Question on the restricted product topology

Question

Let $I$ be an index set and $G_i$ a collection of compact topological groups. Suppose that for each $i \in I$ there are compact (proper) subgroups $H_i \subseteq G_i$. We say that $G =\{(a_i) \in \prod G_i \,:\, a_i \in H_i \text{ for almost all $i \in I$}\}$ is the restricted product of $G_i$ with respect to the $H_i$.

We can give $G$ a topology by declaring the collection $\mathcal{B} = \{B \,:\, B = \prod U_i \text{ with $U_i$ open in $G_i$ and $U_i \neq H_i$ only finitely many times}\}$ as a basis.

I must be missing something really basic here, but I can't see why this is closed under arbitrary unions. For example, consider the sets \begin{align*} W_i = G_i \times \prod_{j\neq i} H_j. \end{align*} Then $W_i \in \mathcal{B}$, but $W = \bigcup_{i\in I} W_i = \prod_iG_i \not\subseteq G$. What is going wrong here?

Answer 1

$W$ is not $\prod_{i\in I}G_i$: if $x\in W$, there is at most one index $i\in I$ such that $x_i\notin H_i$. In more detail, $x$ must be in $W_i$ for some $i\in I$, and therefore $x_j\in H_j$ for each $j\in I\setminus\{i\}$.

Moreover, a base for a topology need not be closed under arbitrary unions: one gets the topology by closing it under arbitrary unions.

Answer 2

You're just using the standard base of the product topology on $\prod_i G_i$, restricted to the subspace $G$. So $G$ has the subspace topology wrt that product topology, nothing more or less.