Prove that for every positive integer n, and sets A and B1, B2, ..., Bn, A - ∩(i=1 to n) Bi = ∪(i=1 to n) (A - Bi).
Question
Prove that for every positive integer n, and sets A and B1, B2, ..., Bn, A - ∩(i=1 to n) Bi = ∪(i=1 to n) (A - Bi).
Solution
Sure, let's prove this statement using the principle of mathematical induction.
Base Case (n=1): For n=1, the statement simplifies to A - B1 = A - B1, which is obviously true.
Inductive Step: Assume the statement is true for some positive integer k. That is, we assume that A - ∩(i=1 to k) Bi = ∪(i=1 to k) (A - Bi) is true.
We need to prove that the statement is also true for k+1. That is, we need to prove that A - ∩(i=1 to k+1) Bi = ∪(i=1 to k+1) (A - Bi).
We know that ∩(i=1 to k+1) Bi = (∩(i=1 to k) Bi) ∩ B(k+1) and A - B ∩ C = (A - B) ∪ (A - C) for any sets A, B, and C.
So, A - ∩(i=1 to k+1) Bi = A - (∩(i=1 to k) Bi ∩ B(k+1)) = (A - ∩(i=1 to k) Bi) ∪ (A - B(k+1)).
By the inductive hypothesis, (A - ∩(i=1 to k) Bi) = ∪(i=1 to k) (A - Bi).
Therefore, A - ∩(i=1 to k+1) Bi = ∪(i=1 to k) (A - Bi) ∪ (A - B(k+1)) = ∪(i=1 to k+1) (A - Bi).
This completes the inductive step, and hence, the statement is true for all positive integers n.
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