Re: Cardinality and injection


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).



> 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.

No they define two different things.

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

Defintion 1) is the standard definition and is certainly enough.

The trick is to have one definition for each concept, and
then prove some theorems. E.g.

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 gentile subset of B if every member x in A is member x in
B ,
and card(A) does not equal card (B)

3) A is a well bred subset of B if every member x in A is member x in
B ,
and subcard(A) does not equal subcard (B)

Then we can ask if every proper subset is a gentile subset (no)
and whether A is a proper subset if and only if it
is a well bred subset (yes). We could now use definition 3) rather
than definition 1), but why bother?

-William Hughes

.

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