Proof.
Assume \mathcal{X} is locally Noetherian. Choose a scheme U and a surjective smooth morphism U \to \mathcal{X}. As \mathcal{X} is locally Noetherian we see that U is locally Noetherian. By Properties, Lemma 28.5.5 this means that |U| is a locally Noetherian topological space. Since |U| \to |\mathcal{X}| is open and surjective we conclude that |\mathcal{X}| is locally Noetherian by Topology, Lemma 5.9.3. This proves (1). If \mathcal{X} is quasi-compact and locally Noetherian, then |\mathcal{X}| is quasi-compact and locally Noetherian. Hence |\mathcal{X}| is Noetherian by Topology, Lemma 5.12.14.
\square
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