Sets: questions with answers

Both words use exactly the letters C, A, T and R, so X = Y — this is the chapter's own Example 23. The words have different LENGTHS and list their letters in a different ORDER, and neither of those affects a set: repetition and order are both irrelevant to which objects belong.

Well-definedness is about whether membership can be decided, not about what kind of thing the members are. The members here are not numbers and it is still a set — the chapter's own examples include the rivers of India and the boys in your class. Whether anyone has counted them, or could write them all down, does not enter into it either.

The dots have to be resolved before the set can be counted: the elements go 4, 4 + 3, … up to 34, which is 11 elements. Counting only the four printed numerals ignores what the dots stand for, and 34 − 4 is a difference rather than a count — it would only be right if the step were 1.

The empty set φ has no elements. The sets {φ} and {0} each have one element, but their elements differ: the empty set and the number zero.

Replacing 9 with 13 changes which numbers belong to A. Reversing the order and repeating an existing entry leave the set unchanged. Equal numbers of elements do not guarantee equal sets.

Every element of A ∩ B is already an element of A, so uniting the two adds nothing to A. There is no cancellation law for sets: A ∩ B = A ∩ C does NOT give B = C, because intersecting with A destroys every distinction outside A. Take A = {1}, B = {1, 2}, C = {1, 3}, where both intersections are {1} and yet B ≠ C.

Every set is a subset of itself, and the empty set is a subset of every set — including of φ itself, so the non-empty proviso is wrong. The last option reverses the empty-set rule: φ ⊂ A holds for all A, while A ⊂ φ forces A to be empty. Note that in this chapter ⊂ allows equality; a PROPER subset is the extra condition A ≠ B.

Each end has to be read from its own circle: filled means included and takes a square bracket, hollow means excluded and takes a round one. Here the two ends are drawn differently, so [3, 8) is right and both uniform choices of bracket are wrong at exactly one end.

The chapter notes that an interval contains infinitely many points. Counting the whole numbers in (4, 9] gives 5, but that counts only a handful of its members; the interval also contains every fraction and every irrational number between its ends. The difference of the end points is the interval's LENGTH, which is a size, not a count.

The region between the circle and the rectangle is not empty space: it holds exactly the elements of U that are not in A, which is {6, 10, 14}. That region is what makes the complement readable off the picture at all — A′ is precisely what is drawn there. The set {4, 8, 12} is A itself, and {4, 6, 8, 10, 12, 14} is the whole of U.

The two circles cross, and each has a region lying outside the other. One circle drawn wholly inside another means every element of the inner set is in the outer one; circles that cross mean some elements are shared and some are not; circles drawn apart mean none are shared. All three are drawn in the same rectangle, so the geometry is the only thing that differs and the only thing that carries the meaning.

A − B′ is the odd one out: being outside B′ means being inside B, so A − B′ = A ∩ B, which is the opposite of A − B. The other two are standard rewritings: A − B = A ∩ B′ is the definition in symbols, and A − (A ∩ B) removes from A exactly the part it shares with B. Note that A − B = B − A is also false in general — it is listed here as a distractor, and it holds only when both differences are empty.

Everything inside the rectangle and outside both circles is shaded, so an element is shaded when it lies in neither. The expression A′ ∪ B′ names a different region: it keeps the operation instead of swapping it when the complement moves inside the bracket, and De Morgan’s laws say (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

A and A′ divide U between them, so their union is the whole of U. A ∩ A′ = φ, since nothing is both inside and outside A. A ∪ A = A by the idempotent law, which gives U only in the special case A = U. And A′′ = A, so A ∩ A′′ = A ∩ A = A.

Disjoint means A ∩ B = φ. Here A ∩ B = {7}, so No — they both contain 7. Being unequal is far weaker than being disjoint: two sets can differ and still share elements. Both sets holding numbers of the same kind settles nothing either way, and disjointness is decided by the two sets alone, without any reference to a universal set.