I am stuck on Theorem 2.2.8 (Tychonoff's Theorem) in Dudley's Real Analysis and Probability.
The theorem is here:
Let $(K_i , T_i )$ be compact topological spaces for each $i$ in a set $I$. Then the Cartesian product $\Pi_i K_i$ with product topology is compact.
Since the proof uses ultrafilter, I write a definition of ultrafilter that Dudley used (p. 35 in his book); A Filter $\mathcal F$ in a set $X$ is called an ultrafilter iff for all $Y\subset X$, either $Y\in \mathcal F$ of $X-Y\in \mathcal F$.
The proof is here:
Let $\mathcal U$ be an ultrafilter in $\Pi_i K_i$. Then for all $i$, $p_i(\mathcal U)$ is an ultrafilter in $K_i$, since for each set $A\subset K_i$, either $p_i^{-1}(A)$ or its complement $p_i^{-1}(K_i-A)$ is in $\mathcal U$. So by
Theorem 2.2.5 : A topological space $(S, \mathcal T)$ is compact iff every ultrafilter in $S$ converges; that is, for any $x\in S$ every neighborhood of $x$ belongs to the ultrafilter,
$p_i(\mathcal U)$ converges to some $x_i\in K_i$. For any neighborhood $U$ of $x\equiv \{x_i \}_{i\in I} $, by definition of product topology, there is a finite set $F\subset I$ and $U_i \in \mathcal T$ for $i\in F$ such that $x\in \bigcap \{ p_i^{-1}(U_i) : i\in F \}\subset U$. For each $i\in F$, $p_i^{-1}(U_i)\in \mathcal U$, so $U\in \mathcal U$ and $\mathcal U\longrightarrow x$. So every ultrafilter converges and by Therem 2.2.5 again, $\Pi_i K_i$ is compact.
My questions:
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Why can we use Theorem 2.2.5 for $p_i(\mathcal U)$? It looks like we need to show that $p_i(\mathcal U)$ is closed to do this because of Theorem 2.2.2. Theorem 2.2.2 says that a closed subset of compact set is also compact. Note that $K_i (\supset p_i(\mathcal U))$ is compact by the assumption.
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How did Dudley use the definition of product topology to get the $F$ and $U_i$ that satisfy the condition? By the definition of product topology, every $p_i^{-1}(U_i)$ is open set in $\Pi_i K_i$. Thus $\bigcap \{ p_i^{-1}(U_i)$ is also open (since $F$ is finite). Since $U$ is a neighborhood, we can take open set in $U$, but can it be $\bigcap \{ p_i^{-1}(U_i) : i\in F \}\subset U$ for any neighborhood $U$? Is it always possible to say that there is such an finite set $F$?
Best Answer
We can apply 2.2.5 to $p_i(\mathcal{U})$ because it is an ultrafilter on $K_i$ and $K_i$ is compact. Once you accept that $p_i(\mathcal{U})$ is an ultrafilter on $K_i$, 2.2.5 says it has to converge to some $x_i \in K_i$.
If $A \subseteq K_i$, either $p_i^{-1}[A]$ is in $\mathcal{U}$, so $p_i[p_i^{-1}[A]] \in p_i(\mathcal{U})$ and as $p_i[p_i^{-1}[A]] \subseteq A$ by definition, the superset property of filters says that $A \in p_i(\mathcal{U})$ as well, or $(\prod_i K_i) - p_i^{-1}[A] \in \mathcal{U}$ (as $\mathcal{U}$ is an ultrafilter) but then $p_i[ (\prod_i K_i) - p_i^{-1}[A] ] \subseteq K_i - A$ and by the same reasoning, $K_i - A \in \mathcal{U}$. As $A$ is arbitrary, $p_i(\mathcal{U})$ is an ultrafilter on $K_i$.
The definition of the product topology is used in the fact that the collection of all sets of the form $\bigcap \{p_i^{-1}[U_i] \mid i \in F\}$, where $F \subseteq I$ is finite and all $U_i$ are open in $K_i$ for all $i \in F$, is a base for the product topology and so suffice to determine the convergence of filters: a filter converges to $x$ iff all basic open sets that contain $x$ are in that filter. This is a simple general fact about filter convergence. Any neighbourhood $U$ of $x$ contains such a basic neighbourhood by the definition of the product topology. It follows in essence from the fact that the product topology is the minimal topology such that all projections $p_i$ are continuous: we need these finite intersections to be open for that and we can stop there for minimality reasons.