I am going through the proof in this book of Tychonoff theorem.

First I understand proof of Alexander subbase theorem. Then the proof of Tychonoff theorem uses that. However, I am unable to follow the last step. Concretely: We have defined $\mathcal U_i = \{ O : O \text{ open in } X_i \text{ and } \pi_i^{-1}O \in \mathcal U\}$ where $\mathcal U$ is a cover of $X$ consisting only of sets in a subbase $S = \{ \pi_i^{-1}O : O \text{ open in } X_i \}$ of topology on $X$. Then assume that none of the $\mathcal U_i$ cover $X_i$ and choose $x_i \in X_i \setminus \bigcup \mathcal U_i$. Let $x = (x_i)_i$. How does it follow that $x \in X \setminus \mathcal U = \prod_i X_i \setminus \bigcup \mathcal U = (\bigcup \mathcal U)^c $?

I have $\pi_i^{-1}\mathcal U_i \subseteq \mathcal U$ by definition of $\mathcal U_i$ and therefore $\bigcup \bigcup_i \pi_i^{-1}\mathcal U_i \subseteq \bigcup \mathcal U$ and therefore $\left ( \bigcup \bigcup_i \pi_i^{-1}\mathcal U_i \right )^c \supseteq \left (\bigcup \mathcal U \right )^c$ but to finish the proof it's needed that $\left ( \bigcup \bigcup_i \pi_i^{-1}\mathcal U_i \right )^c \subseteq \left (\bigcup \mathcal U \right )^c$.

Thank you for help.


If $x = ( x_i )_{i} \in \bigcup \mathcal{U}$ then there must be a $U \in \mathcal{U}$ such that $x \in U$. Now by choice of covering, there is an $i$ and an open $O \subseteq X_i$ such that $U = \pi_i^{-1} O$; in particular $O \in \mathcal{U}_i$. Note that $x_i$ was chosen so that $x_i \notin \bigcup \mathcal{U}_i$, but since $x = (x_i)_i \in U = \pi_i^{-1} O$ it must be that $x_i \in O$!


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