The Empty Set, Finite and Infinite Sets, Equal Sets

If A is a subset of B, every element of A is also in B. Consequently, an element that is outside B cannot be in A; otherwise it would be in B as well. Therefore every element of B′ is an element of A′, which proves B′ ⊆ A′. Taking complements reverses the direction of inclusion. Option A expresses this correctly: the larger set B leaves a smaller complement, while the smaller set A leaves a larger complement.

A−B is the portion of A lying outside B. If its cardinality is zero, that portion contains no elements, so no element of A can lie outside B. Therefore every element of A belongs to B, which is exactly the definition of A⊆B. The condition n(A)>0 only confirms that A is nonempty; it does not imply B⊆A or disjointness.

The union contains all elements of both sets, so it must contain at least as many elements as the larger set. Therefore n(A ∪ B) ≥ max(n(A), n(B)) = 109. This lower bound is attainable if the smaller set B is completely contained in A. In that case A ∪ B = A and its cardinality is 109. Hence the minimum possible value is 109, option A.

The intersection consists of all elements that belong to both sets. The order is irrelevant in set notation, so \(B=\{3,2,1\}\) is the same set as \(\{1,2,3\}\). Every element 1, 2, and 3 is common to A and B. Therefore, \(A\cap B=\{1,2,3\}\), making option A correct.

The union of two sets contains every distinct element appearing in either set, but a common element is counted only once. Here, \(A\cup B=\{1,2,3,4,5,6\}\), so its cardinality is 6. Equivalently, \(n(A\cup B)=4+3-1=6\), because 4 belongs to both sets. Option B incorrectly counts 4 twice; options C and D do not represent the complete union.

This is the distributive law of intersection over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). An element belongs to the left side when it is in A and also in at least one of B or C. That means it is either in both A and B or in both A and C, which is exactly the right side. Hence option A is correct.

The power set of any set contains the empty set. For the empty set itself, the power set is \(\mathcal{P}(\varnothing)=\{\varnothing\}\), which has exactly one element. Therefore, if \(\mathcal{P}(A\cap B)=\{\varnothing\}\), the original set must be empty. Hence, \(A\cap B=\varnothing\). Notice that \(\{\varnothing\}\) is a set containing the empty set, whereas \(\varnothing\) contains no elements.

Use each possible y from B and calculate x = 2y - 1. For y = 1, x = 1, giving (1,1); for y = 2, x = 3, giving (3,2); and for y = 3, x = 5, giving (5,3). All three calculated x-values belong to A, and every y-value belongs to B. Therefore, the valid ordered pairs are (1,1), (3,2), and (5,3), so their number is 3 and option C is correct.

Check each possible value of \(x\) from A. For \(x=0\), the equation becomes \(y^2=1\), so \(y=-1\) or \(y=1\), giving two pairs: \((0,-1)\) and \((0,1)\). For \(x=1\), \(y^2=0\), so \(y=0\), giving \((1,0)\). For \(x=2\), \(y^2=-3\), which is impossible. Thus there are three valid ordered pairs, so option B is correct.

The equation \(x+y=0\) is equivalent to \(y=-x\). We must ensure that the first component belongs to A and the second belongs to B. For \(x=-1\), \(y=1\); for \(x=0\), \(y=0\); and for \(x=1\), \(y=-1\). These give \((-1,1),(0,0),(1,-1)\). Values \(x=-2\) and \(x=2\) would require 2 and -2, neither of which is in B. Therefore, the answer is 3, option B.