**Proof.**
First proof. It is clear that (5), (6), and (7) are equivalent. It is clear that (4) and (8) are equivalent as every quasi-compact open is a finite union of standard opens. The implication (7) $\Rightarrow $ (4) follows from Lemma 10.26.4. Assume (4) holds. Let $\mathfrak p, \mathfrak p'$ be distinct primes of $R$. Choose an $f \in \mathfrak p'$, $f \not\in \mathfrak p$ (if needed switch $\mathfrak p$ with $\mathfrak p'$). Then $\mathfrak p' \not\in D(f)$ and $\mathfrak p \in D(f)$. By (4) the open $D(f)$ is also closed. Hence $\mathfrak p$ and $\mathfrak p'$ are in disjoint open neighbourhoods whose union is $X$. Thus $X$ is Hausdorff and totally disconnected. Thus (4) $\Rightarrow $ (2) and (3). If (3) holds then there cannot be any specializations between points of $\mathop{\mathrm{Spec}}(R)$ and we see that (5) holds. If $X$ is Hausdorff then every point is closed, so (2) implies (6). Thus (2), (3), (4), (5), (6), (7) and (8) are equivalent. Any profinite space is Hausdorff, so (1) implies (2). If $X$ satisfies (2) and (3), then $X$ (being quasi-compact by Lemma 10.17.8) is profinite by Topology, Lemma 5.22.2.

Second proof. Besides the equivalence of (4) and (8) this follows from Lemma 10.26.2 and purely topological facts, see Topology, Lemma 5.23.8.
$\square$

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