Is the following language recursively enumerable? Let $L =\{  &lt;M&gt; | $ the amount of words $w\in\Sigma^*$ that $M$ does not halt on is finite $\}$.

I would like to prove that $L\notin RE$. 

I can show that $\overline{L}\notin RE $ that is $L\notin CoRE$. 

I do this with showing a reduction: $\overline{L_{acc}} \leq \overline{L} $ and since $\overline{L_{acc}}\notin RE$  then $\overline{L}\notin RE$. 

The reduction goes as follows: 


input: $&lt;M&gt;,&lt;w&gt;$
output: $&lt;M_w &gt;$


where $&lt;M_w&gt;$ is a TM that, on an input $x$ ignores it and performs:


Run $M$ on $w$
If $M$ rejected get yourself into an inf loop 
If $M$ accepted then accepts as well.


Now if $&lt;M&gt;&lt;w&gt;\in\overline{L_{acc}}$ then $w\notin L(M)$, thus $M$ either rejects $w$ or does not end its calculation on it, but in both cases $M_w$ will not end its calculation on all inputs. Otherwise, if $&lt;M&gt;&lt;w&gt;\notin \overline{L_{acc}}$ then $w\in L(M)$ and then $M_w$ will halt (and accept) its calculation on every input $x$. 

So I have that $\overline{L_{acc}} \leq \overline{L} $ and thus $\overline{L}\notin RE$, that is $L\notin CoRE$. 

But I am at a loss on how I can prove $L\notin RE$? I tried a reduction from $\overline{L_{acc}}$, but this idea does not work because $M$ might not halt on $w$.  I also tried from $L_{\Sigma^*}$ or from $L_d$, but could not find a reduction that will prove me that $L\notin RE$.

Does someone have an idea for such a reduction? Or perhaps another method to prove $L\notin RE$?
I'll provide some hint. Try building a reduction $\overline{L_{acc}} \leq L$
changing your argument as follows:

... where $&lt;M_w&gt;$ is a TM that, on an input $x$ it performs:


Run $M$ on $w$ for $|x|$ steps
If $M$ rejected within $|x|$ steps ...
If $M$ accepted within $|x|$ steps ...
If $M$ did not halt within $|x|$ steps ...


When $M$ does not accept $w$, you want the above to diverge only on finitely many $x$s.

When $M$ accepts $w$, you want the above to diverge on infinitely many $x$s.



Alternatively, you can try to exploit the Rice-Shapiro theorem.