Answer:
Since, the function has been proven to be both surjective and injective, it is therefore said to be bijective and as such the question has been proved.
Step-by-step explanation:
Let A be the set of subsets of [n] that consists of an even number of elements, and let B be the set of subsets of [n] that consists of an odd number of elements.
Let's establish a bijection from A to B.
First of all,we have to establish a function that is both surjective and injective so that it is bijective.
Let the function be "f"
To prove the "f" is injective, let A1 and A2 be two subsets and consider f(A1)=f(A2)
From that, we get 2 options;
Either; A1 - {n} =f(A1) = f(A2) = A2 - {n}
Or AI u {n} =f(A1) = f(A2) = A2 u {n}
In both cases above, we can conclude that A1 = A2 and therefore, "f" is injective.
To prove that "f" is surjective, let B be an element of the range of "f" (a subset of odd size).
If B contains "n", then B−{n} is a subset of even size that maps to B under "f". Also, if B does not contain n, then B u {n} is a subset of even size that maps to B under "f".
Since everything in the image has something in the domain that maps to it, we can say that "f" is surjective.
Since, the function has been proven to be both surjective and injective, it is therefore said to be bijective and as such the question has been proved.
