Power Set and Subsets

The set has two odd elements, 1 and 3, and two even elements, 2 and 4. A subset has an even sum when it contains either zero or two odd elements. The number of choices is (choose 0 of 2 odd elements)(2^2 choices for even elements) plus (choose 2 of 2 odd elements)(2^2 choices for even elements), giving 4 + 4 = 8.

Every element of A is odd. The sum of a subset is odd exactly when the subset contains an odd number of elements. Therefore, select either one element or three elements: C(4,1) + C(4,3) = 4 + 4 = 8. The empty set and all two-element or four-element subsets have even sums, so they are not counted.

The elements of A are the numbers 1, 2, and 3, whereas the elements of B are the singleton sets {1}, {2}, and {3}. Braces are mathematically significant: 1 is not the same object as {1}. Thus A and B are not equal. Also, 1, 2, and 3 are not elements of B, while {1}, {2}, and {3} are not elements of A, so neither set is a subset of the other.

The power set P(A) consists of all subsets of A. Since A has the two elements {1} and {2}, its subsets are ∅, {{1}}, {{2}}, and {{1}, {2}}. Therefore {{1}} is an element of P(A). Notice that {1} is an element of A, not a subset of A, and 1 is neither an element of A nor a subset of A.

A two-element subset that does not contain 5 must be formed entirely from the remaining set {1, 2, 3, 4}. We therefore choose any two of these four elements. The number of choices is C(4,2) = 4×3/2 = 6. The value 10 would count all two-element subsets of A and would incorrectly include the four subsets containing 5.

The elements 2 and 5 are compulsory, so they have no choice. Element 7 is forbidden and therefore is also fixed as absent. The remaining four elements, 1, 3, 4, and 6, can each either be included or excluded independently. Hence the number of allowed subsets is 2⁴ = 16, so option C is correct.

The governing definition is P(A) = the set of all subsets of A. Here A has two elements: ∅ and {∅}. The set {{∅}} contains the single element {∅}, which is an element of A; hence {{∅}} is a subset of A and therefore belongs to P(A). Option A is correct. Option B is a membership statement rather than the required subset, and 1 and {1} contain objects unrelated to A, so they are not subsets of A.

For a finite set with n distinct elements, each element has two independent choices when forming a subset: it may be included or excluded. Therefore, the total number of subsets is 2ⁿ. Here n = 3, so the number is 2³ = 8. These include the empty set, the three one-element subsets, the three two-element subsets, and A itself.

If a finite set has n elements, its total number of subsets is 2ⁿ. A proper subset includes every subset except the set itself, so the number of proper subsets is 2ⁿ − 1. Given 2ⁿ − 1 = 31, we obtain 2ⁿ = 32 = 2⁵. Therefore n = 5, making option B correct. The empty set is included among the proper subsets.

The distinct letters occurring in the word “math” are m, a, t, and h. Therefore the verbal description of set A gives exactly \(A=\{m,a,t,h\}\), which is the same as set B. In set theory, the order in which elements are written does not matter, and repeated elements would also be written only once. Hence A and B have precisely the same elements, so \(A=B\). A proper-subset statement would be false because neither set has an extra element.

The positive odd integers smaller than 8 are 1, 3, 5, and 7. No other positive odd integer satisfies the condition: 9 is not less than 8, while 0 and negative odd integers are not positive. Thus the roster form of B is \(B=\{1,3,5,7\}\), exactly the same as A. Since equal sets contain all and only the same elements, the correct relation is \(A=B\), not a proper-subset relation.

Every element of B, namely 1 and 2, is also an element of A. Therefore \(B\subseteq A\). However, A also contains 3 and 4, which are not in B, so the two sets are not equal. Consequently B is a proper subset of A, written \(B\subset A\) when the symbol denotes a proper subset. Option A reverses the inclusion, option C claims equality, and option D is false because the intersection is \(\{1,2\}\), not empty.

The prime factorization of 12 is \(12=2^2\times3\). Its distinct prime factors are therefore 2 and 3. A set does not record multiplicity, so the repeated factor 2 is listed only once; we do not write \(\{2,2,3\}\). Hence \(A=\{2,3\}=B\). Option C is false because equal sets are not proper subsets of one another, and option D is false because 12 is not an element of either set.

Set B has exactly one element, and that element is the set \(\{1,2\}\) itself. Since A is the set \(\{1,2\}\), A is therefore an element of B, so \(A\in B\). This does not mean A equals B: A has two numerical elements, whereas B has one set-valued element. Also, B is not a subset of A because its element \(\{1,2\}\) is not the number 1 or 2, and 1 is not directly an element of B.

Because both p and r belong to A, every element of B belongs to A; hence \(B\subseteq A\). The subsets of B are \(\varnothing\), \(\{p\}\), \(\{r\}\), and \(\{p,r\}\). Each of these contains only elements from A, so each is also a subset of A. This is the transitive property of inclusion: if \(C\subseteq B\) and \(B\subseteq A\), then \(C\subseteq A\). Therefore all subsets of B satisfy the requirement.

A set C is a subset of A when every element of C is also an element of A. The empty set is a subset of every set, so option A qualifies. Both 0 and 2 are in A, making option B a subset, and 1 is in A, making option C a subset. In option D, the only element is the empty set \(\varnothing\), not the number 0. Since \(\varnothing\notin A\), the set \(\{\varnothing\}\) is not a subset of A.

If a finite set A has n elements, then each element has two choices when forming a subset: it is either included or excluded. Therefore the power set has \(|P(A)|=2^n\) elements. Here \(2^n=16=2^4\), so n=4. Thus A contains 4 elements. The number 16 describes the total number of subsets, not the number of original elements. For comparison, sets with 2 and 3 elements have power sets of sizes 4 and 8 respectively.

The elements of \(P(A)\) are exactly the subsets of A. Option B, \(\{2,3\}\), is a subset because both 2 and 3 belong to A; therefore it is an element of \(P(A)\). Option A is only the number 2, an element of A but not a subset of A. Option C is not a subset because 4 is absent from A. Option D contains the set \(\{1\}\) as its element, rather than the number 1, so it is not a subset of A. The distinction between membership and subset notation is essential.

If \(A\subseteq B\), every element of A belongs to B. If at the same time \(B\subseteq A\), every element of B belongs to A. Thus neither set contains an element absent from the other, so they have exactly the same elements and must be equal: \(A=B\). This is the standard two-inclusion method for proving equality of sets. The conclusion says nothing about whether the sets are finite or infinite. Their intersection is actually the common set itself, not the empty set.

A subset must contain only elements that are themselves elements of A. The elements of A are the number 1, the set {2}, and the number 3. Option B contains {2} and 3, so every element of that option belongs to A; hence it is a subset. In options A and D, the number 2 appears, but 2 is not an element of A. In option C, the element {1, 2} is not in A. The distinction between 2 and {2} is essential.