Re: Cardinality and injection


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:
> > > > > > > > > > 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.
> > >
> >
> > I defined
> >
> > Let G(x) = x+1 if x is an odd integer,
> > x-1 if x is an even integer,
> > x otherwise
> >
> >
> > [Note that G is a single valued function defined on every real.
> > True, (x+1) <> (x-1) but this does not matter because G(x)
> > is only ever equal to one or the other. We let
> > G(x) induce f and f'.]
> >
> > Now f is G restricted to the odd integers and
> > f'' is G restricted to the even integers.
> > Therefore (by your definitions) f = f''.
> >
> > The problem is that you think there is some way to
> > have f = f''' even though the two functions have different domains.
> > This is not possible. f mapping {1,2,3,4...} to f'' = {1,1/2.1/3,...}
> > is not the same function as f'' mapping {1,1/2.1/3,...} to
> > {1,2,3,4...}.
>
> Let me fist correct what you want to say :
>
> You are saying that:-
> f mapping {1, 2 , 3 , 4 ,...} to {1, 1/2 , 1/3 ,...}
> is not the same function as f'' mapping {1,1/2,1/3,...} to
> {1,2,3,4...}.
>
> My answer to that is Why?

Because if g is the same function as h if and only if
the domains of g and f are the same, say X, and for
every x in X g(x) = h(x) (which implies that the ranges
are the same). The domain of f is A, and the domain
of f'' is B. Therefore f can only equal f'' if A = B.

>
> f which is mapping A={1,2 , 3 , 4 ,...} to B= {1, 1/2 , 1/3 ,...} is
> f(x)= 1/x
> f'' mapping B={1,1/2,1/3,...} to A= {1,2,3,4...} is also f"(x) = 1/x
>

To specify a function you must specify domain and rule.
Here you are specifying the rule as "1/x" , which has
the same form (although differnt meaning) for f and f''.
But you do not specify domains.


> Now f:A ->B is exactly the same as f":A->B , because both gives the
> same results

f'' is defined on B. f'' is not defined on A.

> I can generate B from A by any of the two functions
>
> Also f:B->A is exactly the same as f":B->A , both gives exactly the
> same results

f is definied on A, f is not defined on B.

> I can generate A from B by any of the two functions.
>


To compare f and f'', either A must equal B or we must
have some way of extending f to B and f'' to A (e.g. make
A and B both subsets R, Let G be a function on R and let
f and f'' be the restrictions of G).

> How do I define " first order similarity" between two functions?
>
> Given X is the dom of f, and Y is the range of f
> Given X is the dom of f", then if Y is the range of f" then f = f"
>
> This is clearly applicable for f(x)=1/x , and f"(x)=1/x
>

No, if f mapps from
A={1,2 , 3 , 4 ,...} to B= {1, 1/2 , 1/3 ,...}
and f'' maps from
B= {1, 1/2 , 1/3 ,...} to A={1,2 , 3 , 4 ,...}
the domains are not the same.


> It is clear that f(x)=f"(x) , were x is the same domain.
>
> G(x) present between sets O and E is what is different from f and f"
> above , that's why I call it
> Only second order reflexive set, which follows of coarse "second order
> similarity" of sets.
>
> Again I repeat every f that is a first order reflexive function IS also
> a second order reflexive
> function , but not the opposite, ie not every second order reflexive
> function is also a first
> order reflexive function.
>
> I know what are you thinking about , you think that the similarity
> between f and f" is
> due to a function like G, ie the similarity between f and f" is not
> genuine , it is fixed
> so you think that f and f" are in reality fixed to be similar by the
> concept of G.
>
> So according to your interpretation of coarse you will hold the
> converse of what I am
> saying, that is: Every second order reflexive function is also a first
> order reflexive function
> and NOT every first order reflexive function is also a second order
> reflexive function.
>
> Which is clearly wrong.
>
> This is wrong the similarity between f and f" is not due to a backround
> of G.
>
> By now since I defined what I mean by f=f" , I think the confusion you
> are having about it
> will resolve.
>
> Their is no confusion in what I am saying.
>
> It is G that is fixed to be similar , and the similarity definition for
> G is of coarse different
> from that of f , this is clear , that's why I call it second order
> reflexive function.
>
> In reality I think there is no propblem with what I am saying since I
> defined it clearly
> The confusion arrives if one want to impose his definitions on a
> backround of mine,
> of coarse that will lead to confusion.
>
>
> However I want to remind you about something you wrote in a previous
> post, about
> an index set I and a set of pairs { (a_i,b_i): i in I } you said that
> the set of all a_i
> and the set of all b_i are essentially identical if a_i <> b_i for the
> same i.
>

No. The statement was

such that a_i not equal to b_j for
any i not equal to j.

I.e. the exact opposite of what you state, the sets are
essentially identical if a_i <> b_j for different values
of the index. (In which case there is a simple bijection
between A and B, just flip the ordered pairs).

-William Hughes

.

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