Re: Relative Cardinality



David Kastrup wrote:
> mueckenh@xxxxxxxxxxxxxxxxx writes:
>
> > Relative Cardinality
> >
> > Given two finite or infinite sets A and B with elements a e A and b e
> > B. The union of these sets does exist. If the elements can be put into
> > an order < (not necessarily a well-order) such that in this order
> > 1) there are all elements a e A and b e B
> > 2) there are never two elements b,b' e B without an element a e A
> > between them with respect to <
> > then the cardinality Card(B) of B is not larger than the cardinality
> > Card(A) of A:
> > Card(B) =< Card(A).
>
> So Card({1,3}) =< Card({2}).
> And card({1}) =< Card({}).
>
> Great. Do you even check your ideas with trivial examples?

Of course, by posting them here. Someone will certainly find the error
if there is one. I did not check the finite case, because it is not so
interesting. I have to correct my theorem:

Given two finite or infinite sets A and B with elements a e A and b e
B. The union of these sets does exist. If the elements can be put into
an order < (not necessarily a well-order) such that in this order
1) there are all elements a e A and b e B
2) there are never two elements b,b' e B without an element a e A
between them with respect to < then the cardinality Card(B) of B is at
most by one larger than the cardinality Card(A) of A: Card(B) =<
Card(A) + 1.

> Apart from that, cardinality of a set is a property of the number of
> elements, and not of their values. So orderedness is not fundamental
> to cardinality.

Of course order is not fundamental, but if an order can be established,
then my definition is a sharp criterion to determine whether other
criteria are meaningful.

> Bijections are

leading to false results and, therefore, they are worthless.

Regards, WM

.

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