ok, prove that the natural numbers, the even natural numbers, and the odd natural numbers all have the same cardinality.
so from this, I guess it is saying that
l N l = l 2N l = l 2N+1 l
now, I need to prove that.
so from what I know, having bijection for all 3 implies that they all have the same cardinality. So that is my main problem. how do I show they are all bijective?
I think it is pretty obvious that 2N is injective to N, because numbers in 2N will always hit something in N(since N contains everything). Also, same for 2N + 1. but that only proves injective.
to prove bijective, if you can find injective AND surjective, then it proves bijective, and therefore, would prove cardinality.
another way is to prove that A ---> B is injective and B ----> A is injective, then that would also prove bijective, which means it is cardinal.
for the first method, I dont see how it is surjective, and for the second method, I dont see how A ----> B is injective. If 2N is B, and N is A, then I can see how B ----> A is injective....but that is all?
any tips or aything guys?
so from this, I guess it is saying that
l N l = l 2N l = l 2N+1 l
now, I need to prove that.
so from what I know, having bijection for all 3 implies that they all have the same cardinality. So that is my main problem. how do I show they are all bijective?
I think it is pretty obvious that 2N is injective to N, because numbers in 2N will always hit something in N(since N contains everything). Also, same for 2N + 1. but that only proves injective.
to prove bijective, if you can find injective AND surjective, then it proves bijective, and therefore, would prove cardinality.
another way is to prove that A ---> B is injective and B ----> A is injective, then that would also prove bijective, which means it is cardinal.
for the first method, I dont see how it is surjective, and for the second method, I dont see how A ----> B is injective. If 2N is B, and N is A, then I can see how B ----> A is injective....but that is all?
any tips or aything guys?
