Which of the following would be a valid partition of the set of all strings A^* over the alphabet A = \{a, b\}?

Question

BlackTom AI · Accepted Answer

To tackle this question, I will examine what constitutes a partition of A* for A = {a, b} and then evaluate each option.
Option a: {{ε}, {all strings starting with a}, {all strings starting with b}}
- First, note that ε is included in the first block, which is appropriate since ε does not start with a or b. 
- The second block contains all strings that begin with the letter a, regardless of what comes after. The third block contains all strings that begin with the letter b. Every nonempty string over {a,b} begins with either a or b, so these three blocks cover all of A*. 
- The blocks are pairwise disjoint: ε is not in any of the “starting with a” or “starting with b” sets, and a string cannot start with both a and b. 
- Therefore, the collection forms a partition of A* into three nonempty, disjoint sets, whose union equals A*.
Option b: {{aa}, {ab}, {ba}, {bb}}
- This is a partition of only a finite subset of A*, namely the strings of length 2. 
- A* contains strings of all lengths, as well as ε, and also many longer strings like aaa, abb, etc. Since these are not included in the blocks, the union of these blocks does not equal A*. Thus, this is not a partition of A*.
Option c: There is no way of partitioning the set, because it's an infinite set.
- The size of A* being infinite does not prevent partitioning. Infinite sets can indeed be partitioned into infinitely many disjoint nonempty blocks (for example, A^n blocks). The statement is incorrect because infinitude does not preclude forming a partition.
Option d: {A^n : n ∈ ℕ}
- Here, each block A^n is the set of all strings of length n over {a,b}. These blocks are disjoint, since a string has a unique length, and their union over all n ∈ ℕ is precisely A*, since every string has some finite length n. 
- Therefore, this collection forms a partition of A* into infinitely many blocks, one for each natural number n. 
- In summary, this option presents a valid partition by grouping strings by their length, and it is a standard example of a partition of A*.
- Across the options, there are two distinct correct-style partitions: one based on the first symbol (option a) and one based on string length (option d). The other options fail to cover all of A* or to be disjoint, or they rest on a false claim about infinite partitions.

Which of the following would be a valid partition of the set of all strings A^* over the alphabet A = \{a, b\}?Single choice

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