To determine which statement about the set of all recursively enumerable (RE) languages is correct, we'll analyze each option individually. Understanding the properties of RE languages in the theory of computation will help us make an informed decision.

Understanding Recursively Enumerable Languages

Recursively Enumerable Languages (RE):

A language is recursively enumerable if there exists a Turing machine that will accept any string in the language but may not halt for strings not in the language.

RE languages are also known as Turing-recognizable languages.

They are a broader class than recursive languages, which are languages for which a Turing machine will halt on all inputs, accepting those in the language and rejecting those not in the language.

Analyzing Each Option

Option A: The set of all recursively enumerable languages is closed under complementation.

Explanation:

Closure under complementation means that if a language $ L $ is recursively enumerable, then its complement $ \overline{L} $ is also recursively enumerable.

Fact about RE Languages:

The class of recursively enumerable languages is not closed under complementation.

This is because the complement of a recursively enumerable language may not be recursively enumerable.

Counterexample:

Let $ L $ be a known non-recursive recursively enumerable language.

Then, $ \overline{L} $ is not recursively enumerable.

Conclusion:

Option A is incorrect.

Option B: The set of all recursively enumerable languages is closed under intersection.

Explanation:

Closure under intersection means that if $ L_1 $ and $ L_2 $ are recursively enumerable languages, then their intersection $ L_1 \cap L_2 $ is also recursively enumerable.

Fact about RE Languages:

The class of recursively enumerable languages is closed under intersection.

Proof Sketch:

Given Turing machines $ M_1 $ and $ M_2 $ that recognize $ L_1 $ and $ L_2 $ respectively.

Construct a new Turing machine $ M $ that simulates both $ M_1 $ and $ M_2 $ in parallel (e.g., alternate steps between them) on the input string.

$ M $ accepts the input if both $ M_1 $ and $ M_2 $ accept.

If the input is in $ L_1 \cap L_2 $, both machines will accept, and so will $ M $.

If the input is not in $ L_1 \cap L_2 $, at least one machine may loop indefinitely or reject, so $ M $ may not accept.

Conclusion:

Option B is correct.

Option C: The set of all recursively enumerable languages is a subset of the set of all recursive languages.

Explanation:

Relationship between RE and Recursive Languages:

Recursive languages are a subset of recursively enumerable languages, not the other way around.

Every recursive language is recursively enumerable, but not all recursively enumerable languages are recursive.

Fact:

There exist recursively enumerable languages that are not recursive.

These are languages for which membership is semi-decidable: acceptance is guaranteed for strings in the language, but non-membership may result in non-termination.

Conclusion:

Option C is incorrect.

Option D: The set of all recursively enumerable languages is an uncountable set.

Explanation:

Countability:

A set is countable if its elements can be put into a one-to-one correspondence with the natural numbers.

The set of all Turing machines is countable because Turing machines can be described by finite strings over a finite alphabet, and there are countably many such strings.

Since each recursively enumerable language is recognized by some Turing machine, the set of recursively enumerable languages is also countable.

Contrast with All Languages:

The set of all possible languages over an alphabet (the power set of all strings) is uncountable.

Therefore, there are languages that are not recursively enumerable.

Conclusion:

Option D is incorrect.

Final Answer

After analyzing all options, we find that only Option B is correct:

The set of all recursively enumerable languages is closed under intersection.

Answer: Option B