If n(U)=80, n(A)=40 and n(B)=30, find the maximum value of n(AuB)complement and minimum value of n(AuB)complement

to show that the set of all even integers is denumerable, we prove that there is a bijection from to , where , the set of all even integers.

proof. note that the set of even integers can be put into a list, that is, . we can intuitively see that all the even integers can be listed, allowing for a way to count all the even integers (however, we will never finish counting). we can be more rigorous in our argument by finding an explicit bijection. we can think about how to recreate the list we have made.

let the function be defined by  

for example, , , , , and so on. clearly this function generates this list. now we show that is a bijection. to do this, we must show that is injective and surjective.

to show that is injective, we must show that for any , implies .

suppose that such that . then we have four cases:

case 1: if are both odd, then

case 2: if are both even, then

case 3: if is even and is odd, then

  .

but since is a positive even integer, we have and since is a positive odd integer, we have .

we inferred that is true in this case, but the left- and right-hand inequalities actually do not agree. this is a contradiction, so case 3 cannot happen.

case 4: if is odd and is even, then by similar argument as in case 3, case 4 cannot happen.

hence is injective.

now we show that is a surjection. to do this, we must show that for any , there is such that . we must assert that there is an element in the domain that has as its image. so suppose we have . here, is even.

case 1: if , then and , since is even.

case 2: if , then so and , since is odd.

case 3: if then .

so is also a surjection.

therefore, is a bijection because it is injective and surjective.

it follows that , the set of all even integers, is denumerable.