It is proposed to partition a set of positive integers into two disjoint subsets $ A$ and $ B$ subject to the conditions

[b]i.)[/b] 1 is in $ A$
[b]ii.)[/b] no two distinct members of $ A$ have a sum of the form $ 2^k \plus{} 2, k \equal{} 0,1,2, \ldots;$ and
[b]iii.)[/b] no two distinct members of B have a sum of that form.

Show that this partitioning can be carried out in unique manner and determine the subsets to which 1987, 1988 and 1989 belong.
1. **Initialization and Base Case:**
   - We start by placing \(1\) in set \(A\). This is given by condition (i).
   - Next, we need to ensure that no two distinct members of \(A\) or \(B\) have a sum of the form \(2^k + 2\) for \(k = 0, 1, 2, \ldots\).

2. **Constructing the Sets:**
   - Place \(2\) and \(3\) in set \(B\) because \(1 + 2 = 3\) and \(1 + 3 = 4\) are not of the form \(2^k + 2\).
   - Place \(4\) in set \(A\) because \(4\) does not form a sum \(2^k + 2\) with \(1\).

3. **Inductive Step:**
   - Assume that up to \(n-1\), we have successfully partitioned the numbers into sets \(A\) and \(B\) such that the conditions are satisfied.
   - Consider the number \(n\). We need to determine whether \(n\) should be placed in \(A\) or \(B\).

4. **Prohibition Check:**
   - Suppose \(A\) and \(B\) contain numbers \(a\) and \(b\) respectively such that \(n + a\) and \(n + b\) are prohibited sums.
   - Without loss of generality, assume \(a > b\). Then \(n + a = 2^k + 2\) for some \(k\).
   - Since \(a < n\), we have \(n \geq 2^{k-1} + 2\).
   - Similarly, \(n + b = 2^m + 2\) for some \(m < k\). This implies \(m \leq k-1\).

5. **Contradiction and Placement:**
   - Given \(n \geq 2^{k-1} + 2\) and \(b > 0\), it leads to a contradiction if both \(n + a\) and \(n + b\) are prohibited.
   - Therefore, we can always place \(n\) in one of the sets \(A\) or \(B\) without violating the conditions.

6. **Uniqueness:**
   - To ensure uniqueness, consider the smallest number larger than \(n\) of the form \(2^k + 2\).
   - If \(n \geq 2^{k-1} + 2\), then \(2^k + 2 - n < n\) has already been placed.
   - Thus, \(n\) must be placed in the opposite set of \(2^k + 2 - n\).

7. **Determining the Subsets for 1987, 1988, and 1989:**
   - Find the smallest \(2^k + 2\) larger than 1987, 1988, and 1989. This is \(2048 + 2 = 2050\).
   - Calculate the opposites:
     - For 1987: \(2050 - 1987 = 63\)
     - For 1988: \(2050 - 1988 = 62\)
     - For 1989: \(2050 - 1989 = 61\)
   - Determine the subsets of 63, 62, and 61:
     - 63 is in \(B\) (since 64 is in \(A\))
     - 62 is in \(A\)
     - 61 is in \(B\)

8. **Conclusion:**
   - Therefore, 1987 and 1989 are in \(B\), and 1988 is in \(A\).

The final answer is \( \boxed{ 1987 \in B } \), \(1988 \in A\), \(1989 \in B\).