Re: Cardinality and injection


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) .


>
> 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.

the definition you mentioned is definition by intention, mine is by
extention.

Zuhair


>
>
> -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, ...
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  • Re: Cardinality and injection
    ... > William Hughes wrote: ... >> zuhair wrote: ... > a proper subset of at least one of them, then subcard A subcard B. ...
    (sci.math)
  • Re: Cardinality and injection
    ... > zuhair wrote: ... >> William Hughes wrote: ... >> These are also set which has subcard A = subcard B ... a proper subset of at least one of them, then subcard A subcard 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)