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A language A, even(A) is the subset of A consisting of those strings in A of even length:

even(A) = { x∈A | |x| is even}

I need to use closure properties show that if A is regular, then even(A) is also regular.

Isn't that a definition called "A language is called a regular language if some finite automaton recognizes it", could prove even(A) is regular? How to use closure properties to prove this question?

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  • $\begingroup$ It's hard to provide a hint without giving it away completely, but you may want to use a correspondence between the regular languages and the even regular languages. $\endgroup$ – reinierpost Jan 27 at 18:34
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Could you make a FA that accepts all and only even-length strings? What do you know about the intersection of two regular languages?

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You need to use one intersection operation. It is a known closure property that if two languages $A, B$ are regular, then the intersection $A \cap B$ will also be regular. In this case, $A$ is the given regular language. The other one, $B = \textrm{Even}(\Sigma^*)$, is the language of all even-length strings. The language you want to prove is the language containing all the strings in $A$ AND having even length [in $B$].

It is easy to show that $B$ is regular as well. Its regular expression is (..)*, where the period is the metacharacter that matches any character in the alphabet.

Your language is equivalent to determining that $B$ is regular, and then stating that $A \cap B$ is also regular. However, determining the regular expression of the intersection, will take doubly exponential time to do so.

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Be careful, $L_1 \subseteq L_2$ with $L_2$ regular doesn't always mean that $L_1$ is regular. For one example, pick $L_1 = \{a^n b^n \colon n \ge 1\}$ (usually the first language proven non-regular), $L_2 = \mathcal{L}(a^* b^*)$ (defined by a regular expression, thus regular).

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