Sets

A collection is a set when membership has an objective rule. “Beginning with J”, “in this class”, “below 100”, “written by…”, “even”, and “questions in this chapter” are checkable. “Most talented/best/most dangerous” has no stated measure, so different people may choose differently.

Compare each number with the listed elements of A. Membership is exact: a number not listed is not an element.

List only values satisfying the rule. Sets do not repeat elements, so each letter appears only once even if the word repeats it.

Describe the common generating rule and any endpoint. For the finite lists, restrict the index so no extra terms are included.

Apply the restriction to the correct number system. In (ii), list all integers strictly between −2 and 4.5. In letter collections remove duplicates and exclude vowels for (vi).

Compare the defining property with every listed element. The set of divisors includes 1 and 6; prime divisors exclude 1.

An odd number cannot be divisible by 2. No number can be both below 5 and above 7. Distinct parallel lines do not meet. But 2 is an even prime, so set (ii) contains 2.

A set is finite if it has a fixed finite count. The natural-number sequence and positive integers above 100 continue without end; the others have bounded lists.

A line can be shifted to infinitely many heights while remaining parallel to x-axis. There are 26 English letters. Multiples of 5 continue forever. At a fixed time the living animals form a finite collection. A circle through the origin can have infinitely many centres/radii.

Order does not affect a set. In (ii), 12 is missing and 18 is extra. In (iii), both list exactly the positive even numbers through 10. In (iv), the second list contains numbers such as 15 that are not multiples of 10.

Factor x²+5x+6=(x+2)(x+3), giving roots −2,−3. For words, collect distinct letters only; repeated letters do not change a set.

Compare distinct members without considering order. B and D each contain 1,2,3,4; E and G each contain −1 and 1.

A is a subset of B only if every element of A belongs to B. One counterexample (such as a circle not of radius 1 in iv) is enough to show “not a subset”.

(i) The two sets are equal. (iii) 2 is missing on the right. In (v), a is an element, but the set {a} itself is not listed as an element. In (vi), the left set is {2,4}, and both divide 36.

In A={1,2,{3,4},5}, the set {3,4} is one element. For a subset, each member must itself be an element of A. Also, ∅ is a subset of every set, but is not automatically an element.

A set with n elements has 2ⁿ subsets. Include the empty set and the whole set, as well as every combination in between.

Use a round bracket for an excluded endpoint (< or >) and a square bracket for an included endpoint (≤ or ≥).

Translate each bracket separately: square means include that endpoint; round means exclude it.

A universal set must contain every element in the set being discussed. All right and isosceles triangles are triangles in that plane.

A universal set must contain every element of A∪B∪C={0,1,2,3,4,5,6,8}. Only option (iii) contains all of them.

The union contains every element appearing in either set, with duplicates written once. For (iii), combine 1–5 with the positive multiples of 3; this also includes 6, 9, 12, and so on.

Every member of A already belongs to B, so joining A adds no new elements to B.

Combine the listed elements and remove repeated values.

The intersection keeps only elements common to both sets. The two ranges in (iv) do not overlap, and no set shares an element with ∅.

Evaluate parentheses first, then retain common elements. For example B∪C={7,9,11,13,15}; intersecting with A keeps 7,9,11.

Even and odd naturals do not overlap. The only even prime is 2; every other prime is odd.

Two sets are disjoint exactly when their intersection is empty.

A−B keeps elements in A that are not in B. The order matters, so generally A−B differs from B−A.

Check shared elements. In (iii) and (iv), one set contains only numbers congruent to 2 mod 4 and the other only numbers congruent to 3 mod 4; no values coincide.

A complement is taken inside U. First form a union or difference when requested, then list elements of U left out.

Remove the members of each subset from the universal set U.

Take the complement relative to N, not all integers. Solve the simple equations/inequality first, then exclude their solution sets. For primes, 1 is included in the complement because 1 is not prime.

A∪B={2,3,4,5,6,7,8}; its complement is {1,9}. A∩B={2}; its complement is every other element of U. Computing both sides gives matching sets.

Draw U as a rectangle and A and B as overlapping circles. Shade: (i) outside A∪B; (ii) outside A and outside B; (iii) all of U except A∩B; (iv) outside A or outside B. Parts (i)=(ii) and (iii)=(iv).

A triangle not in A has no angle different from 60°, so all three angles are 60°. Such a triangle is equilateral.

A and its complement together cover U; the empty set’s complement is U; a set and its complement do not overlap; U′ is empty.

Factor the quadratic: (x−2)(x−6)=0. Then check inclusion element by element.

Membership and subset are different relations. For (iii), subset inclusion is transitive. For (vi), if x were in A, A⊂B would force x∈B, contradicting the given. Counterexamples disprove the others.

Take x∈B. If x∈A, then x∈A∩B=A∩C, so x∈C. If x∉A, then x∈A∪B=A∪C, and since x∉A it must be in C. Thus B⊂C; symmetry gives C⊂B.

If A⊂B, no element of A is outside B, so A−B=∅. If that difference is empty, every A-element is in B. When A⊂B, adding A to B changes nothing and the common elements of A and B are exactly A. Conversely, A∪B=B forces A⊂B; A∩B=A also forces every A-element into B.

If x∈C−B, then x∈C and x∉B. Since A⊂B, every A-element would be in B; therefore x∉A. So x∈C−A.

An element of A either belongs to B (then it is in A∩B) or does not (then it is in A−B), giving the first identity. In the second, A already includes its elements; B−A adds exactly the B-elements missing from A, so the result is A∪B.

A∪(A∩B)=(A∪A)∩(A∪B)=A∩(A∪B)=A. Also A∩(A∪B)=(A∩A)∪(A∩B)=A∪(A∩B)=A.

Intersecting any set with ∅ gives ∅, so the two intersections match even when B and C differ.

A=A∩(A∪X)=A∩(B∪X)=(A∩B)∪(A∩X)=A∩B. Similarly B=A∩B. Hence A=B.

A∩B={2}, B∩C={3}, and A∩C={1}, all nonempty. No element belongs to all three, so the triple intersection is empty.