Proving the Equivalence of Languages over \Sigma: A Mathematical Approach

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SUMMARY

The discussion centers on proving the equivalence of languages over the alphabet \Sigma, specifically addressing the condition that lim n -> inf of L^n = \Sigma* if and only if (\Sigma ∪ {\lambda}) ⊆ L. Participants explore the implications of including the empty string in language L and the limitations of using induction for proving the statement. A critical point raised is that if L does not contain the empty string, the sequence L^n may not converge as expected, leading to disjoint languages.

PREREQUISITES
  • Understanding of formal languages and automata theory
  • Familiarity with the concept of limits in mathematical analysis
  • Knowledge of induction principles in mathematical proofs
  • Basic comprehension of set theory and language operations
NEXT STEPS
  • Study the properties of formal languages and their closure under operations
  • Learn about the role of the empty string in language theory
  • Research mathematical induction techniques in formal proofs
  • Explore the concept of disjoint languages and their implications in automata theory
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This discussion is beneficial for students and researchers in theoretical computer science, particularly those focusing on formal languages, automata theory, and mathematical proofs.

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Homework Statement


If L is a language over \Sigma then lim n -> inf of Ln = \Sigma* iff (\Sigma\cup{\lambda})\subseteq L

Also Lk = {x1x2...xk | x1, x2, ...xk \in L}


Homework Equations





The Attempt at a Solution


I started by saying there is an w is an element of sigma then it is also an element of L so I might use induction but I really don't even know if I started right. Thanks.
 
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What is the definition of limit you use in the expression \lim_{n\to\infty} L^n?

If L does not contain the empty string \epsilon, then it is not true that L \subset L^2 \subset L^3 \subset \cdots, because the shortest string in L^n has length n times the length of the shortest string in L. In this case, you therefore cannot use \lim_{n\to\infty} L^n = \bigcup_{n=0}^\infty L^n as a definition. You can even construct a case (say, L = \Sigma) where the L^n are pairwise disjoint!
 

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