Open Sets and a Basis for the Product Topology

The product topology on $X \times Y$ is defined to be that topology whose basis consists of sets of the form $U \times V$, where $U$ is open in $X$ and $V$ is open in $Y$. According to a lemma in my book, the topology generated by a basis is in fact just the collection of all arbitrary unions of basis elements.

I realize that what is to follow is simple, but I just want my reasoning to corroborated, as I am studying topology on my own. If $\mathcal{O}$ is open in the product topology, then $\mathcal{O} = \bigcup_{i \in I} U_i \times V_i = \left(\bigcup_{i \in I} U_i \right) \times \left(\bigcup_{i \in I} V_i \right)$. Since $\bigcup_{i \in I} U_i$ is a union of open sets in $X$, it, too, must be some open set in $X$, call it $U$; the same goes for $\bigcup_{i \in I} V_i$. Hence, $\mathcal{O} = U \times V$. So, open sets in the product topology really amount to cartesian products of open sets in the factor spaces, and so we don't have to concern ourselves with arbitrary unions.

Does this seem right?

• The union of products is in general not the product of unions. Consider $\bigl((0,1) \times (0,1)\bigr) \cup \bigl((1,2)\times (1,2)\bigr)$ for an example. – Daniel Fischer Nov 23 '16 at 12:06
• @DanielFischer What am misunderstanding about the "general result" found here: proofwiki.org/wiki/Cartesian_Product_of_Unions In this result, there are two different indexing sets, but couldn't we take them to be the same? – user193319 Nov 23 '16 at 12:10
• That's a one-way result. You can always write the product of unions as the union of products, but in general the other way doesn't work. Also, when you write the product of unions as a union of products, the index set of the union of products becomes the product of the index sets of the factors. If one takes $I = J$ in your link, the index set of the union of products becomes $I^2 = I\times I$. – Daniel Fischer Nov 23 '16 at 12:30
• To be more specific, why can't I write $\bigcup_{i \in I} U_i \times V_i = \bigcup_{(i,i) \in I \times I} U_i \times V_i = (\bigcup_{i \in I} U_I) \times (\bigcup_{i \in I} V_i )$? – user193319 Nov 23 '16 at 12:32
• Look at examples. Generally, one has $(U_1 \cup U_2) \times (V_1 \cup V_2) = U_1 \times V_1 \cup U_1 \times V_2 \cup U_2 \times V_1 \cup U_2 \times V_2 \supsetneqq U_1 \times V_1 \cup U_2 \times V_2$. – Daniel Fischer Nov 23 '16 at 13:34