In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
If x ∈ A and A ∈ B, then x ∈ B
False. Example: Let A = {1} and B = {{1}, 2}. Here x = 1 ∈ A and A ∈ B, but 1 ∉ B. The element 1 is not in B; B contains {1} and 2 as elements.
If A ⊂ B and B ∈ C, then A ∈ C
False. Example: Let A = {1}, B = {1, 2}, C = {{1, 2}, 3}. Here A ⊂ B and B ∈ C, but A ∉ C. The set C contains {1, 2} and 3 as elements, not {1}.
If A ⊂ B and B ⊂ C, then A ⊂ C
True. Proof: Since A ⊂ B, every element of A is in B. Since B ⊂ C, every element of B is in C. Therefore, every element of A is in C. Hence, A ⊂ C.
If A ⊄ B and B ⊄ C, then A ⊄ C
False. Example: Let A = {1, 2}, B = {2, 3}, C = {1, 2, 4}. Here A ⊄ B (since 1 ∈ A but 1 ∉ B) and B ⊄ C (since 3 ∈ B but 3 ∉ C), but A ⊂ C (every element of A is in C).
If x ∈ A and A ⊄ B, then x ∈ B
False. Example: Let A = {1, 2}, B = {2, 3}, x = 1. Here x = 1 ∈ A and A ⊄ B (since 1 ∈ A but 1 ∉ B), but x = 1 ∉ B.
If A ⊂ B and x ∉ B, then x ∉ A
True. Proof: Since A ⊂ B, every element of A is in B. If x ∉ B, then x cannot be in A (otherwise x would be in B). Therefore, x ∉ A.
Explanation
This question tests understanding of set membership (∈) versus subset relation (⊂). Key distinction: A ∈ B means A is an element of B, while A ⊂ B means every element of A is in B. For false statements, counterexamples using small sets demonstrate the logical flaw. For true statements, proof follows from definitions of subset and element relations.