$A_{TM} = \{<M,w> | $ M  is a TM and M accepts w $\}$$H_{TM} = \{<M,w> | $ M  is a TM and M halts on w $\}$I thought that $L = H_{TM} \cap \overline{A_{TM}} \in R$But I saw the proof for why $L \in RE \setminus R$But I don't understand why this is wrong:$H_{TM} \in RE$ and $\overline{A_{TM}} \in Co-RE $We know that $RE \cap Co-RE = R$So why is this wrong $L = H_{TM} \cap \overline{A_{TM}} \in RE \cap Co-RE = R$ ?
Why is my proof wrong for $L = H_{TM} \cap \overline{A_{TM}} $
turing machines;undecidability
$RE$ and $co-RE$ are sets of sets of strings. $H_{TM}$ and $\overline{A_{TM}}$ are sets of strings. You correctly have that $H_{TM}\in RE$ and $\overline{A_{TM}}\in co-RE$. However, you cannot then conclude that$$H_{TM}\cap\overline{A_{TM}}\in RE\cap co-RE$$For a simple example, let $$A=\{\{1\},\{2\}, \{1, 3\}, \{2, 3\}\}\text{ and }B=\{\{1\},\{3\}, \{1, 2\}, \{2, 3\}\}$$So $A$ will serve as $RE$ and $B$ will serve as $co-RE$. Then we have $A\cap B=\{\{1\},\{2,3\}\}$. Now let $x=\{1,3\}\in A$ and $y=\{2,3\}\in B$ (here $x$ serves as $H_{TM}$ and $y$ serves as $\overline{A_{TM}}$). Then $x\cap y=\{3\}\notin A\cap B$.  