Re: Desc set theory - Projections of product spaces

persres@xxxxxxxxxxxxxx wrote:

I am reading Descriptive set theory by moschovakis.
Given a polish space X. Take the product space X * w,
with the natural product topology and w with the
discrete topology. Given (a subset of the product
space) P \subset X * w, its projection on X is
defined as
= { x \in X : (\exists n \in w) (x,n) \in P }.

Consider the class of closed sets PI_1,0 (in the
borel hierarchy). The set of all projections of
sets in PI_1,0 is supposed to be the unions of
sets in PI_1,0.

(That is the class Sigma_2,0 (also called F-sigma sets)
can be formed by taking the projections of sets of
PI_1,0.According to Moschovakis)

So, it looks like projections of sets in X*w, somehow
work out just like performing union operations on
sets in X. If anybody can understand what I am
talking about please explain.

Your second to last sentence isn't correct, since
the projection of an open set doesn't have to be
an open set (it will be a G_delta set, however),
and recall that any union of open sets is open
(so the projection of an open set doesn't have
to be a union of open sets).

Let E be a closed set in X * w. Then we can write
E = E_1 union E_2 union E_3 union ... (countable union),
where each E_k is the k'th horizontal slice of E
in X * w (k being an element of w). Although projections
don't have to map closed sets to closed sets, it will
be the case that the projection of each of the E_k
sets is closed in X. You can prove this directly
using the fact that a set is closed if and only
if it contains all its limit points (or using some
other characterization of closed sets in a metric
space) or by applying the tube lemma (see [1]).

[1] http://en.wikipedia.org/wiki/Tube_lemma

Since the function-image of a union (countable or not,
in fact) is the union of the function-images, it follows
that the projection of E is the union of the projections
of the E_k sets, hence a countable union of closed sets
(i.e. an F_sigma set).

To show that every F_sigma set in X is the projection
of some closed set in X * w, let F be an F_sigma set
in X. Then we can write
F = C_1 union C_2 union C_3 union ... (countable union),
where each C_k is a closed set in X. Now just form the
desired closed set E in X * w by putting a copy of C_k
at the k'th horizontal level of X * w. That is, define
C to be
(C_1 x {1}) union (C_2 x {2}) union (C_3 x {3}) union ...
Finally, show this is actually a closed set. Unless
I'm missing something, I think you can simply observe
that the complement of C is the union of the sets
U_k x {k}, where U_k = X - C_k is open in X and {k}
is open in w, and thus a union of open sets in X * w.

Incidentally, the same result holds for a projection
from R^2 to R, namely the collection of all projections
of closed sets in R^2 is the collection of all F_sigma
sets in R. For one direction, use the fact that a projection
is a continuous function and the fact that R^2 is a
countable union of compact sets. [Any closed set in R^2
is thus a countable union of compact sets, the continuous
image of a compact set is compact (hence closed), and the
function-image of a union is the union of the function
images.] For the other direction, just use the same
method I used for X * w and note that a closed subset
of R x w will also be a closed subset of R x R = R^2.

Dave L. Renfro
.

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