Hi all,

Define:
x is finite <-> ER Ay (y non empty subset of x -
Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))).

were R is a strict total ordering on x.
i.e R is asymmetric,transitive and total binary relation on x.

In words: x is said to be finite if and only if there exist a strict
total order R on x such that every non empty subset y of x is
 R_bi-ended.

y is R_bi-ended iff Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))

So the above definition will be reduced to

x is finite iff ER Ay (y non empty subset of x -> y is R_bi-ended).

***********
e : epsilon membership
E: existential quatifier
A: universal quantifier

Zuhair

On Thu, 24 Apr 2008 Zaljo...@gmail.com wrote:

In words: x is said to be finite if and only if there exist a strict
total order R on x such that every non empty subset y of x is
 R_bi-ended.

y is R_bi-ended iff Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))

Huh?  Let me guess, rather thatdecipherthewriter'scramp.

A set S is finite when there is a linear order for S
with for all nonnul A subset S, min A and max A in A.

A set S is finite when there's a linear order
for S with for all A subset S, A is compact.

e : epsilon membership
E: existential quatifier
A: universal quantifier

's e t' is 's in t' while 'set' is set, like set up.

Riddle of the day.  Do you have to be a member to get into a set?

----

In words: x is said to be finite if and only if there exist a strict
total order R on x such that every non empty subset y of x is
R_bi-ended.

y is R_bi-ended iff Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))

Huh? Let me guess, rather thatdecipherthewriter'scramp.

e : epsilon membership
E: existential quatifier
A: universal quantifier

's e t' is 's in t' while 'set' is set, like set up.

On Thu, 24 Apr 2008 Zaljohar@gmail.com wrote:

In words: x is said to be finite if and only if there exist a strict
total order R on x such that every non empty subset y of x is
R_bi-ended.

y is R_bi-ended iff Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))

Huh? Let me guess, rather thatdecipherthewriter'scramp.

A set S is finite when there is a linear order for S
with for all nonnul A subset S, min A and max A in A.

A set S is finite when there's a linear order
for S with for all A subset S, A is compact.

A set is finite iff it can be given a Hausdorff

e : epsilon membership
E: existential quatifier
A: universal quantifier

's e t' is 's in t' while 'set' is set, like set up.

Riddle of the day. Do you have to be a member to get into a set?

----

R is total on x <-> Amn((mex nex) -> mRn or nRm)

y is R_bi-ended <-> Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))

Define: x is finite <-
ER Ay (y non empty subset of x -> (y is R_bi-ended & R is total on
y)).

would that definition be equivalent to the standard definition of x is
finite , I mean Tarski's definition of equinumerousity to a natural
number?

On Apr 25, 1:12 am, William Elliot <ma...@hevanet.remove.com> wrote:
On Thu, 24 Apr 2008 Zaljo...@gmail.com wrote:

Define: y is R_bi-ended iff exists m exists n ( m in y and
n in y and notexist c ( c in y and c R m ) and
notexist c ( c in y and n R c ) )

Define: R is total on y iff
forall m forall n ((m in y and n in y) implies (m R n or n R
m))

Define: x is finite iff
exists R forall y (y non empty subset of x implies
(y is R_bi-ended and R is total on y)).

Is that clear to you now?

Riddle of the day. @Do you have to be a member to get into a set?

On Apr 24, 6:56 pm, Zaljo...@gmail.com wrote:

It seems to me your definition entails that {0} is not finite.

It does? How do you read his cramedtogetherstuff?

Do you play around with your newfangled proposals for at even just a
few seconds before you post them?

On Apr 24, 6:56 pm, Zaljo...@gmail.com wrote:

R is total on x <-> Amn((mex nex) -> mRn or nRm)

y is R_bi-ended <-> Emn (mey ney ~Ec(cey cRm) ~Ec(cey nRc))

Define: x is finite <-
ER Ay (y non empty subset of x -> (y is R_bi-ended & R is total on
y)).

would that definition be equivalent to the standard definition of x is
finite , I mean Tarski's definition of equinumerousity to a natural
number?

It seems to me your definition entails that {0} is not finite.

Do you play around with your newfangled proposals for at even just a
few seconds before you post them?

yes few seconds yes.

On Fri, 25 Apr 2008, MoeBlee wrote:
On Apr 24, 6:56 pm, Zaljo...@gmail.com wrote:

It seems to me your definition entails that {0} is not finite.

See my answer to OP where I give a definition using orders
for finite that makes {a} finite.

Do you play around with your newfangled proposals for at even just a
few seconds before you post them?

