Operations under which the class of undecidable languages isn't closed
Yes, let $H$ be binary encoding of the halting problem and $A=0H\cup 1\{0,1\}^{\ast}\cup\{\epsilon\}$, $B=1H\cup0\{0,1\}^{\ast}\cup \{\epsilon\}$, then $AB=\{0,1\}^{\ast}$ (why?)