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Science Forum Index » Logic Forum » Size theory.
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| Guest |
 Posted: Fri May 02, 2008 3:56 pm |
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Hi all,
The aim of this size theory is to find a consistent notion of size in
which the size of a proper subset is always smaller than the size of
its set. It is only a preliminary trial.
So lets add the PRIMITIVE one place function symbole S which denotes
'size' to the list of primitives of ZF, and lets extend ZF with the
following axioms:
Define: Sx<Sy <-> Ez (z is a proper subset of y & Sx=Sz)
Define: Sx>Sy <-> Sy<Sx
Let f:N-->N
Define: f is cruciate <-> [f is bijective & Axy(<x,y>ef -> <y,x>ef)]
were <x,y> is the ordered pair of x and y.
Define: f is the n-th cruciate <->
[ f is cruciate & EnAxy(<x,y>ef -> |x-y|=n) ]
were |x-y| is defined as the absolute value of x minus y
as in customary mathematics.
Examples:
The identity function is the zero-th cruciate function.
[f:N->N , f is cruciate , Axy( <x,y>ef -> |x-y|=1)] <-> f is the 1st-
cruciate from N to N.
which is {<0,1>,<1,0>,<2,3>,<3,2>,<3,4>,<4,3>,.....}
Axiom1: for any two subsets X and Z of N , were SX=SZ
(f is the n-th cruciate from X to Z & y subset of X) ->
Sy=Srange f as restricted to y.
Now lets take the 1st cructiate function from N to N
and lets restrict it to the domain of all evens E,
now let O be the size of all odds
we get SE=SO.
In a similar manner we can get
S{0,3,6,........} = S{1,4,7,.......}=S{2,5,8,....}
Axiom2: SN = S{0,-1,-2,-3,....}
Axiom3: S{1,2,3,...} = S{1,1/2,1/3,.....}
Zuhair |
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| Guest |
| Posted: Sat May 03, 2008 5:35 pm |
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On May 3, 4:06 pm, Zaljo...@gmail.com wrote:
Quote: On May 2, 6:56 pm, Zaljo...@gmail.com wrote:
The aim of this size theory is to find a consistent notion of size in
which the size of a proper subset is always smaller than the size of
its set. It is only a preliminary trial.
Interesting. BTW, Zuhair is hardly the first to desire such a
theory -- Tony Orlow certainly wanted a theory in which
proper subsets have strictly smaller sizes ("bigulosities")
and others have alluded to such a theory.
Quote: Let f:N-->N
Define: f is cruciate <-> [f is bijective & Axy(<x,y>ef -> <y,x>ef)]
Or more simply put, a "cruciate" is a function f:N-->N
such that AneN (f(f(n)) = n). It is a permutation of the
natural numbers that is equal to its own inverse.
Quote: Define: f is the n-th cruciate <-
[ f is cruciate & EnAxy(<x,y>ef -> |x-y|=n) ]
Examples:
The identity function is the zero-th cruciate function.
[f:N->N , f is cruciate , Axy( <x,y>ef -> |x-y|=1)] <-> f is the 1st-
cruciate from N to N.
which is {<0,1>,<1,0>,<2,3>,<3,2>,<3,4>,<4,3>,.....}
I assume Zuhair here means:
{<0,1>,<1,0>,<2,3>,<3,2>,<4,5>,<5,4>,.....}
as what Zuhair wrote isn't even a function.
Quote: Axiom1: for any two subsets X and Z of N , were SX=SZ
(f is the n-th cruciate from X to Z & y subset of X) -
Sy=Srange f as restricted to y.
Now lets take the 1st cructiate function from N to N
and lets restrict it to the domain of all evens E,
now let O be the size of all odds
we get SE=SO.
This looks interesting, but unfortunately I may have found a
contradiction, if I've interpreted Zuhair's rules correctly.
For let X = Z = N\{0} = {1,2,3,4,5,6,...}
Now the first cruciate function from X to Z is:
{<1,2>,<2,1>,<3,4>,<4,3>,<5,6>,<6,5>.....}
and from Axiom 1, we conclude that:
S{1,3,5,...} = S{2,4,6}
S(O) = S(E\{0})
and since Zuhair already proved S(E) = S(O) above, we
conclude from transitivity that S(E) = S(E\{0}).
In other words E has the same size as its proper subset.
This is not necessarily a contradiction, since none of
Zuhair's Axioms 1-3 state that no set may have the same
size as its proper subset. But he stated right at the start of
this thread that his intent is to have a set theory where
proper subsets have proper subsets, even if he doesn't
state this explicitly as a axiom. (Otherwise, all of Zuhair's
axioms are true in ZF with "size" replaced with "cardinality"
for a standard definition of cardinality.)
Notice that this contradiction disappears if we only allow
the domain of the function f to be N, not any proper subset
of N. Then we can prove S(E) = S(O) but not S(O) = S(E\{0}).
Quote: In a similar manner we can get
S{0,3,6,........} = S{1,4,7,.......}=S{2,5,8,....}
Proof: a={0,1,3,4,6,7,.....}
But unfortunately, we can't use a as the domain of our function
f, since only N can be the domain of f in order to avoid the
problem I state above.
Suppose we only allow N as the domain of f. Then what sets
can we prove have equal size? Since Zuhair showed us the
first cruciate, what about the second cruciate?
Now the second cruciate function from X to Z is:
{<0,2>,<1,3>,<2,0>,<3,1>,<4,6>,<5,7>,<6,4>,<7,5>,.....}
This shows us that
S{0,1,4,5,...} = S{2,3,6,7,...}
S{0,3,4,7,...} = S{1,2,5,6,...}
We may think of E and O as equivalence classes of N
(natural numbers) mod 2. Then the second cruciate proves
that the union of any two equivalence classes mod 4,
one even and one odd, has the same size as the other
two classes mod 4.
Zuhair wishes to prove that any two equivalence classes
mod 3 has the same size. Cruciates of N can't prove this.
(Note: I'm wondering in my head whether I can weaken
the requirement X = Z = N to something like 0eX and/or
0eZ in order to allow Zuhair to prove this for mod 3.)
Quote: Now Sa=Sa
we have f= {<0,1>,<1,0>,<3,4>,<4,3>,<6,7>,<7,6>,.....}
and f is the first cruciate bijection function from a to a.
Now let's restrict f to the subset of a , a'={0,3,6,....}
so the range of this restricted function will be
Range(f as restricted to a')= {1,4,7,....}
so according to axiom 1 it follows that
S{0,3,6,........} = S{1,4,7,.......}
Similarily we can prove that
S{1,4,7,.......}=S{2,5,8,....}
from identity theory we can prove that
S{0,3,6,........} = S{1,4,7,.......}=S{2,5,8,....}
Zuhair
Axiom2: SN = S{0,-1,-2,-3,....}
Axiom3: S{1,2,3,...} = S{1,1/2,1/3,.....}
Zuhair- Hide quoted text -
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