Proof.
We have (1) \Rightarrow (2) because \mathop{\mathrm{Spec}}(R) is quasi-compact, see Lemma 10.17.8. We have (2) \Rightarrow (3) because standard opens form a basis for the topology. Proof of (3) \Rightarrow (1). Let U = \bigcup _{i = 1\ldots n} D(f_ i). To show that U is retrocompact in \mathop{\mathrm{Spec}}(R) it suffices to show that U \cap V is quasi-compact for any quasi-compact open V of \mathop{\mathrm{Spec}}(R). Write V = \bigcup _{j = 1\ldots m} D(g_ j) which is possible by (2) \Rightarrow (3). Each standard open is homeomorphic to the spectrum of a ring and hence quasi-compact, see Lemmas 10.17.6 and 10.17.8. Thus U \cap V = (\bigcup _{i = 1\ldots n} D(f_ i)) \cap (\bigcup _{j = 1\ldots m} D(g_ j)) = \bigcup _{i, j} D(f_ i g_ j) is a finite union of quasi-compact opens hence quasi-compact. To finish the proof note that (4) is equivalent to (3) by Lemma 10.17.2.
\square
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