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# Prove the given sets are measurable and have zero area

Prove measurability and establish that the area is zero for each of the following.

1. Any set consisting of a single point.
2. Any set consisting of finitely many points in the plane.
3. The finite union of line segments in the plane.

1. Proof. Since all rectangles are measurable with area equal to where and are the lengths of the edges of the rectangle, a single point is measurable with area 0, since a point is rectangle with
2. Proof. We prove by induction on , the number of points. For the case , the statement is true by part (a). Now, assume it is true for some . Then, we have a set of points in the plane and . Let be a point in the plane. By part (a), and . Thus,

But, , so

(Where Axiom 1 of area guarantees us that cannot be negative.) Thus, . Hence, the statement is true for points in a plane, and thus for all

3. Proof. We again use induction, now on , the number of lines in a plane. For , we let be a set with one line in a plane. Since a line is rectangle and all rectangles are measurable, we have . Further, since a line is a rectangle with either or , and so in either case, . Thus, the statement is true for a single line in the plane, the case .
Assume then that it is true for . Let be a set of lines in the plane. Then by the induction hypothesis, and . Let be a single line in the plane. By the case above, and . Thus, and (since ). Hence, the statement is true for lines in a plane, and so for all

# One correct and one incorrect formula for set complements

1. Given the two formulas:

Identify which one is always correct and which one is sometimes wrong, and prove the result.

2. Give an additional condition for the incorrect formula to be always true.

1. The formula is false in general.
2. Proof. Let

Then,

Thus, the formula does not hold in this counterexample.∎

The formula is correct.
Proof. Let be any element of . This means and , which in turn means and . Since and we have . Then, since , we have . Thus, .
For the reverse inclusion, let be any element of . This gives us and . The first part in turn gives us and . But then we have in neither nor ; hence, . Since , we then have . Therefore, .
Therefore, .∎

3. To make the first formula in part (a) always correct, we add the condition that . The claim is then:
If , then .
Proof. Let be any element in , then and . But, means that or (since this is the negation of which means and , so its negation is or ). So, if , then and hence . On the other hand, if , then as well. Hence, .
For the reverse inclusion, let . Then or . If , then and . Since, , then we know (since every must be in ). Therefore, . On the other hand, if , then since (by our additional hypothesis) we know . Further, since , we know . Therefore, . So, .
Therefore, indeed, .∎

# Prove relationships between complements, unions and intersections for classes of sets

Prove that for a class of sets we have,

Proof. Let be an arbitrary element of . This means that and , which means that is not in for any in the class . Hence, for every we have (since is in no matter what, and is not in for any that we choose, so it must be in ). But, for all means that is in the intersection ; and hence, .
For the reverse inclusion, let be any element in . This means that for every , i.e., and for every . Since for every , we then have . Hence, . Therefore, .
Hence, . ∎

Proof. Let be any element of . This means that and . Further, not in the intersection of the sets means that there is at least one , say , such that . Since and , we know . Then we can conclude . Thus, .
For the reverse inclusion, we let be any element in . This means there is at least one , say , such that , which means and . Since , we know ; therefore, . Hence, .
Therefore,

# Prove that the complement of an intersection is the union of the complements

Prove that .

Proof. First, let be any element in . By definition of complement, this means that and . Since we have either or which implies, coupled with the fact that is in , means or , respectively. Since is in at least one of these, is in the union . Therefore, .
For the reverse inclusion, let be an arbitrary element of . Then, either or .
If , then and ; hence, . Therefore, is in . On the other hand, if , then and ; hence, . This again implies is in . Therefore, .
Hence, .∎

# Prove some more statements about sets and subsets

1. Prove that if and , then .
2. Prove that if and , then .
3. If and then we claim .
4. If and , then .
5. What can you say if and ?

1. Proof. Let be any element in . Then, since . Further, there is some such that (since means that is a proper subset of ). Then, since we have both . Hence, (since every element of is in , and there is at least one element of not in , so ).∎
2. Proof. Let be any element of . Then, since . Further, implies since . Thus, .∎
3. Proof. This was established in part (a) since we didn’t need in that proof. (Since a proper subset of guaranteed there was some in that wasn’t in , and since is a subset, proper or otherwise, of , this is in ; hence, is a proper subset of .)∎
4. Proof. Since we know every element of is in . Hence, if , then .∎
5. If and , then we cannot conclude that . For example, let and . Then, (since is the set and contains this set). However, , but (since contains the set which contains 1, but does not contain the element 1).

# If C is a subset of A and B, then it is a subset of their intersection

Prove that if and , then .

Proof. Let be any element in . Then, since , and since . Therefore, . Hence, .∎

# Prove that if A and B are subsets of C, then so is their union

Prove that if and , then .

Proof. Let be any element in . Then, by definition of union, we have or . If , then since . On the other hand, if , then as well since . Hence, .∎

# Yet more proofs on intersections and unions of sets

Prove that and .

Proof. First, it is clear that (see Exercise 12 of Section I.2.5, or simply note that implies since is in ).
For the reverse inclusion, if is any element of then or . But implies (and ). So, in either case ; hence, .
Therefore, .∎

Proof. From Exericse 12 (Section I.2.5) we know that for any set ; hence, .
For the reverse inclusion, if is any element in , then and ; hence, . Therefore, .
Hence, .∎

# Union and intersection of a set with the empty set

Prove that and that .

Proof. If , then or by definition of union. Since (by definition of ), we must have . Hence, .
On the other hand, if , then by definition of union, so .
Therefore, .∎

Proof. If , then by definition of intersection and . But, by definition of , there is no such that . Hence, must be empty. By uniqueness of the empty set then, we have .∎

# Prove some more relations between sets

Prove that and that .

Proof. If is any element of , then by definition of , we have (since is the set of elements in either or ). Thus, . ∎

Proof. If is any element of , then since is defined to be the set of elements that are in both and . Therefore, . ∎