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I am following this source: http://www.hss.caltech.edu/~kcb/Notes/Kolmogorov.pdf and agree with everything done in sections 1-3.

In section 4, I cannot fill in the detail for for Lemma 4, because I cannot deal with the finite unions.

Also, I tried to apply the result of section 3 directly to $R^\mathbb{N}$ but I could not find a compact class with the desired property. I tried the set of finite-dimensional cylinders where the nontrivial section is compact, but I just can't see that this is a compact class. (I.e. sets of the form $K \times R \times R...$ where $K$ is a compact subset of some $R^n$.)

For some sort of self-containedness, a compact class is one where every sequence in the class has the property that all finite intersections being nonempty implies the whole intersection of the sequence is nonempty.

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1 Answer 1

up vote 2 down vote accepted

$\newcommand{\ms}{\mathscr}$Let $\ms C$ be a compact class. Let $$\ms C_u=\left\{\bigcup\ms F:\ms F\subseteq\ms C\text{ is finite}\right\}\;,$$ and suppose that $\{F_n:n\in\Bbb N\}\subseteq\ms C_u$ is centred (i.e., finite intersections are non-empty); we want to show that $\bigcap_{n\in\Bbb N}F_n\ne\varnothing$.

Let $\displaystyle\Phi=\prod_{n\in\Bbb N}\ms F_n$; a point $\varphi\in\Phi$ is a sequence $\varphi=\langle C_n^\varphi:n\in\Bbb N\rangle$ such that $C_n^\varphi\in\ms F_n$ for each $n\in\Bbb N$. Then

$$\bigcap_{n\in\Bbb N}F_n=\bigcap_{n\in\Bbb N}\bigcup\ms F_n=\bigcup_{\varphi\in\Phi}\bigcap_{n\in\Bbb N}C_n^\varphi\;.$$

(This is just distributivity of intersection over union, though you may have to think about it a bit if you’ve not seen an instance of the infinite distributive law before.)

Thus, $\bigcap_{n\in\Bbb N}F_n\ne\varnothing$ iff there is a $\varphi\in\Phi$ such that $\bigcap_{n\in\Bbb N}C_n^\varphi\ne\varnothing$. The existence of such a $\varphi$ will follow from König’s lemma after we build a suitable tree.

For each $n\in\Bbb N$ and $\varphi\in\Phi$ let $\varphi_n=\varphi\upharpoonright\{k\in\Bbb N:k<n\}$. Viewed as a set of ordered pairs, $\varphi=\bigcup_{n\in\Bbb N}\varphi_n$; if $\varphi$ is viewed as a sequence, $\varphi_n$ is its initial segment of length $n$. Note that $\varphi_n\subsetneqq\varphi_{n+1}$ for each $\varphi\in\Phi$ and $n\in\Bbb N$. Let $$T=\left\{\varphi_n:\varphi\in\Phi,~n\in\Bbb N,\text{ and }\bigcap_{k<n}C_k^\varphi\ne\varnothing\right\}\;;$$

$\langle T,\subseteq\rangle$ is a tree. The root of $T$ is $\varnothing=\varphi_0$ for all $\varphi\in\Phi$. The predecessors of $\varphi^n\in T$ are the $\varphi^k$ with $k<n$. Each $\varphi_n\in T$ has at most $|\ms F_{n+1}|$ children, so $T$ is finitely branching. Finally, for each $n\in\Bbb N$ we know that $\bigcap_{k\le n}F_k\ne\varnothing$, so there must be some $\varphi^n\in T$, and $T$ is therefore infinite. By König’s lemma there must be a $\varphi\in\Phi$ such that $\varphi^n\in T$ for all $n\in\Bbb N$, and therefore

$$\bigcap_{n\in\Bbb N}F_n\supseteq\bigcap_{n\in\Bbb N}C_n^\varphi\ne\varnothing\;,$$

since $\ms C$ is a compact class.

Closing $\ms C_u$ under countable intersections shouldn’t be a problem.

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+1 Nice, Brian! –  amWhy Nov 7 '12 at 18:27
@amWhy: Thanks! That one was fun. –  Brian M. Scott Nov 7 '12 at 18:30

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