Components of an open set in product topology

Question

Suppose $U = U_1 \times U_2 \times \cdots \times U_n $ is an open set in the topological space $X = X_1 \times X_2 \times \cdots \times X_n $, equipped with the box topology. (Note that $n$ is finite here). Then are $U_1, U_2, \cdots U_n$ open?

Answer 1

Suppose all of $A_1,\ldots, A_n$ are subsets of $X_1,\ldots,X_n$ respectively, and are all non-empty (this is needed). Then suppose that $A = \prod_{i=1}^n A_i$ is open in the product topology. Then all $A_i$ are open.

Proof

Fix $p_i \in A_i$ for all $i$. Then to see that $A_k$ is open (for fixed $k \in \{1,\ldots,n\}$, pick $x \in A_k$ and we'll show it is an interior point of $A_k$. Note that $\hat{x} = (p_1, \ldots, x, \ldots p_n) \in A$ ($\hat{x}$ is just $x$ at the $k$'th. coordinate and the fixed point $p_i$ at others) and so there is a basic open subset $\prod_i^n O_i$ (so all $O_i$ are open) in the product such that $\hat{x} \in \prod_{i=1}^n O_i \subseteq A$. This implies $x \in O_k \subseteq A_k$ (because $y \in O_k$, implies $\hat{y} \in \prod_{i=1}^n O_i$, so $\hat{y} \in A$ hence $y \in A_k$). This shows that $x$ is an interior point of $A_k$.

ALternatively, let $p_j$ be the projection of $X = \prod_{i=1}^n X_i$. Then as $A$ is open, write $A = \bigcup\{ O^{(i)}_1 \times \ldots \times O^{(i)}_n : i \in I\}$, for some index set $I$ (an open set is a union of basic open boxes), so all$O^{(i)}_j$ are open. Then

$$A_j = p_j[A] = p_j[\bigcup\{ O^{(i)}_1 \times \ldots \times O^{(i)}_n : i \in I\} = \bigcup_{i \in I} p_j[O^{(i)}_1 \times \ldots \times O^{(i)}_n] = \bigcup_{i \in I} O^{i}_j $$

as unions and function images commute and the last set is open, as union of open sets in $X_j$. So $A_j$ is open for every $j$.

Answer 2

Your statement is correct, and your proof is nearly correct.

Union of product is not product of union, as @positrón0802 has pointed out. Examples include putting two squares on plane, and observe that the union of two square can possible be not a square even if squares have intersection.

For your statement, since $U = \cup_\alpha B_1^\alpha \times \dots B_n^\alpha$, define $V=\cup_\alpha B_1^\alpha$. $V$ is an open set. Now observe that $V=U_1$.