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The set of all recursively enumerable languages is
  • a)
    closed under complementation.
  • b)
    closed under intersection.
  • c)
    a subset of the set of all recursive languages.
  • d)
    an uncountable set.
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The set of all recursively enumerable languages isa)closed under compl...
Recursive Enumerable Language are closed under Union, Intersection, Concatenation and Kleene Closure (but not Complementation). So, we can easily rule out option (A) and option (B) comes to be TRUE. Recursive Language are subset of REL, but option (C) is saying opposite, so Option (C) is FALSE. REL are countable, since set of all Turing Machines are countable. So option (D) is also FALSE. Only Option (B) is CORRECT.
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The set of all recursively enumerable languages isa)closed under compl...
Introduction:
The set of all recursively enumerable languages is a fundamental concept in theoretical computer science. It refers to a collection of languages that can be recognized by a Turing machine, a theoretical computing device. In this question, we are asked to determine the properties of this set.

Closed under complementation:
A language is said to be recursively enumerable if there exists a Turing machine that accepts all the strings in the language. Complementation refers to taking the complement of a language, which means all the strings that are not in the language.

To determine if the set of recursively enumerable languages is closed under complementation, we need to check if the complement of a recursively enumerable language is also recursively enumerable.

Explanation:
The complement of a recursively enumerable language can be recognized by a Turing machine that simulates the Turing machine for the original language and accepts all the strings that are not accepted by the original machine. This is because if a string is accepted by the original machine, it means it belongs to the language, and if it is not accepted, it means it does not belong to the language.

Closed under intersection:
The intersection of two languages refers to the collection of strings that are common to both languages.

To determine if the set of recursively enumerable languages is closed under intersection, we need to check if the intersection of two recursively enumerable languages is also recursively enumerable.

Explanation:
The intersection of two recursively enumerable languages can be recognized by a Turing machine that simulates both Turing machines for the original languages and accepts all the strings that are accepted by both machines. This is because if a string is accepted by both machines, it means it belongs to both languages, and if it is not accepted by either machine, it means it does not belong to the intersection.

A subset of the set of all recursive languages:
Recursive languages are a subset of recursively enumerable languages. A language is said to be recursive if there exists a Turing machine that accepts all the strings in the language and halts on all inputs.

Since a recursively enumerable language can be recognized by a Turing machine that may not halt on all inputs, it is more general than a recursive language.

Uncountable set:
An uncountable set refers to a set that has cardinality greater than that of the natural numbers. In other words, it cannot be put into a one-to-one correspondence with the set of natural numbers.

The set of recursively enumerable languages is countable because each Turing machine can be represented by a finite description (e.g., a string) and the set of all possible finite descriptions is countable. Therefore, the set of recursively enumerable languages is not uncountable.

Conclusion:
Based on the above explanations, the correct answer is option 'B' - the set of all recursively enumerable languages is closed under intersection.
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The set of all recursively enumerable languages isa)closed under complementation.b)closed under intersection.c)a subset of the set of all recursive languages.d)an uncountable set.Correct answer is option 'B'. Can you explain this answer?
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