I'm trying to prove that $\mathrm{Inf}=\{\langle M\rangle\mid L(M)= \infty\} \notin R$ where $R$ is the group of all decidable languages.
I'm trying to make a reduction from The acceptance problem $A_{TM}=\{\langle M,w\rangle\mid $ $M$ accepts $ w\}$ which is not in R.I'm new to these reductions, so I'm not sure if I'm doing it well. and I This is my reduction, I wonder if it is correct.
$f(\langle M\rangle,w)=\langle M'\rangle$.
Given $\langle M\rangle,w$ I'll create a new Turing machine $M'$. For an input $x$: 1.if $x \ne w$, $L(M')=\{1\}$, otherwise if $x=w$ I'll emulate eunning of $w$ on $M$ if $M$ accepts, $L(M')=\Sigma^*$, otherwise $L(M')=\{1\}$, in this way $L(M')=\infty$ iff $M$ accepts $w$.
Is this correct?
\langle
and\rangle
instead of<
and>
in this context. The latter are relation symbols, and not only look different, but also produce the wrong spacing. I fixed it for you (along with some other minor TeX fixes). $\endgroup$