Re: A new definition for Cardinality


Virgil wrote:
In article <1156349750.340494.28400@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:

Daryl McCullough wrote:
zuhair says...

I love this kind of discussion , but still I don't see how can this be
a proove , since it seem to work bothways to me.

since (x,y) is an element of f is always a false statement when f is
an empty set

then I can conclud the following

for all x, for all y : ( x,y) is an element of f implies that x is not
an element of A and y is not an element of B.

If you are starting with A = {} you are quite correct.
But that does not affect what goes on in the much more usual cases in
which A != {}.

Functions of form f:{} -> B are anomolous.

because you said "whatever V is so it is obvious that I can choose V as
" x is not an element of A and y is not an element of B "

This would also be true.

Not only that I can go even more than that

(x,y) is an element of f when f={ } implies that f is not an
injecitve function

A false statement implies any statement,

And at the same time I can state the following

(x,y) is an element of f when f={ } implies that f is an injective
funciton

According to your analysis then the relation { } which exists between A
and B when at least one of them is an empty set , this relation would
be both injective and not injective at the same time,

That depends.

If A != {} and B = {} then AxB = {} but there do not exist any functions
from A to B, since existence of a function from A to B requires that if
a in A then there exists some f(a) in B.

One must be very literal minded when dealing with definitions. It is a
talent that for most of us must be practiced for quite some time in
order to become good at it.

But it is a talent whose observance often irritates those who do not
have it excessively.


and this is
contradictive , and even if it is right then it proves that Cantor's
system will not work, This logic only proves that my system of ranking
cardinality is better than Cantor's since it is the one which can work
bothways weather { } is injective or not. Suprizingly you only proved
the superimacy of my system.

Zuhair

So you see your logic works both ways

I will examin these replies made in this discussion thread, but before
doing that I have a question, and I think if you and the others manage
to answer it , then matters would be clarified to me.

You said a relation R between A and B denoted as A R B is any subset of
the Cartasian product AxB. And you defined the AxB as the set of all
ordered pairs (a,b) were a in A and b in B .

Ok this is fine, now if A or B or both are empty then it is obvious
that AxB = { }

In symboles A R B and A= { } or B= { } ==> R={ }

I don't think there is dispute about that.

Now , you said that " function" is a subset of AxB in which every a in
(a,b) has one and only one b associated with it. that for any two
ordered pairs within f , (a1,b1) , (a2,b2)
if b1<>b2 then a1<>a2 , provided that f should contain all a in A.

Now how would you define a subset of AxB that is not a function.

I think it would be something like a subset of AxB that do not fullfill
the requirement of a function. I imagine that to be something like the
following

"a relation between A and B that is not a function denoted as f* " is
a subset of AxB in which every a in (a,b) do not necessarily have one
and only one b associated with it. that for any two ordered pairs
within f , (a1,b1) , (a2,b2)
if b1<>b2 then it doesn't imply that a1<>a2 , provided that f* should
contain all a in A.

Now you said that when A or B or both are empty , then the subset of
AxB that contains ordered pairs fullfilling the properties present in
the definition of a function is { }
and I agree with you about this. And accordinglly it seems that you
concluded that { } is a function. And it seems that a same rational is
followed to prove that { } is an injection
since the definition of injection is similar to the definition of
function but added to it the statement , that if a1<>a2 then b1<>b2.

so you conclued that the subset of f that has ordered pairs that
fullfills the properties mentioned in the definition of "injection" is
{ } , and thus { } is an injection.

It appears to me that this is the rational you are using.

Put what I want to tell you is that I can used the same rational you
are using to prove that { } is f* (ie. a subset of R that is not a
funciton ) and accordingly { } is not an injection.

This rational behind yours( as I suppose) is paradoxical since it
proves that { } is injective and not injective at the same time.

Let me try it again so that it appears more clearer to you.

Look you said that the Cartasian product between A and B one A or B is
empty , is an empty set, and that empty set is itself the injective
function from A to B if A is empty.

But what I do not understand is the following, if B is empty and A is
not empty, still the AxB = { } , and according to your rational since {
} is trivially injective, then there would seem to be an injection from
A to B when A is not empty and B is empty, Why not?

From the above I think it is obvious that using the defintion of
injection that you've stated is not enough to prove that f:A->B when A
={ } ==> f is an injection.

There should be another kind of logic to prove that.

I only want to say one thing, I am not saying that I am sure that the
statement

f:A->B when A ={ } ==> f is an injection is a false statement.

All of what I am saying is that the definitions you've illustrated for
injection doesn't lead to the statement above being correct, nor it
leads it to be false, Simply speaking only mentioning the definition of
injection and then saying that it is a trivial fact that f:A->B is
injective when A={}, this way is not enough to prove what you are
saying.

I will examine the other replies in that thread and see there validity.

Zuhair

.

Relevant Pages

  • Re: A new definition for Cardinality
    ... Define f* as the subsets of AxB that do not fullfill ... is in reality not an empty set. ... It can be proved that any singlton subset of AxB is an injection. ... So still Cantor's system fails in establishing the Card A = Card B when ...
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  • Re: Definition of functions.
    ... So what we are getting here is that for every non empty set x ... that formula contradicts set theory. ... ,this definition will lead to any non empty set having injection from ...
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  • Re: Definition of functions.
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  • Re: A new definition for Cardinality
    ... If S is a subset of AxB such that there exist and in S with ... When A of AxB is empty then any subset of AxB is a function, ... followed to prove that is an injection ... It only leads to the conclusion that Zuhair cannot read with the literal ...
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  • Re: A new definition for Cardinality
    ... If S is a subset of AxB such that there exist and in S with ... When A of AxB is empty then any subset of AxB is a function, ... followed to prove that is an injection ... Define f* as the subsets of AxB that do not fullfill ...
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