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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