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OK, here is an attempted answer under the assumption that $G$ is locally compact, which can perhaps be refined to give a general answer for Polish groups. A good reference would still be appreciated though. Locally compact Polish spaces are Lindelöf (every open cover has a countable subcover). So if we take a closed subset of $G$, it is a countable union ...

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I just thought that I should mention a result that considerably strengthens André Weil’s result mentioned by François in his answer above. Adam Kleppner, in his 1989 paper Measurable Homomorphisms of Locally Compact Groups, proved that any Borel-measurable homomorphism between locally compact (Hausdorff) groups is continuous. It does not matter whether or ...

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The answer to whether it is "locally h.j.i." is "in most cases". The precise condition is the following : Theroem (C. Riehm, THE CONGRUENCE SUBGROUP PROBLEM OVER LOCAL FIELDS) : Let $G(k)$ be an absolutely quasi-simple algebraic group over a local field and let $U$ be any open subgroup (in the strong topology). Then $U$ is hereditarily just infinite if and ...

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I switch to call $G$ your group because $X$ is a weird notation. The answer is: yes iff $G$ is infinite. I assume you have a proof for $G$ infinite separable. Next you can deal with the general case thanks to the following lemma: if $G$ is an infinite compact group, then it has an infinite separable quotient. Indeed, if the lemma is OK, then just take ...

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Apart from profinite groups, you can take infinite (Tychonoff) product of $SU(2)$ with itself; this has infinitely many irreps of dimension two. But the product $\prod _{n=2}^{\infty} SU(n)$ has only finitely many irreps in each dimension

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This is an addendum to the existing answers, but might still be of interest even if this is an old question. (I apologize for bumping with an answer rather than by reformatting an existing answer; please note that the absence of MathJax is deliberate, and I request that people don't prettify the text.) Among some people working on non-abelian harmonic ...

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