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Let $\{(X_\alpha,T_\alpha):\alpha\in{\lambda}\}$ be an indexed family of topological spaces, let $X=\prod_{\alpha\in{\lambda}}X_\alpha$, and let $T$ be the box topology on $X$. Then for each $\beta\in{\lambda}$, the projection map $\pi_{\beta}:X\to{X_\beta}$ is open.

  • I would like to see a proof of this theorem please.
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    Hint: use the standard basis for the box topology.2012-12-20

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Let $\mathscr{B}=\left\{\prod_{\alpha\in\lambda}U_\alpha:U_\alpha\in T_\alpha\text{ for each }\alpha\in\lambda\right\}\;;$ clearly is $\mathscr{B}$ is a base for $T$, and for each $B\in\mathscr{B}$ we have $B=\prod_{\alpha\in\lambda}\pi_\alpha[B]\;,$ and $\pi_\alpha[B]\in T_\alpha$ for each $\alpha\in\lambda$.

Fix $\alpha\in\lambda$, and let $U\in T$ be arbitrary. There is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. Then

$\pi_\alpha[U]=\pi_\alpha\left[\bigcup\mathscr{B}_U\right]=\bigcup\big\{\pi_\alpha[B]:B\in\mathscr{B}_U\big\}\in T_\alpha\;,$

since $\pi_\alpha[B]\in T_\alpha$ for each $B\in\mathscr{B}$.

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    @Fernando: You’re welcome. $\pi_\alpha[B]\in T_\alpha$ for all $B\in\mathscr{B}$ because each $B\in\mathscr{B}$ is a product of open sets in the factors: that’s part of the definition of $\mathscr{B}$. And $\mathscr{B}$ is the standard base for $T$.2012-12-20