Re: Infinity......

William Hughes wrote:
zuhair wrote:
William Hughes wrote:
zuhair wrote:

<but first he cut out some important stuff. Here replaced>


[William Hughes]

If there is a bijection from A to B then there is a way to
change the name of each of A's elements.
If we adopt the principal that changing the names of the
elements does not change the size of the set
then A and B have the same size.

If you wish to abandon the idea that the existence
of a bijection between two sets means that they are
the same size, they you must also abandon the idea
that changing the name of every element does not
change the size of a set.


Let me tell one thing, I said and I will always say that Cantor's
cardinality is BIASED. and it is biased to bijection.


Being biased to bijection and being biased to
"changing the name of every element in a set
does not change its size" is exactly the same thing.

You are in heavy denial. You do no want to accept
the consequences of having "there is a bijection
from A to B mean that A and B are the same size", and
at the same time you want to say "changing the name of
every element of A does not change its size".

You can't have it both ways.

- William Hughes

In order to be able to change the name of every element of A, by names
of elements from B, in order to do that, you should have size A <= size
B.ohterwize if size A > size B, then
there will be always at least an element in A who's name is not
changed. I think you would agree with me about this point.

There is no difference between saying "there is a bijection from
A to B" and "we can use the elements of B to rename the
elements of A".

Let f be a bijection from A to B. Then for every element of
a we have an element of B f(a). So change the name of a to f(a).
We have changed every the name of every element of a using
elements of B. So size A <= size B.

Of course if there is a bijection from A to B there is a bijection
from B to A. so size B <= Size A.

So if there is a bijection from A to B we have size A = size B.




OK, let us take a clear example to see what are we talking about, to
avoid any misunderstanding.

for example lets take N={0,1,2,3,4,...........} and
M={-1,-2,-3,.......}
Now we want to change every member of N by a member from M to have the
set deltaN
and suppose also we want to change every member of M by a member from N
to have the set delta N.
You tend think that if there exist a bijective mapping between N and M,
then the replacement above can take palce, and the result would be
deltaN=M and deltaM=N.

That's what is in your mind.

I don't agree with that. The mere existance of bijection between N and
M doesn't entail that such a replacement as outlined above can happen.


Such a replacement is simple.
Fore every element x in N change the name to -(x+1).
deltaN is now M. Given an element y in M, change the name
to -(y+1). deltaM is now N.



- William Hughes

All of what you are talking above is irrelevant to what I am presenting
in this thread.

Look William, let me just try understand that replacement prinicipal of
yours.

let A={ 1,2,3,4}
let B= { 5,6}

now f:A->B, f(x)=[floor((x+1)/2)]+4
Now I can replace each x in A by f(x) in B.
I will get deltaA = { 5,5,6,6} = { 5,6} Extensionality.

does that mean that f is bijective?????, it is clear that f is
serjective and not bijective.
Now you might say if I replace 5 istead of 1 then there is no 5 left in
B to replace 2 by it, and the same case applies to 3 in A, if it is
replaced by 6,then there is no 6 left in B to replace 4 in A with it.
and accordingly deltaA={5,2,6,4}. Let me call this last approach
as Replacement without repetition, not like the first replacement.

so you want to say if there exist replacement without repetition of all
members of one set by members of the other set, and vice versa then
these two sets has bijection.

Let me go back to the infinite sets

A={0,1,2,3,4...........}
B={-1,-2,-3,............}

Now you are saying that there exist replacement without repetition from
A to B and from B to A. and this mean that bijection exists between
them, and this bijection is
f:A->B, f(x)=-(x+1). and since they are replaceable then they should
have equal size.

Ok.

But see the same sets above also can be viusalized to be related to
each other by the following function.

f:A->B, f(x)= - floor((x/2)+1)

you see this is a 2-1 serjective function from A to B. Now can you have
replacement without repetition of elements of A by elements of B using
this function. The answer is no.
and accordinglly size A > size B.

Also there exist an n-1 serjective function from B to A were n>1. and
according to it you cannot have replacement without repetition of all
members of B by elements of A. and
accordingly size A < size B.

Now the question the impose itself here, which of these three injective
functions between A and B we should depend upon to estimate the
comparson between size A and size B.

what is the FAIRE criterion according to which we can know the function
which determines the size comparison of A and B.

Cantor and you are simply making an obvious bias by simply saying WE
choose the bijective function.

what you and Cantor are saying is the following when we are confronted
with a situation were there are evidence that sizeA>sizeB and there are
evidence that sizeA=sizeB and there are evidence that sizeA<sizeB, then
we choose size A = size B. That's what Cantor's cardinality is mounting
too. And I call it Biased selection of bijection in preference of it
over all other injective functions, without leaving sets A and B
determines the selection of the injective function.

I replaced that by a more fair notion. that is of "the simplest
function between A and B"
this simplest function might be the bijective, or the non bijective
injective function between A and B. it depends on A and B, they will
decide which is the simplest function between them
and after this we go and say if that simplest function is bijective
then size A = size B. if that simples injective function is not
bijectable from A to B , then size A < size B. if that simples
injective function is not bijectable from B to A , then size A > size
B.

This later approach of mine is the fair approach. Your and Cantor's are
only biased approaches.

Till now you didn't discuss with me this basic point. all of what are
you pointing is irrelevant material to what I am presenting. You are
not in the subject, you are playing around the field, rather than in
the field.

I didn't see any refutation from you on what I am saying. Analyse what
I said above.
if you can see my point, then this is good, but it looks as if you are
not really seeing my point I am freguentlly addressing. The matter has
nothing to do with replacements it has all to do with BIASED and fair
determination of size comparison. you simply missed my point.

Zuhair

.

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