I'm working through John M. Lee's Introduction to Topological Manifolds and am trying to complete exercise 3.25. In this exercise, we are asked to show that if $X_1, \dots, X_n$ are arbitrary topological spaces then $$ \mathcal{B} = \{ U_1 \times \cdots \times U_n : U_i \text{ is an open subset of }X_i, i = 1,2,3,\dots,n \} $$
is the basis for a topology on $X = X_1 \times \cdots \times X_n$.
My attempt so far
To show that $\mathcal B$ generates a topology, we must show
- $\cup_{B \in \mathcal B} B = X$.
- For any $B_1, B_2 \in \mathcal B$ and any $x \in B_1 \cap B_2$, there exists some $B_3 \in \mathcal B$ such that $x \in B_3 \subseteq B_1 \cap B_2$.
To show requirement 1, we can show set containment in both directions.
Suppose $x = (p_1, \dots, p_n) \in X = X_1 \times \cdots \times X_n$. So, $p_i \in X_i$. Then, for each $p_i$ there exists some set $U_i^{p_i}$ open in $X_i$ st $x \in U_i^{p_1}$. So, $x \in U_1^{p_1} \times \cdots \times U_n^{p_n} = B_x$. This holds for any point $x \in X$, so clearly, $X \subseteq \cup_{B \in \mathcal B} B$.
On the other hand, suppose $x = (p_1, \dots, p_n) \in \cup_{B \in \mathcal B} B$. Then there is some $B_x = U_1^{p_1} \times \cdots \times U_n^{p_n} $ such that $p_i \in U_i^{p_i}$ and $U_i^{p_i} \subseteq X_i$. So, certainly, we have $x \in X$. This shows $\cup_{B \in \mathcal B} B \subseteq X$. Thus, the two are equal.
To show requirement 2, suppose we have $B_1, B_2 \in \mathcal B$ and $x \in B_1 \cap B_2$. We can write $x = (p_1, \dots, p_n) $, and express the basis sets as $B_1 = U_1 \times \cdots U_n$, $B_2 = V_1 \times \cdots \times V_n$ where $U_i$'s and $V_i$'s are open sets in $X_i$.
Then, $x \in B_1 \cap B_2$ means that $p_i \in U_i \cap V_i$. Since each $U_i, V_i$ is open, the intersection $W_i = U_i \cap V_i \subseteq X_i$ is itself open in $X_i$. Then, we can write $x \in B_3 = (W_1, \dots, W_n) \subseteq B_1 \cap B_2$.
This shows the second requirement is fulfilled, and so $\mathcal B$ generates a topology (the product topology) on $X$.
My question is whether or not my argument for requirement 2 is valid. I feel like there is something subtle i'm misunderstanding here. Any corrections or suggestions in general would be greatly appreciated!
The proof is ok. Actually, you can simplify the proof of requirement 1 noting that the inclusion $\subseteq$ is obvious as clearly every $B$ is contained an $X$, so the same is true for the union.