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Chapter 6: Problem 17

If \(A(\neq \emptyset)\) is a language and \(A^{2}=A\), prove that \(A=A^{*}\).

Short Answer

Expert verified
If \(A^2 = A\), for every language A, we can prove that \(A = A^*\) by demonstrating that A is a subset of \(A^*\) and \(A^*\) is a subset of A.

Step by step solution

01

Define the Problem

Establish the definition of the Kleene star, and the given condition that \(A^2 = A\). The Kleene closure of a language A, denoted \(A^*\), is the smallest superset of A that contains the empty string and is closed under string concatenation. The given condition means for all strings w and v that are in A, their concatenation (wv) is also in A.
02

Proving \(A \subseteq A^*\)

Prove that A is a subset of \(A^*\). By definition, any language A is a subset of its closure \(A^*\) because \(A^*\) includes all strings from A along with their finite concatenations and the empty string.
03

Proving \(A^* \subseteq A\)

Now, we must prove that \(A^*\) is a subset of A. By definition, \(A^*\) is the smallest superset of A that includes any concatenation of any number of strings from A. Given that \(A^2 = A\), it follows that \(A^n = A\) for any natural number \(n\), since the right hand side of the equation \(A^2 = A\) can be replaced with A repeatedly. This implies that any concatenation of any number of strings in A is also in A, implying \(A^* \subseteq A\).
04

Final Proof

Since we have proven that \(A \subseteq A^*\) and \(A^* \subseteq A\), by definition of equality of sets, we can conclude that \(A = A^*\).

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Most popular questions from this chapter

For \(\mathbf{\Sigma}=\\{0,1\\}\) determine whether the string 00010 is in each of the following languages (taken from \(\mathbf{\Sigma}^{*}\) ). a) \(\\{0,1\\}^{*}\) b) \(\\{000,101\\}\\{10,11\\}\) c) \(\\{00\\}\\{0\\}^{*}\\{10\\}\) d) \(\\{000\\}^{*}\\{1\\}^{*}\\{0\\}\) e) \(\\{00\\}^{*}\\{10\\}^{*}\) f) \(\\{0\\}^{*}\\{1\\}^{*}\\{0\\}^{*}\)

For \(\mathbf{\Sigma}=\\{x, y, z\\}\), let \(A, B \subseteq \mathbf{\Sigma}^{*}\) be given by \(A=\\{x y\\}\) and \(B=\\{\lambda, x\\} .\) Determine (a) \(A B\); (b) \(B A ;\) (c) \(B^{3} ;\) (d) \(B^{+} ;\)(e) \(A^{*}\).

Let \(\mathbf{\Sigma}\) be the alphabet \(\\{0,1\\}\), and let \(A \subseteq \mathbf{\Sigma}^{*}\) be the language defined recursively as follows: 1) The symbols 0,1 are both in \(A\)-this is the base for our definition; and, 2) For any word \(x\) in \(A\), the word \(0 x 1\) is also in \(A\)-this constitutes the recursive process. a) Find four different words-two of length 3 and two of length 5 -in \(A\). b) Use the given recursive definition to show that 0001111 is in \(A\). c) Explain why 00001111 is not in \(A\).

Let \(A=\\{10,11\\}, B=\\{00,1\\}\) be languages for the alphabet \(\mathbf{\Sigma}=\\{0,1\\}\). Determine each of the following: (a) \(A B\); (b) \(B A\); (c) \(A^{3}\); (d) \(B^{2}\).

Given a nonempty language \(A \subseteq \Sigma^{*}\), prove that if \(A^{2}=A\), then \(\lambda \in A\).
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