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I'm quite certain that this should be trivially simple, but it's very late and I'm not that bright at the best of times:

$\{(X_\lambda, \mathcal{U}_\lambda)\,|\,\lambda \in \Lambda\}$ is a family of topological spaces with natural projection

$$pr_\mu\,:\, \prod\limits_{\lambda\in\Lambda} X_\lambda \rightarrow X_\mu, \quad (x_\lambda)_{\lambda\in\Lambda} \mapsto x_\mu.$$

I need to prove that $G \subseteq \prod X_\lambda$ is open in the product topology if and only if $G = \bigcup\limits_{\alpha\in A} G_\alpha$, and each $G_\alpha = \prod\limits_{\lambda \in \Lambda} G_{\alpha \lambda}$, with each $G_{\alpha\lambda}\in \mathcal{U}_\lambda$ and $G_{\alpha\lambda} = X_\lambda$ for all but finitely many $\lambda$. The product topology is then taken to be the topology induced by $\{pr_\mu\,|\,\mu\in\Lambda\}$.

Then I need to show that each $pr_\mu$ is an open mapping.

I appreciate that this is basically just writing out the definition of the product topology, and it feels like it's very close to clicking in my brain, but I keep getting jumbled and the devil Time ruthlessly pushes ever forward.

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Since this is, as you say, basically the definition of the product topology, it would help if you tell us what definition of the product topology you're using; otherwise it's hard to say what's to be proved based on what. – joriki Jun 8 '11 at 17:16
I stated that the product topology is defined to be that induced by the canonical projections. – Hargrove Jun 8 '11 at 17:18
I see. That was a bit confusing because you first stated what you want to prove about the product topology, then introduced its definition using "then". It would be clearer to define the product topology first, or at least change "then" to "here". – joriki Jun 8 '11 at 17:23
up vote 2 down vote accepted

The product topology is the one induced by the projections. That means that a subbase for the topology is given by all sets of the form $pr_{\mu}^{-1}[O]$ where $O$ is any open subset of $X_{\mu}$, and $\mu$ ranges over the index set $\Lambda$. Note that this set is just the product of $O$ in coordinate $\mu$ and $X_{\lambda}$ for all other coordinates $\lambda \neq \mu$. This means that all finite intersections of such sets form a base, and these are exactly the sets that are products of open sets in finitely many coordinates, and the whole space in all others. (The subbase and base remarks are always true for topologies induced by maps, here we just apply them to projections.) As these sets form a base, any open set can be written as a union of them.

As to showing that $pr_{\mu}$ is open, it suffices to show that basic open sets (as described above) have open images, as unions are preserved by function images, and this is quite clear: a projection of such a set is either the whole space or the non-trivial open factor. This does not depend on there being finitely many non-trivial factors etc. and also holds for the box topology on the product set.

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