Is the following fuction computable? I'm trying to show that $K_1 \le_1 K$ where $K$ is the diagonal halting set $\{x : \varphi_x(x) \downarrow\}$ and $K_1=\{x: \exists y \,\, \varphi_x(y) \downarrow\}$, then I defined the function
$$\psi(x, y)= \begin{cases} 0 \, \, if \, \, x \in K_1 \\ \uparrow \, if \, x \notin K_1\end{cases}$$
By the Parameter Theorem there is a one to one computable function $f$ such that $\varphi_{f(x)}(y) = \psi(x, y)$ and it's easy to show that $x \in K_1 \leftrightarrow f(x) \in K$. All above works if and only if $\psi$ is partial computable, I think that function is partial computable because $K_1$ is r.e. Is this true? I mean, Can I always define a function computable by 
$$\psi_A(x, y)= \begin{cases} 0 \, \, if \, \, x \in A \\ \uparrow \, if \, x \notin A\end{cases}$$
for each r.e. set $A$?
You're using the terminology of recursion theory. Let me rephrase the argument in terms of computer programs.


$K$ is the set of all programs $x$ which halt when run on $x$ as input.
$K_1$ is the set of all programs $x$ which halt on some input.


Given a program $x$, consider the following program $f(x)$:


For $n=1,2,3,\ldots$: run program $x$ on inputs $1,\ldots,n$ for $n$ steps, and halt if $x$ halts of any of them halt.


The program $f(x)$ halts iff $x \in K_1$. In particular, $x \in K_1$ iff $f(x) \in K$. Since $f$ is computable, this gives a reduction from $K_1$ to $K$.