Likely not as he has often multiple posts correcting his post even
before he gets a reply. Now on the other hand, my new and improved
finite which knows singletons are finite, swears up and down that empty
set is infinite.

That is useful for proving inf empty = oo and sup empty = -oo. ;-)

Whoops, another way will have to be found for proving that as

On Fri, 25 Apr 2008, MoeBlee wrote:

It seems to me your definition entails that {0} is not finite.

It does?  

See my answer to OP where I give a definition using orders
for finite that makes {a} finite.

Do you play around with your newfangled proposals for at even just a
few seconds before you post them?

Likely not as he has often multiple posts correcting his post even before
he gets a reply.  Now on the other hand, my new and improved finite which
knows singletons are finite, swears up and down that empty set is
infinite.

That is useful for proving inf empty = oo and sup empty = -oo. ;-)

----

On Apr 25, 10:39 pm, William Elliot <ma...@hevanet.remove.com> wrote:

On Fri, 25 Apr 2008, MoeBlee wrote:
It seems to me your definition entails that {0} is not finite.

It does?  

Not just {0}, but any singleton.

Here's how (using Zuhair's definitions):

Definition:
R is total on x <-> AmexAnex(<m n>eR  v <n m>eR)

Definition:
y is R_bi-ended <-> EmeyEney(~Ecey <c m>eR & ~Ecey <n c>eR)

Definition:
x is finite <-> ERAy(y is a non-empty subset of x -> (y is R_bi-ended
& R is total on y))

Suppose {z} is finite.
So, let Ay(y is a non-empty subset of {z}-> (y is R_bi-ended & R is
total on y)).
So {z} is R_bi-ended and R is total on {z}.
So let me{z} and ne{z} such that ~Ece{z} <c m>eR & ~Ece{z} <n c>eR.
So m=z=n. So ~<z z>eR.
But, since R is total on {z}, we have <z z>eR, which is a
contradiction.

So {z} is not finite.

MoeBlee

.





See my answer to OP where I give a definition using orders
for finite that makes {a} finite.

Do you play around with your newfangled proposals for at even just a
few seconds before you post them?

Likely not as he has often multiple posts correcting his post even before
he gets a reply.  Now on the other hand, my new and improved finite which
knows singletons are finite, swears up and down that empty set is
infinite.

That is useful for proving inf empty = oo and sup empty = -oo. ;-)

----- Hide quoted text -

- Show quoted text -

On Apr 25, 10:39 pm, William Elliot <ma...@hevanet.remove.com> wrote:

On Fri, 25 Apr 2008, MoeBlee wrote:
It seems to me your definition entails that {0} is not finite.

It does?  

Not just {0}, but any singleton.

Here's how (using Zuhair's definitions):

Definition:
R is total on x <-> AmexAnex(<m n>eR  v <n m>eR)

Definition:
y is R_bi-ended <-> EmeyEney(~Ecey <c m>eR & ~Ecey <n c>eR)

Definition:
x is finite <-> ERAy(y is a non-empty subset of x -> (y is R_bi-ended
& R is total on y))

Suppose {z} is finite.
So, let Ay(y is a non-empty subset of {z}-> (y is R_bi-ended & R is
total on y)).
So {z} is R_bi-ended and R is total on {z}.
So let me{z} and ne{z} such that ~Ece{z} <c m>eR & ~Ece{z} <n c>eR.
So m=z=n. So ~<z z>eR.
But, since R is total on {z}, we have <z z>eR, which is a
contradiction.
So {z} is not finite.

MoeBlee

.





See my answer to OP where I give a definition using orders
for finite that makes {a} finite.

Do you play around with your newfangled proposals for at even just a
few seconds before you post them?

Likely not as he has often multiple posts correcting his post even before
he gets a reply.  Now on the other hand, my new and improved finite which
knows singletons are finite, swears up and down that empty set is
infinite.

That is useful for proving inf empty = oo and sup empty = -oo. ;-)

----- Hide quoted text -

- Show quoted text -

I already corrected that. So what you say is not applicable to my
correction.
According to my modification on the definition of "total', it doesn't
follow that
if R is total on {z} then we have <z z>eR.

On Apr 29, 3:30 pm, Zaljo...@gmail.com wrote:

I already corrected that. So what you say is not applicable to my
correction.
According to my modification on the definition of "total', it doesn't
follow that
if R is total on {z} then we have <z z>eR.

Fine. And my post in response to the other poster was about my remark
that was based on your posts AT THAT TIME. Any subsequent changes you
made to your formulations have no bearing on what I proved based on
your formulations AT THAT TIME.

MoeBlee