Prove the languages |L<M>| = 2 and |L<M>| $\not=$ 2 to be non-Turing recognizable or non-recursively enumerable

Question

I am trying to prove the non-recursively enumerable property of two languages.

$L_2 = \{\langle M \rangle: |L\langle M \rangle| = 2\}$ and
$L_{\not=2} = \{\langle M \rangle: |L\langle M \rangle| \not= 2\}$.

My idea is to use the following two properties:

1. If A $\leq_m$ B and A is not Turing recognizable then B is not Turing-recognizable.

2. A $\leq_m$ B if and only if $A^c \leq_m B^c$.

For language $L_{2}$, If I can show that $A_{TM}$ is mapping reducible to $L_{\not=2}$, according to property 2 we could get that $\overline{A_{TM}}$(the complement of $A_{TM}$) is mapping reducible to $L_{2}$, and for the fact that $\overline{A_{TM}}$ is not Turing recognizable, we could get that $L_{=2}$ is not Turing recognizable.

For language $L_{\not=2}$, If I can show that $\overline{A_{TM}}$ is also mapping reducible to $L_{\not=2}$, and for the fact that $\overline{A_{TM}}$ is not Turing recognizable, we could get that $L_{\not=2}$ is not Turing recognizable.

However, I don't know how to prove the mapping reducible relationships between them. Or my direction could be totally wrong.

Could somebody give me the hints or possible solutions?

Answer 1

If I can show that $A_{TM}$ is mapping reducible to $L_{=2}$, ...

That is a nice idea. Given $\langle M, w \rangle$, can you construct a TM $M_1$ that accepts only one word, the word $w$ if and only if $M$ accepts $w$?

Then, can you change $M_1$ so that it accepts another word that is different from $w$?

You can also show that $A_{TM}$ is mapping reducible to $L_{\not=2}$ by creating a TM $M_3$ which accepts 2 words if $M$ does not accepts $w$ and 3 words otherwise.