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:
> > > > > > > > > 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.

Wrong. O and E are second order identical sets that are not first order
identical sets, because x+1 <> x -1.

You didn't read my post carefully, G(x) is not a first order reflexive
function
so that O and E can be first order identical sets, it is a second order
reflexive
function, G(x) is called first order reflexive if and only if f = f"
so your G(x)= f(x) , f:E->O , f(x)=x+1 , and G(x)= f"(x)= x-1 f":O->E
Now f<>f" => G(x) is second order reflexive function that is not a
first order
reflexive function.

So E and O are second order identical sets that are not first order
identical sets.

I think you understood things upside down,

if A and B are first order identical sets then they are always a second
order identical sets
but not the converse, ie if A and B are second order identical sets
then it doesn't imply
that A and B are always a first order identical sets.

I can even form a third, fourth, etc....... order reflexive functions
and consequentely
third, fourth, etc... order identical sets.



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

Wrong the example you made is the best example for such a pair

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

No, you didn't read it well, and you were confused, you can say
the above is not well presented and you might be right but certainly
it is not confused.

>
> 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.

hmmm........................
>
> 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.


That's the problem with you , what you have made now is what I really
was talking about, this subcardinality is another way of looking at
"proper subset"
in terms of extention , and not intention only, definitions by
extention
are also important, especially in mathematics, while definitions
by intention are the ones which logicians should care for.


>
> I think it is about time you abandoned subcardinality.
>
> - William Hughes

May be because I demonstrated no additional usefullness of the term ,
but
it is a nice game to play with.It is a way of looking at things from a
different
angle, tell me that 4=2+2 and 4=3+1 are not useful and are only
repetitions
of the same thing 4.

I think subcardinality is interesting really, but what is the benifit
of a game
with me as the only player? This will make me abandon it sure, because
the game is not appealing to mathematicians to play, it is a waste of
time
to them.

In reality I can add even a further property for subcardinality to
fulfill , in order
to almost completely order subcardinality, by simply saying.

if card A = card B and subcard A <> subcard B.

then subcard A < subcard B if f:A ->B , f(x) = x ,f is injective but
not bijective.
but if f is not injective , then this means that there is A . b that
is an infinite proper
subset of both A and B. and here subcard A <> subcard B ( complete
ordering
not established for such a case )

It is a pitty that I am the only player of the game of subcardinality.


Zuhair...........

.

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