Show that the class of decidable languages is closed under the operations of: Complementation, Concatenation, and Intersection

Question

I understand that the problem is asking for a proof for the specified Turing machine. My issue derives from not being able to understand what type of structure this problem is proposing, as indicated below:

---

We say that a class C of languages is closed under an operation ⊕(L₁,L₂,…,L_k) if for any L₁,L₂,…,L_k∈C, the language ⊕(L₁,L₂,…,L_k) is in C.

---

Can someone please explain to me what this means?

Answer 1

Suppose that C is the class of all objects that have some property (e.g. the property of being a decidable language). Also suppose that you have an operation · which takes two of these objects and returns another object (the number two is just an example). We say that C is closed under · when: given two objects x and y in C, applying the operator to them gives an object which is also in C: x · y \in C. (This is just the definition given in your question).

For example, the set B of objects that have the property "it is a boolean value" { true, false } is closed under complementation. For any value b from B, the complementation ! b has the property "it is a boolean value" so it is also in B. So the set B of all boolean values is closed under complementation. In this case, a proof could be a table listing every combination of inputs to the operation and verify the output in each case.

Another example: the set S of all objects that have the property "it is a string of one or more letters from A-Z" is closed under concatenation: given any two strings x and y in S, the concatenation x + y is also a string, so it must be contained in the set S of objects that have the property "it is a string of one or more letters A-Z". In this case, the proof might be an argument to show that the output satisfies the property based on the definition of the concatenation operation on strings.

The question in the title asks you to show (prove) that the class of decidable languages is closed under the complementation operation, and under the concatenation operation, and under the intersection operation. That is: to show that the concatenation of two decidable languages is a decidable language, that the intersection of two decidable language is a decidable languages, etc.

Answer 2

A set is closed under an operation on the elements of that set if and only if the result of applying the operation to elements of the sets is guaranteed to produce another element of the set. So, the set of natural numbers (nonnegative integers) is closed under addition, since adding any two natural numbers produces another natural number. The set of natural numbers is not closed under subtraction since some differences (e.g. 10 - 20) are not natural numbers.

To show the three things you're asked to do, you can reason as follows:

1. Complementation: because decidability means you can answer yes or no to the question of membership in finite time for all candidate strings, the set of decidable languages must be closed under complementation since you can just swap the answers "yes" and "no" to get a decider for the complement of any decidable language.
2. Concatenation: if you have deciders for two decidable languages then you can get a decider for the concatenation of those languages by nondeterministically guessing in the first decider when you have read the part of the string from the first language, and then checking whether the remainder is a string in the second language. Nondeterministic TMs are no more powerful than regular TMs so no issue here.
3. Intersection: similar to concatenation, except you just run both deciders on the input string and answer yes if both answered yes, no otherwise.

We have described constructions which show that applying these operations to decidable languages result in decidable languages. That means the set of decidable languages is closed under these operations.