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Science Forum Index » Logic Forum » Two definitions for finite set.
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| Guest |
 Posted: Mon Apr 21, 2008 2:04 pm |
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Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
Definition 2)
X is finite iff X has a first member, a last member such that every
set that succeed the first member but precede or is the last member
should have an immediate predecessor in X, and every set that precede
the last member but succeed or is the first member should have an
immediate successor in X.
Zuhair |
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| george |
| Posted: Mon Apr 21, 2008 7:18 pm |
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Guest
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On Apr 21, 8:04 pm, Zaljo...@gmail.com wrote:
Quote: Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
You can't legitimately use successors or predecessors in a
definition of finite AT ALL. There are PLENTY of finite sets WITH
NO NATURAL *order* WHATEVER. The whole POINT about sets
is that they DON'T have order; they're NOT lists. |
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| William Elliot |
| Posted: Mon Apr 21, 2008 11:35 pm |
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Guest
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On Mon, 21 Apr 2008 Zaljohar@gmail.com wrote:
Quote: Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
This definition doesn't fly unless you give X an order.
Quote: Definition 2)
X is finite iff X has a first member, a last member such that every
set that succeed the first member but precede or is the last member
should have an immediate predecessor in X, and every set that precede
the last member but succeed or is the first member should have an
immediate successor in X.
In addition to having the same problem as Definition 1,
this definition is too complex.
S is finite iff for all A subset S, (|A| = |S| --> A = S) |
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| Guest |
| Posted: Tue Apr 22, 2008 2:11 am |
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On Apr 21, 9:35 pm, William Elliot <ma...@hevanet.remove.com> wrote:
Quote: On Mon, 21 Apr 2008 Zaljo...@gmail.com wrote:
Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
This definition doesn't fly unless you give X an order.
Of course.
All finite sets are totally orderable!
Ax ( x is finite -> ER ( R is a total ordering on x )
Even more , all finite sets are well orderable
Ax ( x is finite -> ER ( R is a well ordering on x ).
So there is no problem.
let me write definition 1 in symbols:
X is finite <-> ER AY(Y subset_of X ->[ Emn( ~Ec(ceY & cRm) ~Ec(ceY &
nRc)
Az( (zeY & (mRz or z=n)) -> Eu(ueY ~Ec(ceY & uRcRz))
Az( (zeY & (zRn or z=m)) -> Eu(ueY ~Ec(ceY & zRcRu)) ] )
Zuhair
Quote:
Definition 2)
X is finite iff X has a first member, a last member such that every
set that succeed the first member but precede or is the last member
should have an immediate predecessor in X, and every set that precede
the last member but succeed or is the first member should have an
immediate successor in X.
In addition to having the same problem as Definition 1,
this definition is too complex.
S is finite iff for all A subset S, (|A| = |S| --> A = S)
What has this Dedekindian definition of 'finite set' has to do with
the definitions presented above, it is not even equivalent to the
standard definition of 'finite set'
Zuhair |
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| Guest |
| Posted: Tue Apr 22, 2008 2:18 am |
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On Apr 22, 5:11 am, Zaljo...@gmail.com wrote:
Quote: On Apr 21, 9:35 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Mon, 21 Apr 2008 Zaljo...@gmail.com wrote:
Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
This definition doesn't fly unless you give X an order.
Of course.
All finite sets are totally orderable!
Ax ( x is finite -> ER ( R is a total ordering on x )
Even more , all finite sets are well orderable
Ax ( x is finite -> ER ( R is a well ordering on x ).
So there is no problem.
let me write definition 1 in symbols:
X is finite <-> ER AY(Y subset_of X ->[ Emn( ~Ec(ceY & cRm) ~Ec(ceY &
nRc)
Az( (zeY & (mRz or z=n)) -> Eu(ueY ~Ec(ceY & uRcRz))
Az( (zeY & (zRn or z=m)) -> Eu(ueY ~Ec(ceY & zRcRu)) ] )
Sorry I forgot a braket
X is finite <-> ER AY(Y subset_of X ->[ Emn( ~Ec(ceY & cRm)
~Ec(ceY & nRc)
Az((zeY & (mRz or z=n)) -> Eu(ueY ~Ec(ceY & uRcRz))
Az((zeY & (zRn or z=m)) -> Eu(ueY ~Ec(ceY & zRcRu)) ) ] )
Zuhair
Quote:
Zuhair
Definition 2)
X is finite iff X has a first member, a last member such that every
set that succeed the first member but precede or is the last member
should have an immediate predecessor in X, and every set that precede
the last member but succeed or is the first member should have an
immediate successor in X.
In addition to having the same problem as Definition 1,
this definition is too complex.
