abc please show work dont copy existing answer i dont think its correct

(a) Consider the following faulty "proof" that any language L is decidable, which has multiple fatal errors. 1 Identify two of these errors by the lines in which they appear, and explain what is wrong with them. (For a few extra points, identify and explain more than two errors.) i. Because is countable, so is L. ii. Therefore, there exists an ordered (finite or infinite) list of L 's elements, as L= {x1,x2,x3,} iii. We use this to define a Turing machine D, defined as: " D(x) : A. Check whether x=x1,x=x2,x=x3, and so on. B. If any of these checks hold, then accept; otherwise, reject." iv. It is clear by inspection that D decides L, because D accepts every string xL and rejects every string x/L. v. Therefore, L is decidable. (b) Prove that any infinite language L (even a decidable one) contains an undecidable subset. For example, the (trivially decidable) language contains an undecidable language (i.e., subset). Hint: Can you view a subset of L as an infinite binary sequence? (c) Let L1,L2,L3, be countably many languages over possibly different alphabets, i.e., Lii where each i is some finite alphabet that may vary with i. Prove that L=L1L2L3 is countable. This should hold even if the "combined alphabet" =123 is infinite. 2 For example, we might have 1={0,1},2={0,1,2}, and i={0,1,2,,i} in general. Then the combined alphabet ={0,1,2,}=N is the infinite set of natural numbers, so L is a set of strings whose "characters" may be any natural numbers