Closure and inverse projection Given a product $X=\prod_{k\in K}X_k$ of topological spaces, is one or both of the following true?

$\operatorname{cl}(\pi_k^{-1}(U_k)) \subseteq \pi_k^{-1}(\operatorname{cl}(U_k))\quad $ 
  or $\quad \operatorname{cl}(\pi_k^{-1}(U_k)) = \pi_k^{-1}(\operatorname{cl}(U_k))$ .

$U_k$ is an open set in $X_k$.
 A: I will write the proof when we have a product of two spaces and $\pi_k$ is the first projection but the proof is similar in the general case. Let $(x,y) \in cl(\pi_1^{-1}(U_1))$. Then we can write $(x,y)=\lim (x_i,y_i) $ with $x_i= \pi_k(x_i,y_i) \in U_k$. Now $\pi_1(x,y) =\lim \pi_1(x_i,y_i) \in cl(U_1)$ proving the first inclusion.
Now let $(x,y) \in \pi_1^{-1} (cl(U_1))$. Then $x \in cl(U_1)$ so we can write $x =\lim x_i$ with $x_i=\pi_1(x_i,y_i) \in U_1$ for all $i$. Now $(x,y)=\lim (x_i,y) \in cl(\pi_1^{-1}(U_1))$. This proves the second part.
Note: we cannot always use sequences. In general we have to use nets in this proof.
Note that  a point $x$ in the closure of  a set $U$ iff there is a net in $U$ converging to $x$.  
Proof of second part without using nets:
Let $(x,y)$ belong to RHS. If possible, suppose $(x,y) \notin$ LHS. Then there is an open set $V$ containing $(x,y)$ such that $V \cap \pi_1^{-1}(U_1)=\emptyset$. Let $V_0=\{z:(z,y) \in V\}$. Then $V_0$ is open because $z \to (z,y)$ is continuous. Also $x \in V_0$ Since $x \in cl(U_1)$ there exists some point $z$ in $V_0\cap U_1$. Now $(z,y) \in V \cap \pi^{-1}(U_1)=\emptyset$, a contradiction. 
A: In general for product spaces for $A_i \subseteq X_i$ for all $i$:
$$\operatorname{cl}\left(\prod_{i \in I} A_i\right) = \prod_{i \in I} \operatorname{cl} A_i$$
(See my answer here for a proof.) And your statement (the $=$ version) is just a special case where all $A_i=X_i$ except for $i=k$, where you have $A_k=U_k$. Openness is irrelevant.
A: The first one is clearly true because $\pi_k^{-1}({\rm cl}(U_k))$ is a closed set containing $\pi_k^{-1}(U_k)$
A: Let $(X\,\, S)$ and $(Y\,\, T)$ be two topological spaces.
Let $B\subseteq Y$ be arbitrary (not necessarily open).
I'll show the a bit harder inclusion, that
$$ c(q(B))\supseteq q(c(B)) $$
where $c$ is the closure operation, and $q$ operates
on subsets of $Y$ as the inverse of the projection
$X\times Y\rightarrow Y.$
Proof:
$$(x\,\,y)\in q(c(B))\,\, \Leftarrow:\Rightarrow\,\, 
y\in c(B)\,\,\Leftarrow:\Rightarrow $$
$$ \forall_{V\in T}
(y\in V\Rightarrow V\cap B\ne\emptyset)\,\, \Leftarrow:\Rightarrow $$
$$ \forall_{U\in S}\forall_{V\in T}\,((x\in U)\wedge(y\in V))\Rightarrow
(U\times V)\cap(X\times B)\,\ne\emptyset\,\,\Leftarrow:\Rightarrow $$
$$ (x\,\,y)\in c(q(B))  $$
Actually, I have proved more, namely the equality
$$ c(q(B)) = c(q(B)) $$
Great!
