# Closure properties of undecidable languages

I know that the decidable are close under: complementation, union, intersection and concatenation?

What about the undecidable languages? I think they are close under complementation, but not under concatenation, union and intersection. But I don't know how to prove it...

My prove

"Lets say L is undecidable lang. Because L is undecidable lang we know one of the following:

1. there is a word w1 from L that there is no TM that stop for this word

2. there is a word w2 in L that stopping in reject state in every TM

3. there is a word w3 Not from L that stopping in accept state in every TM

L' is the complementation lang of L

If 2, the word w2 that stopping in reject state, will stop in the accept state for every TM, but w2 not in L' so L' is Undecidable.

If 3, the word w3 that stopping in accept state, will stop in the reject state for every TM, but w2 is in L' so L' is Undecidable.

If 1, so all the TM are not stopping for w1 so L' is Undecidable???? "

• @YuvalFilmus you can look at my proof for complementation Nov 6 '15 at 19:50
Let $K$ be some undecidable language. Define languages $A_0,A_1,B_0,B_1$ as follows: $$A_0 = \{ 0 w : w \in K \}, A_1 = \{ 1 w : w \in K \}, \\ B_0 = A_0 \cup \{ 1 w : w \in \{0,1\}^* \}, B_1 = A_1 \cup \{ 1 w : w \in \{0,1\}^* \}.$$ See if you can use these languages to show that the collection of undecidable languages is not closed under concatenation, union or intersection.
As for complementation, try to use the fact that if you can decide a language $L$ then you can decide its complement $\overline{L}$ (why?).