Re: Cardinality and injection


William Hughes wrote:
> zuhair wrote:
> > William Hughes wrote:
> > > zuhair wrote:
> > > > William Hughes wrote:
> > > > > zuhair wrote:
> > > > > > William Hughes wrote:
> > > > > > > zuhair wrote:
> > > > > > > > William Hughes wrote:
> > > > > > >
> > > > > > >
> > > > > > > <snip>
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > I write
> > > > > > >
> > > > > > >
> > > > > > > > > if A is a subset of B and A and B have the same cardinality,
> > > > > > > > > then subcard A <> subcard B, unless A=B
> > > > > > >
> > > > > > >
> > > > > > > You reply
> > > > > > >
> > > > > > > > No A do not necessarily equal B.
> > > > > > > >
> > > > > > > > see set A = 1,2,3,4,5,...............
> > > > > > > > set B = 1,1/2,1/3,1/4,1/5,................
> > > > > > > >
> > > > > > > > These are also set which has subcard A = subcard B
> > > > > > > >
> > > > > > > > Also set A = 1,2,3,4,5,............
> > > > > > > > and set B = -1,-2,-3,-4,-5,..............
> > > > > > > >
> > > > > > > > subcard A = subcard B.
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > Do you see the problem. (Hint, is A a subset of B
> > > > > > > in either example?)
> > > > > > >
> > > > > > >
> > > > > > > Now go back and answer the questions in the post.
> > > > > > >
> > > > > > > -William Hughes
> > > > > >
> > > > > > I am not understanding anything of what you are saying or hinting at.
> > > > >
> > > > > I stated a theorem, for which one of the conditions was that
> > > > > A was a subset of B. You replied with two counterexamples,
> > > > > but in neither counterexample was A a subset of B.
> > > > >
> > > > > >
> > > > > > All of what I wanted to say is that if their are two sets A and B
> > > > > >
> > > > > > having an intersectional set A.B that is infinite and at the same time
> > > > > >
> > > > > > a proper subset of at least one of them , then subcard A <> subcard B.
> > > > > >
> > > > > > And if sets A is a subset of set B and subcard A <> subcard B.
> > > > > >
> > > > > > then A is a proper subset of B.
> > > > > >
> > > > > > Subcardinality can define Properness of subsethood and not
> > > > > >
> > > > > > subsethood itself.
> > > > > >
> > > > > > Given two sets were their subcardinalities are "<>" doesn't mean
> > > > > >
> > > > > > that one should be a subset of the other.
> > > > > >
> > > > > > But if we know that one is a subset of the other then we know
> > > > > >
> > > > > > that one of them should be a proper subset of the other
> > > > > >
> > > > > > That's all
> > > > > >
> > > > >
> > > > > And as I point out this is useless. Knowing that A and B
> > > > > do not have the same subcardinalities does not tell you anything
> > > > > unless you know A is a subset of B. And if A is a subset of B
> > > > > and A and B have the same cardinality, they have the same
> > > > > subcardinality if and only if A=B. But the only way to show that
> > > > > they have the same subcardinality is to show that A=B. So the final
> > > > > result is that if you can show that A=B you know that A=B.
> > > > >
> > > > > You have
> > > > >
> > > > > Let A be a subset of B. Then card(A) <= card(B)
> > > > >
> > > > > (i) if card(A) < card(B) then subcard(A) < subcard(B)
> > > > > and A is a proper subset of B
> > > > >
> > > > > (ii) if card(A) = card(B) A is a proper subset of B
> > > > > unless subcard(A) <> subcard(B), but the only way
> > > > > to show this is to show that A=B is false.
> > > >
> > > >
> > > > if card(A) = card (B) A is a proper subset of B unless
> > > > subcard (A) = subcard (B) .
> > >
> > > In which case A=B. (Note A is a subset of B)
> > >
> > > >
> > > >
> > > > >
> > > > > How does this differ from:
> > > > >
> > > > > Let A be a subset of B. Then card(A) <= card(B)
> > > > >
> > > > > (i) if card(A) < card(B) A is a proper subset of B
> > > > >
> > > > > (ii) if card(A) = card(B) A is a proper subset of B
> > > > > unless A=B
> > > >
> > > >
> > > > No , A= B means that A contains exactly the same members
> > > > as B. My definition mentions nothing of that sort, it says
> > > >
> > > > (ii) if card (A) = card (B) A is a proper subset of B unless
> > > > subcard A = subcard B.
> > > >
> > > > subcard A = subcard B doesn't imply that A=B.
> > >
> > > If A is a subset of B it does. (To say that subcard(A) = subcard(B) is
> > > to say there exists a bijection f, from A to B, with f its own inverse.
> > > so f(A) = B and f(B)=A. If A is a proper subset of B then since f(B)=A
> > > f(A) must be a proper subset of A, and hence a proper subset of B.
> > > But this says that f(A)=B is a proper subset of B. So A cannot be a
> > > proper subset of B. If A is a subset of B, A=B)
> > >
> > >
> > > -William Hughes
> >
> > subcard A = subcard B and A is a subset of B => A = B This is a
> > correct statement.
> >
> > because
> >
> > 1)if A and B are finite sets then A cannot be a subset of B.
> >
> > 2)if A and B are infinite , then we have two possiblities were subcard
> > A = subcard B
> >
> > 2:A) A can be bijected to B with a function that is the same as it's
> > converse(inverse)
> > were A cannot be a subset of B which implies that A <> B.
> >
> >
> >
> > 2:B) A can be bijected to B with a function that is the same as it's
> > converse(inverse)
> > were A can be a subset of B which implies that A = B
> >
> > So since the first two possiblities are overrolled then we are only
> > left with the third
> > one .
> >
> >
> > Ok , now I got your point. I think what you want to say is that the
> > defintion of subcardinality
> > is useless, since it only show's us what we know before hand. I don't
> > think you found
> > it a contradictive definition. You think it is only non benificial or
> > non practical definition.
> >
>
> Exactly. Note that I said subcardinality was useless, not
> that it was contradictory (If you want to define things
> precisely, you need to firm up the definition of converse
> and the meaning of f=f'', but it's probably not worth the effort).

