# (infinite) product of topological groups is topological group

Any finite product of topological groups is a topological group with the direct product group structure and the product topology.

I was wondering why "finite"-ness was needed. My proof is as follows:

Let $$\{G_i\}$$ be a family of topological groups, with $$\mu_i:G_i \times G_i \rightarrow G_i$$, $$n_i:G_i \rightarrow G_i$$ be the multiplication and inverse map. Then define $$\prod G_i$$ with the direct group product structure and product topology.

The map $$\mu = (\mu_i): \prod (G_i \times G_i) \rightarrow \prod G_i$$ is continuous, and $$n =(n_i) : \prod G_i \rightarrow \prod G_i$$ is continuous too. (*) As $$\prod G_i \times \prod G_i \cong \prod (G_i \times G_i)$$, the multiplication map is continuous.

In $$(*)$$ I used the fact that if $$f_i:X _i \rightarrow Y_i$$ is a family of continuous maps, then $$(f_i): \prod X_i \rightarrow \prod Y_i, \quad \text{ where } (f(x))_i = f_i(x_i)$$ is a continuous map. Let $$\prod U_i$$ be an open basis element $$(U_i =Y_i$$ for all but finitely many $$i$$), then $$(f_i)^{-1}(\prod U_i) = \prod f_i^{-1}(U_i)$$ is also an open basis element of $$\prod X_i$$.

Note: $$i$$ ranges in an arbitrary index set $$I$$.

• There are many possible topologies on the infinite product, different from the finite case. – reuns Sep 21 '17 at 5:40

• Thanks! By the way, what does the $\alpha < \kappa$ notation mean? – Bryan Shih Sep 21 '17 at 5:25
• I'm using a cardinal number $\kappa$ as index set. $i \in I$ is also possible. – Henno Brandsma Sep 21 '17 at 6:08