Borel sets of countable collection of separable spaces?

[sto]

Let
R1, R2, R3, ... be a countable collection of separable metric spaces.
R := R1 x R2 x .... be their product, equipped with the product metric.
p_k:R -> R_k be the projection p_k(x) = x_k for any x in R.

Define the sigma field S (resp S_k) of Borel sets in R (resp R_k) to be
the sigma field generated by the open subsets of R (resp R_k).

Define the product sigma field S1 x S2 x ... to be the sigma field
generated by the family of cylinder sets { p_k^-1(E_k) : k in 1,2,3,...,
E_k subset R_k, E_k open}.

Does S = S1 x S2 x ....?

I think it does. If A is any open subset of R, then for any
x=(x_1,x_2,...) in A, there exists some e > 0 such that the open ball
B(x,e) is contained in A. Because each R_k is separable we can, for
each k, choose some q_k in R_k from a countable dense subset of R_k such
that the distance D( x_k, q_k ) < e / 2. Letting q = (q_1,q_2, ...), it
follows that x in B(q,e/2) subset B(x,e) subset A. Since there are only
countably many such balls, we conclude that every open subset of R is a
countable union of open balls. Furthermore, because of the product
metric on R, each open ball in R is a (countable) cartesian product of
open balls in R_k. Consequently, each open ball B(q,e) can be written
as the countable intersection of the sets p_1^-1( B_1(q_1,e) ),
p_2^-1(B_2(q_2,e)),... Because each of the sets p_k^-1(B_k(q_k,e)) is
contained in the family of cylinder sets { p_k^-1(E_k) : k in 1,2,3,...,
E_k subset R_k, E_k open}, and because this family generates S1 x S2
x..., we conclude that the product sigma field S1 x S2 X ... contains
all the open sets and therefore contains S.

On the other hand, because each set in the family { p_k^-1(E_k) : k in
1,2,3,..., E_k subset R_k, E_k open} is open in R, and S contains all
the open subsets of R, it follows that S contains S1 x S2 x ...

Is this proof correct? Or at least is the statement S = S1 x S2 x ...
correct?

Thanks,
-sto

[Rupert]

On Nov 5, 5:51 pm, sto <s...@address.invalid> wrote:
> Let
>   R1, R2, R3, ... be a countable collection of separable metric spaces.
>   R := R1 x R2 x .... be their product, equipped with the product metric.

Would you be able to clarify what's the product metric for a product
of an infinite family of metric spaces?

[sto]

If D_k:R_k -> [0,infty) is the metric on R_k, and N is the set of
natural numbers, then I would think the product metric D:R->[0,infty) is
given by D(x,y) = sup { D_k(x_k , y_k ) : k in N}

[William Elliot]

That's the uniform sup metric doesn't generate the usual (Tyconov)
product topology. For that use, D(x,y) = sum(j=1,oo) dj(xj,yj)/2^j.
where dj(xj,yj) is the metric min{ 1, Dj(xj,yj) }.

[Rupert]

On Nov 5, 5:51 pm, sto <s...@address.invalid> wrote:

> Let
>   R1, R2, R3, ... be a countable collection of separable metric spaces.
>   R := R1 x R2 x .... be their product, equipped with the product metric.
>   p_k:R -> R_k be the projection p_k(x) = x_k for any x in R.
>
> Define the sigma field S (resp S_k) of Borel sets in R (resp R_k) to be
> the sigma field generated by the open subsets of R (resp R_k).
>
> Define the product sigma field S1 x S2 x ... to be the sigma field
> generated by the family of cylinder sets { p_k^-1(E_k) : k in 1,2,3,...,
> E_k subset R_k, E_k open}.
>
> Does S = S1 x S2 x ....?
>
> I think it does.  If A is any open subset of R, then for any
> x=(x_1,x_2,...) in A, there exists some e > 0 such that the open ball
> B(x,e) is contained in A.  Because each R_k is separable we can, for
> each k, choose some q_k in R_k from a countable dense subset of R_k such
> that the distance D( x_k, q_k ) < e / 2.  Letting q = (q_1,q_2, ...), it
> follows that x in B(q,e/2) subset B(x,e) subset A.  Since there are only
> countably many such balls,

Don't think this is correct. The product of a countable family of
countable sets isn't necessarily countable.

[Rupert]

On Nov 5, 5:51 pm, sto <s...@address.invalid> wrote:

I think you can probably fix the proof by showing that you only need
to worry about finitely many values of k.

[Herman Rubin]

Quite useless. Supposing d_k, the metric on the k-th space, is
bounded by 1, something like sup (d_k)/2^k or sum d_k/2^k) would
work. Convergence in the product topology is by convergence in
each coordinate separately.

This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558