Yes I agree with you on that, In reality I have been thinking about
that term
"essentially identical" although it is informal yet I had the deep
sense that
sets O={ 1,3,5,7,...................} and E= {2,4,6,8,...............}
are essentially
identical in a way???

When I hear essentially identical it looks like "Can be identicalized"

So let me define an identicalizing function G.

Now for any two sets A and B , define G(x)=f(x) if x belongs to A
and G(x) = f"(x) if x belongs to B. were G is a ONE VALUED FUNCTION.

suppose that f is injective from A to B and f" is injective from B to
A.

Then we call G a bijective function that is the same as its converse.

Now if we call any function that is the same as its converse(inverse) a
"reflexive function"

Accordingly if f<>f" then G is not as reflexive as for example f(x)=x
or f(x)= -1 or f(x)=1/x

are.

We fixed G to be reflexive so G is second order reflexive cuntion
ONLY.

Lets G defined above an " essentially reflexive" or " second order
relfexive "

weather f = f" or f <> f.

So what is their is:

f(x)=x or f(x)= -1 or f(x)=1/x ( f=f") is a kind of a first order
reflexive functions, AND of

coarse they are also essentially reflexive or second order reflexive (
these fucntion when

they biject two sets the two

bijected set are called first order identical sets ), but G is "only
essentially reflexive"

or "second order reflexive that is not at the same time first order
reflexive" if f <> f".

In Summary:


1)For any two sets A and B , define G(x)=f(x) if x belongs to A
and G(x) = f"(x) if x belongs to B. were G is a one valued function.

2)G is a second order reflexive function , and A and B are called
second order
identical sets, if f:A->B is injective and f":B->A is injective.

3)A and B are second order identical sets that are at the same time
first order
identical sets if f = f" , because G is a second order reflexive
function
that is at the same time a first order reflexive function.

4)A and B are second order identical sets that are not at the same time
first
order identical sets if f <> f", because G is a second order reflexive
function that
is not at the same time a first order reflexive function.

Now O and E defined above belong to 4), so G between them is only
second order
relfexive function .

Now the beautiful result would be the following:

second order identical sets( essentially identical ) never have a
proper intersectional set that is infinite.

( a proper intersectional set between A and B " A.B" is an interseconal
set AB that
can be a proper subset of either A or B )

This leads to if A and B are essentially identical then A cannot be a
subset of B

and B cannot be a subset of A.

In that manner I can modify the definition of subcard A < subcard B by
increasing

the exception made on it from being first order identical sets to
second order identical

sets.

Accordingly the definition of subcard A < subcard B would be changed
to:

subcard A < subcard B if their exist an injective function from A to B
that is not a bijection , EXCEPT when A and B are essentially identical
sets
(second order identical sets).

In which case subcard A<> subcard B and card A = card B would imply

that there is ALWAYS a non empty intersectional set AB which is ALWAYS
infinite and ALWAYS a proper intersectional set.( hint: G is a one
valued function).


Zuhair



>
>
>
> > Any how I want to ask you a question:
> >
> > If A and B are finite sets , tell me what is the difference between the
> > following two definitions:
> >
> > 1) A is a proper subset of B if every member x in A is member x in B ,
> > and
> > B is not the same set as A.
> >
> > 2) A is a proper subset of B if every member x in A is member x in B,
> > and
> >
> > card A < card B.
> >
> > Do you think that these definitions are identical.
>
> No they define two different things.
>
> > Do you think that
> > the concept of
> > cardinality used in 2) constitute an extra-unnecessary definition of
> > proper subset , and definition 1) is enough.
>
> Defintion 1) is the standard definition and is certainly enough.
>
> The trick is to have one definition for each concept, and
> then prove some theorems. E.g.
>
> 1) A is a proper subset of B if every member x in A is member x in B
> ,
> and B is not the same set as A.
>
> 2) A is a gentile subset of B if every member x in A is member x in
> B ,
> and card(A) does not equal card (B)
>
> 3) A is a well bred subset of B if every member x in A is member x in
> B ,
> and subcard(A) does not equal subcard (B)
>
> Then we can ask if every proper subset is a gentile subset (no)
> and whether A is a proper subset if and only if it
> is a well bred subset (yes). We could now use definition 3) rather
> than definition 1), but why bother?
>
> -William Hughes

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