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Let $(G,\mathcal T)$ be an infinite abelian group and $\Bbb T$ be the circle group. Why there are infinitely many continuous homorphisms $f:G\to \Bbb T$?

Is there a simple proof without using Pontryagin-Van Kampen theorem?

Minimus Heximus
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    In the compact case, it follows from Peter-Weyl. For general locally compact abelian groups, as far as I know the existence of any nontrivial characters is essentially as hard to prove as the full statement of Pontryagin duality. – Eric Wofsey Mar 19 '15 at 14:50
  • Consider, for example, $G = \prod \mathbb{Z}_2 / \bigoplus \mathbb{Z}_2$. I don't think you can write down a nontrivial character of $G$ without something like the ultrafilter lemma. (The ultrafilter lemma is equivalent to the statement that the product of compact Hausdorff spaces is compact Hausdorff, which we need for the discrete-compact version of Pontryagin duality.) – Qiaochu Yuan Mar 19 '15 at 16:11

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