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Kernel Papers

- Union: \(A \cup BB := \{s : s \in A \text{ or } s \in B\}\)
- Concatenation: \(A \cdot B := \{ab : a \in A \text{ and } b \in B \}\)
- Star: \(A^* := \{ a_1 a_2 .... a_k : k \geq 0 \text{ and } a_i \in A\)

Closure means that if you take an element (or several) from a set and apply an operation, the result is another element of the original set. With that in mind, regular languages have the following properties:

**Theorem**: The set of regular languages is closed under union.

Proof by construction: Let \(A, B\) be regular languages. By the equivalence of regular languages to deterministic finite automata, there exists DFAs \(M_A, M_B\) which respective accept \(A, B\). Intuitively, we want to build a new machine that will run both \(M_A, M_B\) machines in parallel, and accept if either \(M_A\) or \(M_B\) accepts. Define a new machine \(M_{AB}\) with states given by the cross product of \(Q_A \times Q_B\), start state \(q_{AB} = (q_A, q_B)\), accept states \((F_A \times Q_A, F_B \times Q_B)\), and transition function \(\delta_{AB}((Q_A, Q_B), w) = (\delta_A( Q_A, w), \delta_B (Q_B, q))\).

**Theorem**: The set of regular languages is closed under concatenation.

Proof by construction: Let \(A, B\) be regular languages. By the equivalence of regular languages to non-deterministic finite automata, there exists NFAs \(M_A, M_B\) which recognize \(A, B\). Intuitively, we want to build a new machine that, given string \(ab\) where \(a \in A\) and \(b \in B\), will sequentially run \(M_A\) until \(a\) is consumed and then run \(M_B\) until \(b\) is consumed. However, we don’t know a prior where \(a\) ends and \(b\) begins, so we’ll use the NFA’s non-determinism to run machine \(M_B\) after each character has been consumed; if any of those machines accept the remaining string, then the new machine accepts.