I have the following problem. Prove that $L_1=\{\langle M\rangle| \exists w \in \Sigma^*: M(w)\downarrow\}$ is undecidable (where $M(w) \downarrow$ means M halts on input w).
I have come up with this reduction ($\langle M\rangle w\longmapsto \langle M'\rangle$) wich i know is wrong but i can't figure out why.
M' on input y:
deletes y from the tape
writes w on the tape
executes M on w
i tried to show the correctness as follow: $$\langle M\rangle w \in L_{halt} \Rightarrow M(w) \downarrow \Rightarrow M'(y) \downarrow \Rightarrow \langle M'\rangle\in L_1$$
$$\langle M\rangle w \notin L_{halt} \Rightarrow M(w) \uparrow \Rightarrow M'(y) \uparrow \Rightarrow \langle M'\rangle \notin L_1$$
Now, I know that the proof i wrote is wrong since it's the proof used to demostrate that $L_{\Sigma^*}=\{\langle M\rangle|\forall w \in \Sigma^*, M(w) \downarrow\}$ is undecidable,but i don't know where i am wrong.
This is correct. Your reduction gives machine that behaves equally on any input, and for such machines membership in $L_1$ and $L_{\Sigma^*}$ is equal.
More precisely, your reduction maps $L_{halt}$ to machines that halts on every input (let it be $L'$) and $\overline{L_{halt}}$ to machines that doesn't halt at any input (let it be $L''$).
As $L' \subseteq L_1 \cap L_{\Sigma^*}$ and $L'' \subseteq \overline{L_1} \cap \overline{L_{\Sigma^*}}$, this is reduction both to $L_1$ and $L_{\Sigma^*}$.