The nonhalting set is productive and the halting set is creative

The set $K$ is computably enumerable: dovetail over all indices $e$, decode each one using [L3], and simulate the decoded machine on input $e$. Enumerate $e$ exactly when that simulation halts. By [L2], this lists precisely the numbers in $K$.

For each $e$, build a coded machine $Q_e$ that on input $x$ simulates the enumeration of $W_e$ and halts exactly when the number $x$ appears in that enumeration. Let $p(e):=<mi\; mathvariant="normal">⌜</mi>Q_e<mi\; mathvariant="normal">⌝</mi>$. The effective machine coding makes $p$ total computable; productivity in [L1] does not require this particular witness to be injective.

Assume $W_e\subseteq \mathbb{N}\setminus K$. If $p(e)\in W_e$, then by the definition of $Q_e$ the computation $Q_e(p(e))$ halts, so $p(e)\in K$, which contradicts $W_e\subseteq \mathbb{N}\setminus K$. Therefore $p(e)\mathrm{\notin }{W}_e$. But then the simulation inside $Q_e$ never sees $p(e)$ enter $W_e$, so $Q_e(p(e))$ diverges and hence $p(e)\mathrm{\notin }K$. Thus $p(e)\in (\mathbb{N}\setminus K)\setminus W_e$.

Step 2.1 shows that $p$ is a productive function for $\mathbb{N}\setminus K$. By [L1], the complement of $K$ is productive. Together with step 1.1, [L1] gives that $K$ is creative.