**Proof.**
For (1) write $K = k(x_ j; j \in J)$. Suppose that $A$ is a $k$-algebra, and $I \subset A$ is an ideal of square zero. Let $\varphi : K \to A/I$ be a $k$-algebra map. Let $a_ j \in A$ be an element such that $a_ j \mod I = \varphi (x_ j)$. Then it is easy to see that there is a unique $k$-algebra map $K \to A$ which maps $x_ j$ to $a_ j$ and which reduces to $\varphi $ mod $I$. Hence $k \subset K$ is formally smooth.

In case (2) we see that $k \subset K$ is a colimit of étale ring extensions. An étale ring map is formally étale (Lemma 10.146.2). Hence this case follows from Lemma 10.146.3 and the trivial observation that a formally étale ring map is formally smooth.

In case (3), write $K = \mathop{\mathrm{colim}}\nolimits K_ i$ as the filtered colimit of its finitely generated sub $k$-extensions. By Definition 10.41.1 each $K_ i$ is separable algebraic over a purely transcendental extension of $k$. Hence $K_ i/k$ is formally smooth by cases (1) and (2) and Lemma 10.136.3. Thus $H_1(L_{K_ i/k}) = 0$ by Lemma 10.152.6. Hence $H_1(L_{K/k}) = 0$ by Lemma 10.132.9. Hence $K/k$ is formally smooth by Lemma 10.152.6 again.
$\square$

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