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18.727, Topics in Algebraic Geometry (rigid analytic geometry)\\
Kiran S. Kedlaya, fall 2004 \\
More corrections on Berkovich spaces
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I'll get this right eventually, I promise.
(Again, many thanks to Brian Conrad for clarifications; remaining
errors are of course not his fault!)

The definition of a quasi-separated rigid space is not what I said in
class. It is (just as for schemes): 
$X$ is quasi-separated if $X \to \Max K$ is quasi-separated,
and a morphism $X \to Y$ is quasi-separated if the diagonal
$\Delta_{X/Y}: X \to X \times_Y X$ is a quasi-compact morphism
(not a locally closed immersion). Oh, and a morphism is quasi-compact
if the inverse image of every quasi-compact open is quasi-compact.

Re the notion of a net: the restriction of a quasi-net to an open subset
means take elements of the quasi-net contained in the open, rather than
intersecting all of them with the open. So my example with the unit squares
is not a net; you can fix it by adding in the closed edges and
the corners.

More serious functor: the rigid-to-Berkovich functor is backwards from what
I've been saying! Mea culpa. It turns Hausdorff strictly analytic spaces
into quasiseparated rigid spaces, and in \emph{that} direction is fully
faithful.
It is true, though, that it induces an equivalence between
paracompact strictly analyic spaces and quasi-separated rigid spaces
with an admissible affinoid covering of finite type. (``Finite type'' means
that any element of the covering meets only finitely many others.
E.g., the covering of the open unit disc by closed discs around the
origin is not of finite type, but there is an admissible covering 
of finite type by closed annuli.)
The point here is that quasi-separatedness is needed in order to glue
meaningfully (otherwise the attaching maps on the Berkovich affinoids
are not uniquely determined by their rigid counterparts, so the cocycle
condition falls apart) and the finite type condition is needed in order
to build a quasi-net.
(Maybe with a more sensible definition of Berkovich's spaces you could
get past the finite type issue?)
Anyway, see Theorem 1.6.1 of Berkovich's IHES paper.

Upshot: the Berkovich category neither contains nor is contained in the
category of rigid spaces, but they share the rigid spaces which are ``not
too pathological'', which the spaces you typically encounter in practice
will be.

And one fun thing you might want to try (suggested by Berkovich in his
ICM talk): compute the Gelfand spectrum of $\ZZ$. (Remember, this means
you allow the normal triangle inequality, and you don't impose any upper
bound on the seminorms.)

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