'd guess that it would be equivalent to Countable
Choice, but I await Butch's response.

I'd guess it would be weaker than Countable Choice, but I wouldn't
know.