Is it possible to make in injection
X*X --> P(X)
(*= carthesian product, P(X) is the powerset of X, that is, the set of all subsets from X)
without the choice-axiom?
I tried to construct such an injection but it seems somewhat hard
there's ofcourse an injection X--> P(X) given by x -> {x}
But X*X--> P(X) doesn't work with <x,y> -> {x,y}
U need a way to distuinguish between <x,y> and <y,x> somehow.
Yes, I know this
This gives a natural injection: X*X--> P(P(X))
(As you said, by using Kuratowski's definition of an ordered pair: by sending <x,y> -> {{x},{x,y}})
I was hoping to even get one step further and construct an injection X*X -> P(X) wich should be possible for any infinite set X.
But I suspect it's impossible to construct such an injection without the Choice-axiom. Was just a curiosity.
Sometimes you can't make it:for example, the case of X has three element.
If X has four element, then it is very easy since they have the same cardinal, every injection satisfy our requirement.
If X has more elements, |X|>4:
Since every set can be well ordered. w.l.o.g. suppose (X, <) is well ordered set.
define the mapping F:X*X to P(X) as following:
F(x,x)={x};
F(x,y)={x,y} if x,y are distinct, and x<y;
F(x,y)=the complement of {x,y}, if x,y are distinct ,and x>y.
Well yes, but then you've actually assumed the choice-axiom.Since every set can be well ordered. w.l.o.g. suppose (X, <) is well ordered set.
The choice axiom is equivalent to the wellordering-theorem of Zermelo. And the set X is not necessarily a ordered set!
And I like to know whether we can construct such an injection without assuming the choice-axiom
And I suspect this is impossible, since it seems the only way to determine the difference between <x,y> and <y,x> is having some wellorder on X
If |X| is less than or equal to 4, it is very easy.
Othwise,Classify the element into two groups:
for the element of the form (x,x) in X*X, associate with {x}.
for the element pair (x,y) and (y,x) wherex,y are distinct,in X*X, associate with pair ,no matter the detail on how they are associated in the sysmetric pair, since we are focus on "injection mapping".