Proof.
It is clear that (2) implies (3) and that (3) implies (1). Assume $K$ is in $D_ c^-(X_{\acute{e}tale}, \Lambda )$. Say $H^ i(K) = 0$ for $i > b$. By induction on $a$ we will construct a complex $\mathcal{F}^ a \to \ldots \to \mathcal{F}^ b$ such that each $\mathcal{F}^ i$ is a finite direct sum of $j_{U!}\underline{\Lambda }$ with $U \in \mathop{\mathrm{Ob}}\nolimits (X_{\acute{e}tale})$ affine and a map $\mathcal{F}^\bullet \to K$ which induces an isomorphism $H^ i(\mathcal{F}^\bullet ) \to H^ i(K)$ for $i > a$ and a surjection $H^ a(\mathcal{F}^\bullet ) \to H^ a(K)$. For $a = b$ this can be done by Lemma 59.76.4. Given such a datum choose a distinguished triangle
\[ \mathcal{F}^\bullet \to K \to L \to \mathcal{F}^\bullet [1] \]
Then we see that $H^ i(L) = 0$ for $i \geq a$. Choose $\mathcal{F}^{a - 1}[-a +1] \to L$ as in Lemma 59.76.4. The composition $\mathcal{F}^{a - 1}[-a +1] \to L \to \mathcal{F}^\bullet $ corresponds to a map $\mathcal{F}^{a - 1} \to \mathcal{F}^ a$ such that the composition with $\mathcal{F}^ a \to \mathcal{F}^{a + 1}$ is zero. By TR4 we obtain a map
\[ (\mathcal{F}^{a - 1} \to \ldots \to \mathcal{F}^ b) \to K \]
in $D(X_{\acute{e}tale}, \Lambda )$. This finishes the induction step and the proof of the lemma.
$\square$
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