S is finite iff for all A subset S, (|A| = |S| --> A = S)
What has this Dedekindian definition of 'finite set' has to do with
the definitions presented above, it is not even equivalent to the
standard definition of 'finite set'
Zuhair |
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| William Elliot |
| Posted: Tue Apr 22, 2008 2:45 am |
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Guest
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On Mon, 21 Apr 2008, george wrote:
Quote: On Apr 21, 8:04 pm, Zaljo...@gmail.com wrote:
Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
You can't legitimately use successors or predecessors in a
definition of finite AT ALL. There are PLENTY of finite sets WITH
NO NATURAL *order* WHATEVER. The whole POINT about sets
is that they DON'T have order; they're NOT lists.
Every set can be given a (partial) order, the trivial order or equality.
The order of an ordered set, can be extended to a linear order.
The question in consideration is: how is an order for X defined? |
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| William Elliot |
| Posted: Tue Apr 22, 2008 11:08 pm |
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Guest
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On Tue, 22 Apr 2008 Zaljohar@gmail.com wrote:
Quote: On Apr 21, 9:35 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Mon, 21 Apr 2008 Zaljo...@gmail.com wrote:
Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
This definition doesn't fly unless you give X an order.
Of course.
All finite sets are totally orderable!
(X,<=) is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
X is finite iff for every liner order of X, every subset Y of X has a
first member, a last member such that every member of Y other than the
first member has an immediate predecessor in Y, and every member of Y
other than the last member has an immediate successor in Y.
X is finite iff every linear order of X is well ordered.
Quote: Ax ( x is finite -> ER ( R is a total ordering on x )
Ok, prove it.
Quote: Even more , all finite sets are well orderable
Ax ( x is finite -> ER ( R is a well ordering on x ).
So there is no problem.
let me write definition 1 in symbols:
Please don't.
Quote: X is finite <-> ER AY(Y subset_of X ->[ Emn( ~Ec(ceY & cRm) ~Ec(ceY &
nRc)
Az( (zeY & (mRz or z=n)) -> Eu(ueY ~Ec(ceY & uRcRz))
Az( (zeY & (zRn or z=m)) -> Eu(ueY ~Ec(ceY & zRcRu)) ] ) |
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| Guest |
| Posted: Wed Apr 23, 2008 12:55 pm |
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On Apr 22, 9:08 pm, William Elliot <ma...@hevanet.remove.com> wrote:
Quote: On Tue, 22 Apr 2008 Zaljo...@gmail.com wrote:
On Apr 21, 9:35 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Mon, 21 Apr 2008 Zaljo...@gmail.com wrote:
Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
This definition doesn't fly unless you give X an order.
Of course.
All finite sets are totally orderable!
(X,<=) is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
X is finite iff for every liner order of X, every subset Y of X has a
first member, a last member such that every member of Y other than the
first member has an immediate predecessor in Y, and every member of Y
other than the last member has an immediate successor in Y.
X is finite iff every linear order of X is well ordered.
Ax ( x is finite -> ER ( R is a total ordering on x )
Ok, prove it.
Even more , all finite sets are well orderable
Ax ( x is finite -> ER ( R is a well ordering on x ).
So there is no problem.
let me write definition 1 in symbols:
Please don't.
Oops I wrote the definition wrongly
The correct symbolization is:
Definition 1)
X is finite iff every subset Y of X has a first member, a last member
such that every member of Y other than the first member has an
immediate predecessor in Y, and every member of Y other than the last
member has an immediate successor in Y.
In symbols this is:
X is finite <-> ER AY ( Y subset_of X -> Emn( ~Ec(ceY & cRm)
~Ec(ceY & nRc)
Az((zeY & ~z=m) -> Eu(ueY & uRz &~Ec(ceY & uRcRz))) &
Az((zeY & ~z=n) -> Eu(ueY & zRu &~Ec(ceY & zRcRu))) ).
Definition 2)
X is finite iff X has a first member, a last member such that every
set that succeed the first member but precede or is the last member
should have an immediate predecessor in X, and every set that precede
the last member but succeed or is the first member should have an
immediate successor in X.
In symbols this is:
X is finite <-> ERmn( ~Ec(ceX & cRm)
~Ec(ceX & nRc)
Az((mRz or z=n) -> Eu(ueX & uRz &~Ec(ceX & uRcRz)))
Az((zRn or z=m) -> Eu(ueX & zRu &~Ec(ceX & zRcRu))) )
Now I claim that these two definitions are equivalent to the standard
definition of
X is finite.
The standard definition of X is finite is the following:
X is finite <-> ER ( R is a well ordering on X & converse(R) is a well
ordering on X)
Zuhair
Quote:
X is finite <-> ER AY(Y subset_of X ->[ Emn( ~Ec(ceY & cRm) ~Ec(ceY &
nRc)
Az( (zeY & (mRz or z=n)) -> Eu(ueY ~Ec(ceY & uRcRz))
Az( (zeY & (zRn or z=m)) -> Eu(ueY ~Ec(ceY & zRcRu)) ] )- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text - |
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