2021-08-27
Compacta are compact Hausdorff spaces. In the first weeks of my MAT5865 subject, prof. Chico suggested we should attempt to prove that for an indexed family \((X_i, i\in I)\) , any function \(f\) \[ f:\prod_{i\in I} X_i \to \mathbb R \] factors through (and therefore depends on) only a countable subproduct. That is, there is a \(J\subseteq I\) with \(|J|\leq\aleph_0\) such that the following composite equals \(f\) . \[ \prod_{i\in I} X_i \xrightarrow\pi \prod_{j\in J} X_j \xrightarrow{f_J} \mathbb R \]
I asked for help and he suggested applying Stone-Weierstrass, and this is me filling in the gaps etc.
Let \(\Pi_J\) , \(J\subseteq I\) , denote the set \(\prod_{j\in J}X_j\) . This will make notation a bit lighter in the future. We now define the minimal set of coordinates on which a function in \(\Pi_I\to\mathbb R\) depends on:
If \(\mathfrak J\) is a collection of index sets \(J\) such that \(f\) factors through the canonical projection \(\pi_J:\Pi_I\to\Pi_J\) , then \(\bigcap\mathfrak J\) is another such index set. And this we shall prove below.
It is surprisingly simple: take the cospan given by \(\Pi_I\)
shooting the projection arrows at \(\Pi_J\) for \(J\in\mathfrak J\) ; the associated pushout
is isomorphic to the quotient of the disjoint union of \(\Pi_J\) by the equivalence relation of
coming from the same object. Well, two tuples are in the same class
exactly when they are the projection of the same point, and therefore
they must equal on the coordinates they share.
Therefore \(\Pi_{\bigcap\mathfrak J}\) is the pushout of that cospan! But \(X\) with the associated \(f_J\) factorizations of \(f\) make a comutative diagram and by universal property of the pushout, there is a unique \(f_{\mathfrak J}\) making that diagram comute. But one readily sees that function factorizes \(f\) through \(\Pi_{\bigcap\mathfrak J}\) as was desired.
It follows there is a minimal set of indices \(\chi_f\) such that \(f\) factors through \(\Pi_{\chi_f}\) via the projections.
Consider the set \(\Xi\) of
functions \(g:\Pi_I\to\mathbb R\)
such that \(\chi_g\) is finite. If we
add two functions or multiply them, it is obvious that they will still
factor through a finite product as well: \(\chi_{g + h}\subseteq\chi_g\cup\chi_h\) and
likewise for subtraction and multiplication. Therefore, \(\Xi\) is a subalgebra of the algebra of
real functions over \(\Pi_I\) .
Moreover, the constant functions factor through the empty (and therefore finite) product! So \(\lambda p. x\) is in \(\Xi\) for every \(x\in\mathbb R\) . We therefore have, through the Stone-Weierstrass theorem, that \(\Xi\) is dense on the space \(\Pi_I\to X\) of continuous functions (with the topology of uniform convergence) if and only if “ \(\Xi\) vanishes nowhere and separates points”
Vanishing nowhere means that there are no points for which all functions in \(\Xi\) are zero. Since we have every constant function this is trivial, the “constantly one” function never vanishes, and thus neither does \(\Xi\) . As for separating points, it suffices to show that given two distinct points there is a function with different values on them.
Well, if the points differ there is a coordinate in which they differ, say, \(k\) . Then \(\pi_k(x) \not = \pi_k(y)\) ; since the space \(X_k\) is compact and Hausdorff, it is regular and thus separates points, there is a function \(s:X_k\to\mathbb R\) taking \(\pi_k(x)\) to a different value from that of \(\pi_k(y)\) ’s. This function can be lifted into a function on \(\prod_iX_i\) that depends only on the variable coordinate \(k\) . And, of course, it separates \(x\) from \(y\) .
So, this proves that \(\Xi\) is a dense subalgebra in the topology of uniform convergence of those continuous functions.
Take a uniformly converging sequence of functions \(f_n:\prod_iX_i\to\mathbb R\) , it is clear to see that \[ \chi_{\lim_n f_n} \subseteq \bigcup_n\chi_{f_n} \] Why? Well, if \(f=\lim_n f_n\) then the value of \(f\) at a given point depend at most on the values of \(f_n\) at that same point as it is given by a limit. But we know that those values depend only on the points coordinates on \(\chi_{f_n}\) ; so \(f\) must depent at most on that big union.
It, of course, could depend on even less than that, say you have a function \(g\) and and \(f_n = \frac 1 n g\) for \(n\) greater than zero. The point is that it might converge to a constant function, but it cannot depend on more than the coordinates the sequence dependend on.