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Folland's book on Fourier analysis proves a result about products of compact groups, except I have no idea where his proof uses compactness (and not just local compactness). Can someone enlighten me?

Here's the result. Assume $G = \prod_i G_i$ is a product of infinitely many compact abelian groups, and $\chi : G \to S^1$ is a character (continuous homomorphism) of the product. For each $i,$ define $\chi_i : G_i\to S^1$ as $\chi_i(g_i) = \chi(..., 1, g_i, 1, ...),$ e.g. just plug in $\chi$ by setting all the components outside the $G_i$ factor to 1.

Then each $\chi_i$ is still a homomorphism. Next, take some small open set $V$ of $S^1$ containing $1$ and containing no nontrivial subgroups of $S^1.$ Then $\chi^{-1}(V)$ is open and contains 1, meaning it contains some $\prod_i V_i$ where $V_i\subseteq \chi_i^{-1}(V)\subseteq G_i$ is open and contains 1, and all but finitely many $V_i = G_i$. If $V_i = G_i,$ then $\chi_i(G_i) = \chi_i(V_i)\subseteq V.$ But $\chi_i(G_i)$ is a subgroup of $S^1,$ and hence must be the trivial subgroup. Since $V_i=G_i$ happens for all but finitely many indices, we conclude.

Where is compactness used? Folland proves every proposition before and after this for locally compact abelian groups, but for this one he only uses compact as the hypothesis. Is this a typo, or is compactness being used in a way I do not see?

(This is proposition 4.9 on page 99 of the book.)

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    For one thing, if you only have that each $G_i$ is locally compact, there's no guarantee that $G$ is locally compact. Let $G_i = \mathbb{R}, i \in \mathbb{N}$. Then each $G_i$ is locally compact, but $G$ is not locally compact since the the sequence $(e_j)$ has no convergent subsequence. So if you don't suppose compactness on each $G_i$ you don't necessarily stay in the realm of LCAs – perpetuallyconfused Aug 31 '22 at 19:22
  • @perpetuallyconfused Ah, that was probably it--a silly oversight of mine to make. –  Aug 31 '22 at 20:57

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