Let $$A = \{x \in R : |x + 1| < 2\}$$ and $$B = \{x \in R : |x - 1| \geq 2\}$$. Then which one the following statements is NOT true?

We have $$A = \{x \in \mathbb{R} : |x + 1| < 2\}$$ and $$B = \{x \in \mathbb{R} : |x - 1| \geq 2\}$$.

For set $$A$$: $$|x + 1| < 2$$ means $$-2 < x + 1 < 2$$, so $$-3 < x < 1$$. Thus $$A = (-3, 1)$$.

For set $$B$$: $$|x - 1| \geq 2$$ means $$x - 1 \geq 2$$ or $$x - 1 \leq -2$$, so $$x \geq 3$$ or $$x \leq -1$$. Thus $$B = (-\infty, -1] \cup [3, \infty)$$.

Now we check each option:

Option A: $$A - B = A \cap B^c$$. We have $$B^c = (-1, 3)$$. So $$A - B = (-3, 1) \cap (-1, 3) = (-1, 1)$$. This matches Option A. $$\checkmark$$

Option B: $$B - A = B \cap A^c$$. We have $$A^c = (-\infty, -3] \cup [1, \infty)$$. So $$B - A = [(-\infty, -1] \cup [3, \infty)] \cap [(-\infty, -3] \cup [1, \infty)] = (-\infty, -3] \cup [3, \infty)$$. This equals $$\mathbb{R} - (-3, 3)$$, not $$\mathbb{R} - (-3, 1)$$. So Option B is NOT true. $$\times$$

Option C: $$A \cap B = (-3, 1) \cap [(-\infty, -1] \cup [3, \infty)] = (-3, -1]$$. This matches Option C. $$\checkmark$$

Option D: $$A \cup B = (-3, 1) \cup (-\infty, -1] \cup [3, \infty) = (-\infty, 1) \cup [3, \infty) = \mathbb{R} - [1, 3)$$. This matches Option D. $$\checkmark$$

The statement that is NOT true is Option B.