# Definition of Product Topology (generated by a subbasis)

Definition (Subbasis for Product Topology):

Let $\mathcal{S}_{\beta}$ denote the collection $$\mathcal{S}_{\beta} = \left\{ \pi_{\beta}^{-1}(U_{\beta}) \ | \ U_{\beta} \text{ is open in} \ X_{\beta}\right\}$$ and let $\mathcal{S}$ denote the union of these collections, $$\mathcal{S} = \bigcup_{\beta \in J}S_{\beta}$$ The topology generated by the subbasis $\mathcal{S}$ is called the product topology.

Since a topology generated by a subbasis is the collection of all unions of finite intersections of subbasis elements, is the following a satisfactory definition of the Product Topology?

$$\mathcal{T}_P = \left\{ \ \bigcup_{\alpha \in I} \left(\bigcap_{\beta \in [1, ..,n]} \pi^{-1}_{\beta}\left(U_{\beta}\right)\right)_{\alpha} \ \ \middle| \ U_{\beta} \text{ is open in some } X_{\beta}\ \right\}$$ and $I$ is an arbitrary indexing set.

As a follow up question, is there any easier way to formally define the product topology on a product space, other than this?

• There are neater definitions, yes, but this one is often the most practical to use. – John Gowers Jan 18 '17 at 23:39

Another way to say it is that open sets in $X = \prod\limits_i X_i$ consist of unions of sets of the form

$$\prod\limits_i U_i$$

where $U_i$ is open in $X_i$, and $U_i = X_i$ for all but finitely many $i$.

Here is a more abstract way: let $\pi_i: X \rightarrow X_i$ be the projection map. For topological spaces $A$ and $B$, let $\textrm{Map}(A,B)$ denote the set of all continuous functions $A \rightarrow B$. Then the product topology is the unique topology on $X$ such that for any topological space $A$,

$$f \mapsto \pi_i \circ f$$

defines a bijection of sets

$$\textrm{Map}(A,X) \rightarrow \prod\limits_i \textrm{Map}(A, X_i)$$

Notation quible: The $n$ depends on $\alpha$ and so do the finite intersections of subbase elements. I'll make the dependence more explicit:

So suppose the $X_\beta, \beta \in B$ are the spacs we take the product of (I don't see you state their index set). So the $O$ is open iff there is some index set $I$ and for every $\alpha \in I$ there is a finite subset $F_\alpha$ of $B$ and for every $\beta \in F_\alpha$ we have an open set $U_\beta \subseteq X_\beta$ and we have $$O = \bigcup_{\alpha \in I} \left(\bigcap_{\beta \in F_{\alpha}} (\pi_\beta)^{-1}[U_\beta]\right)$$

Note that this is just a fancy index-juggling way of saying that all sets of the form $\prod_{\beta \in B} U_\beta$, where all $U_\beta$ are open in $X_\beta$ and the set $\{\beta: U_\beta \neq X_\beta \}$ is finite, form an open base for the topology. Sets of this form are exactly the finite intersections of members from $\mathcal{S}$, as can be easily seen.