Re: Cardinality and injection


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.

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. Do you think that
the concept of
cardinality used in 2) constitute an extra-unnecessary definition of
proper subset , and definition 1) is enough.

And if not, then what is the difference between subcardinality and
that.

Zuhair

.

Relevant Pages

  • Re: Cardinality and injection
    ... >> fmust be a proper subset of A, and hence a proper subset of B. ... > 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 ... > 1) A is a proper subset of B if every member x in A is member x in B, ...
    (sci.math)
  • Re: Cardinality and injection
    ... >> zuhair wrote: ... >a proper subset of at least one of them, then subcard A subcard B. ... The relation "not equals" can already do that ... ...
    (sci.math)
  • Re: Cardinality and injection
    ... > zuhair wrote: ... >> William Hughes wrote: ... >> that one of them should be a proper subset of the other ... subcard A = subcard B. ...
    (sci.math)
  • Re: Cardinality and injection
    ... >> zuhair wrote: ... A was a subset of B. You replied with two counterexamples, ... > a proper subset of at least one of them, then subcard A subcard B. ...
    (sci.math)
  • Re: Cardinality and injection
    ... >> zuhair wrote: ... >>> having an intersectional set A.B that is infinite and at the same time ... >>> that one of them should be a proper subset of the other ... > subcard A = subcard B. ...
    (sci.math)