Two definitions of quasi-separated morphisms which may not be equivalent? 
Definition. A topological space $X$ is said to be quasi-separated if the intersection of two quasi-compact open subsets $U,V\subset X$ is quasi-compact.

In Definition 11.14 from these notes from a course in Algebraic Geometry taught at Bonn some years ago, they define a "quasi-separated morphism of schemes" as one such that the inverse image of any quasi-separated open subset is quasi-separated. On the other hand, on p. 207 of Vakil's Rising Sea, a "quasi-separated morphism of schemes" is defined as one such that the pre-image of an affine open subset is quasi-separated.
My question is: are these definitions equivalent?
Clearly, Bonn's implies Vakil's (as every affine scheme is quasi-separated). But I'm not sure about the converse. Specifically, one can consider the canonical map from the infinite dimensional space with doubled origin to the affine line with doubled origin (induced by $k[x_1]\to k[x_1,x_2,\dots]$). The target of this scheme morphism is a quasi-separated scheme for it is Noetherian. But the source is not quasi-separated (see Example 4 here. So this morphism wouldn't be Bonn-quasi-separated). However, I think this morphism is Vakil-quasi-separated?
 A: No, these are not equivalent. Your example is not Bonn-qs, but it is Vakil-qs: any affine open in the line with doubled origin is of the form $\operatorname{Spec} k[x_1]_{p}$ for $p\in k[x_1]$ a nonzero polynomial. The preimage of this is just $\operatorname{Spec} k[x_1,\cdots,]_p$ which is affine and therefore quasi-separated.
In fact, the Vakil definition matches the standard definition of a quasi-separated morphism as one with the diagonal quasi-compact. By 01KO, a morphism $f:X\to S$ is quasi-separated iff for every pair of affine opens $U,V\subset X$ mapping in to a common affine open $S_0\subset S$, their intersection is a finite union of affine opens of $X$. This is equivalent to Vakil's definition: an open subset of $f^{-1}(S_0)$ is quasi-compact iff it can be written as a finite union of affine opens of $X$.
What's happened here is that the Bonn definition is slightly too strong: your example is quasi-separated in the usual sense because the base change along the open cover $\Bbb A^1\sqcup\Bbb A^1\to S$ gives two copies of $\Bbb A^\infty \to \Bbb A^1$ which is clearly quasi-separated, being affine. As this covering is fpqc and quasi-separated morphisms satisfy fpqc descent, your original map is quasi-separated too.
A: I think the idea for a counterexample I sketched cannot be so and that, in fact, Vakil-quasi-separatedness implies Bonn-quasi-separatedness. I will explain the argument for this implication. First note the following fact:
$$
\tag{*}
\text{If $S$ is a scheme, then $S$ is a quasi-separated topological space}\\\text{ if and only if $S\to\operatorname{Spec}\mathbb{Z}$ is a Vakil-quasi-separated morphism.}
$$
The implication to the left is obvious. The implication to the right follows from the fact that open subsets of a quasi-separated topological space are also quasi-separated.
Now let $f:X\to Y$ be a Vakil-quasi-separated morphism of schemes and let $U\subset Y$ be an open subset that is a quasi-separated topological space. Using the facts:

*

*$f^{-1}(U)\to U$ is Vakil-quasi-separated,

*$U\to\operatorname{Spec}\mathbb{Z}$ is Vakil-quasi-separated, by $(*)$,

*Vakil-quasi-separatedness is stable under composition (see 01KU; note that in the Stacks Project "quasi-separated morphism" means "Vakil-quasi-separated morphism", by 01KO),

we deduce that the composite $f^{-1}(U)\to U\to\operatorname{Spec}\mathbb{Z}$ is Vakil-quasi-separated, and by $(*)$ we conclude that $f^{-1}(U)$ is a quasi-separated topological space and that therefore $f$ is Bonn-quasi-separated.
