Re: Cardinality and injection


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

.

Relevant Pages

  • 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: ... >> subcard A = subcard B. ... > fmust be a proper subset of A, and hence a proper subset of B. ... A is a proper subset of B if every member x in A is member x in B, ...
    (sci.math)
  • Re: Cardinality and injection
    ... > card A <= card B. ... the definition of 'proper subset' makes no mention of cardinality. ... > Then subcard A = subcard B ...
    (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)
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