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Hewitt-Marczewski theorem by mars (March 22, 2010)
- Re: Hewitt-Marczewski theorem by Henno Brandsma (March 22, 2010)

From: Henno Brandsma
Date: March 22, 2010
Subject: Re: Hewitt-Marczewski theorem

In reply to "Hewitt-Marczewski theorem", posted by mars on March 22, 2010:
>Hewitt-Marczewski theorem.
>
>If we have a family of spaces (X_i)_{i in I} such that all
>X_i have a dense subset of size k, some infinite cardinal
>and |I| <= 2^k, then Prod_{i in I} X_i has a dense subset of
>cardinality k as well.
>

Lemma: let k be an infinite cardinal, let D(k) be the discrete topology on that cardinal k,
then D(k)^{2^k} has a dense subset of size k.

Proof: We see 2^k as {0,1}^k as a product of 2 point discrete spaces.
The standard base of 2^k has size k, call it B.
Now define F subset D(k)^{2^k} as follows: (so f: 2^k --> D(k))

(*) f in F iff there exists a finite non-empty subset B(f), where B(f) is
a mutually disjoint family of sets, and such that f | U is constant (with value c(U) in k)
for every U in B(f), and such that f | (2^k \ (\/B(f) ) ) == 0. (0 is in k, as k is a cardinal).

Then F has size k, as we have k many finite pairwise disjoint subfamilies of B (|B| = k, k infinite)
and for each of the members we have k choices for the value on that member.

Claim: F is dense in X:= D(k)^{2^k}.
Proof: The set of basic open sets of this product where the non-trivial factors are singletons
is a base for X. It suffices to show that F meets each member of this base.
To see this, take such a basic open set Prod_{i in 2^k} U_i, and the coordinates where U_i != D(k)
are i_1,...,i_n in 2^k, and there we have the singletons U_{i_j} = d_j in D(k).
As 2^k is a Hausdorff space we can find finitely many pairwise disjoint sets {U_1,..,U_n} from the base B for 2^k
such that d_j is in U_j, for j = 1,..n.
Then define f in X by setting f(x) = d_j for x in U_j, and 0 if it is in none of the U_j.
Then by definition f is in F, (with B(f) = {U_1,...,U_n} ) and f is in U, as i_j is in U_j, so f(i_j) = d_j,
for j = 1..n, which were the only (non-trivial) conditions for being in U.
So F meets U, as required, and F is dense and of size k.
========

Proof of the main theorem (Hewitt-Marczewski-Pondiczery):

Let D_i be a dense set in X_i of size at most kappa.
Note that D:= Prod_i D_i is dense in X:= Prod_{i in I} X_i.

There is a (automatically continuous) surjection f_i from D(k) onto D_i, for every i in I.
Also, fix an injection g: I --> 2^k.

Then F: D(k)^(2^k) ---> D defined by F(f)(i) = f_i ( f( g(i) ) ), i in I, is a well-defined
continuous surjection.
[ f is a function 2^k --> D(k) and F(f) has to be a function from I to \/_i D_i such that f(i) is in D_i
for all i. So we need to define F(f)(i) for all f in D(k)^(2^k) and i in I.
First put i into 2^k via the injection, and then f can be applied, and f(g(i)) is in D(k),
and then f_i can be applied to map into the right D_i. ]

The space D(k)^{2^k} has dense subset of size k, so D has a dense subset of size at most k,
and as D is dense in X, the same holds for X (dense in dense is dense).
QED.

============

I already proved in the other thread that if every X_i has a pair of disjoint non-empty open subsets,
we can show that if |I| > 2^k, X is not separable. So for e.g. non-trivial Hausdorff spaces
this is the best possible result.

Henno

PS: Adaptation of proof as found in standard books like Engelking, General Topology.