# Unsure lines in a proof of Tychonoff's Theorem

Can someone explain some of the later lines in this proof?

I think $\mathcal{T}$ is meant to be $\mathcal{T_i}$ in the third to last line. After that, how is one sure that $x$ can be found in some finite intersection of inverse images of $U_i\in\mathcal{T}_i$?

Why is $p_i^{-1}(U_i)\in\mathcal{U}$ for all $i\in F$? What would go wrong if $p_i^{-1}(K_i\setminus U_i)\in\mathcal{U}$ instead for some $i$?

-

## 2 Answers

You’re right about the typo in the third to last line: that should be $U_i \in \mathcal{T}_i$.

For your first question, remember how the product topology is defined: the sets of the form $\bigcap\{p_i^{-1}[U_i]:i \in F\}$, where $F$ is a finite subset of $I$ and each $U_i \in \mathcal{T}_i$, are a base for the product topology. This means that every open set in the product is a union of sets of this form. Since $x \in U$, and $U$ is open in the product, $U$ is a union of sets of this form, and one of them must contain $x$. Note that this implies that $x_i \in U_i$ for each $i \in F$.

For your second question, remember that $x_i$ was chosen so that $p_i[[\mathcal{U}]] \to x_i$, so every open nbhd of $x_i$ is in $p_i[[\mathcal{U}]]$. In particular, $U_i$ is an open nbhd of $x_i$, so $U_i \in p_i[[\mathcal{U}]]$. This means that there is some $V \in \mathcal{U}$ such that $p_i[V] = U_i$. But then $p_i^{-1}[U_i] \supseteq V \in \mathcal{U}$, so $p_i^{-1}[U_i] \in \mathcal{U}$ (because $\mathcal{U}$ is an ultrafilter).

-
Thanks Brian M. Scott. – Gotye Sep 6 '11 at 0:39

It's part of the construction of the product topology. A basis of open sets in the product topology comes from taking a finite subset $F \subset I$ of the index set, choosing an open set $U_i \subset X_i$ for each $i \in F$, and then forming the product $\prod U_i \times \prod X_j$ where $i \in F$, $j \in I-F$. So any open neighboorhood of a point $x = \{x_i\}_{i \in I}$ will contain a set of the form $\prod U_i \times \prod X_j$ where $x_i \in U_i$ for $i \in F$ (i.e., $x_i \in p_i^{-1}(U_i)$). And of course $x_j \in X_j$ for $j \in I-F$.

-