Projection map is open on box topology.

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.
-
Hint: use the standard basis for the box topology. – Olivier Bégassat Dec 20 '12 at 15:49

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 $B\in\mathscr{B}$.
Thanks for the help!, I have just one question, Why $\pi_\alpha[B]\in{T_\alpha}$? – Fernando Valle Dec 20 '12 at 17:14
@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 $T$. – Brian M. Scott Dec 20 '12 at 17:23