Re: Cardinality and injection


zuhair wrote:
> 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 .

[ Let G(x) = x+1 if x is an odd integer, x-1 if x is an even integer, x
otherwise.
G(x) is a (single valued) bijection that is its own inverse.
This shows that O and E are first order identical sets.

Indeed it is not clear that there exists a pair of second order
identical sets that are not first order identical. ]



The above is rather confused and can be replaced by the following:

Two sets A and B are "essentially indentical" if there exists an
index set I that indexes both A and B, and a set of
ordered pairs { (a_i, b_i) | i in I} such that a_i not equal to b_j for
any i not equal to j.

Then

if A and B are finite they are essentially identical iff they
have the same cardinality

if A and B are infinite

if A and B are disjoint they are essentially identical iff
they have the same cardinality.

If A is a subset of B they are essentially identical iff
A=B.

So if A is a subset of B, A is a proper subset of B iff
A is not essentially identical to B.

Not a particularly impressive result.

Now we know that if
A is a subset of B then A and B have the same subcardinality
iff A=B, thus iff A is essentially identical to B. So
if A is a subset of B, A is a proper subset of B iff
subcard(A) <> subcard(B). Hardly earthshaking.

I think it is about time you abandoned subcardinality.

- William Hughes

.

Quantcast