proof that languages are/are not in RE (probably with mapping reduce)

Question

Given $2$ languages:

Let $u \in \Sigma^*$ (constant word).

$A_u=\{<M> \big{|}\,\, u\in L(M) \text{ and M is TM }\}$

$B_u=\{<M> \big{|}\,\, L(M)=\{u\} \text{ and M is TM }\}$

I already proved those $2$ claimes:

• $A_u \in RE$
• $A_u \notin co-RE$

I need to prove:

• $B_u \notin RE$
• $B_u \notin co-RE$

thanks!

Answer 1

I'm assuming that L(M) means the language recognized by M.
This means M only needs to be a recognizer and not a decider.
Let R be a recognizer for $B_u$
Construct the following machine,

NON-HALT = "On input $<A>$, w
$\quad\quad\quad\quad\quad\quad\quad1. <M>$ = "On input x
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$if x = u
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;$accept
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$otherwise
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;$Run A on w
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;$accept "
$\quad\quad\quad\quad\quad\quad\quad2. $ if R accepts M
$\quad\quad\quad\quad\quad\quad\quad\quad\quad$ accept
$\quad\quad\quad\quad\quad\quad\quad\;\;$ otherwise
$\quad\quad\quad\quad\quad\quad\quad\quad\quad$ reject"

In line 1, $<M>$ is constructed to be a recognizer for $L = \{u\}$ iff A loops on w
So in line 2, if R accepts M, it means A loops on w.
So, if R existed, NON-HALT would be a recognizer for the non-halting problem.
This would be a contradiction that the non-halting problem is not recognizable.
So, $B_u$ must not be recognizable.