Equivalence of closure of sets Let $p_{\alpha_i}: X \rightarrow X_{\alpha_i}$ be the projection mapping.
Defining $V= \bigcap\limits_{i=1}^n p_{\alpha_i}^{-1} (U_{\alpha_i)}$, why is the closure  
$\bar{V}= \bigcap\limits_{i=1}^n p_{\alpha_i}^{-1} (\overline{U_{\alpha_i)}}$.
I was thinking if we have a set $C= \bigcap\limits_{i=1}^nV_i$, 
then $\bar{C}= \overline{\bigcap\limits_{i=1}^nV_i}=\bigcap\limits_{i=1}^n\overline{V_i} $, bu believe that this must be wrong somewhere.
 A: HINT: First, each projection map is continuous, so each set $p_{\alpha_i}^{-1}[\operatorname{cl}U_{\alpha_i}]$ is closed in $X$, and therefore $\bigcap_{i=1}^np_{\alpha_i}^{-1}[\operatorname{cl}U_{\alpha_i}]$ is also closed in $X$. It’s also clear that $V\subseteq\bigcap_{i=1}^np_{\alpha_i}^{-1}[\operatorname{cl}U_{\alpha_i}]$, so 
$$\operatorname{cl}V\subseteq\bigcap_{i=1}^np_{\alpha_i}^{-1}[\operatorname{cl}U_{\alpha_i}]\;,$$
and you have only to prove that 
$$\bigcap_{i=1}^np_{\alpha_i}^{-1}[\operatorname{cl}U_{\alpha_i}]\subseteq\operatorname{cl}V\;.$$
The most straightforward way to do this is to let $x\in\bigcap_{i=1}^np_{\alpha_i}^{-1}[\operatorname{cl}U_{\alpha_i}]$, let $W$ be an arbitrary open nbhd of $x$ in $X$, and show that $W\cap V\ne\varnothing$. Remember that without loss of generality you may assume that $W=\prod_\alpha W_\alpha$, where each $W_\alpha$ is open in $X_\alpha$, and $W_\alpha=X_\alpha$ for all but finitely many indices $\alpha$. You may also want to observe that if we set $U_\alpha=X_\alpha$ for each index $\alpha\notin\{\alpha_1,\ldots,\alpha_n\}$, then $V=\prod_\alpha U_\alpha$.
You may find it helpful to look first at the simple case in which $X=X_1\times X_2$, so that you’re dealing with only two factors; you can draw diagrams for that case that may point you in the right direction.
