# Does every (Abelian) Polish group have a nontrivial locally compact subgroup?

The question is pretty much in the title, suppose that $$G$$ is an (Abelian) nontrivial Polish group, must $$G$$ have a nontrivial locally compact (in the induced topology, hence necessarily closed) subgroup?

There are easy examples without compact subgroups, but all the usual examples of very big non locally compact Polish groups I checked have plenty of locally compact subgroups. However I cannot prove that such a subgroup must exist, so I'm suspecting that there is a counterexample.

No. Here's a counterexample.

Every 3-adic integer $$x\in\mathbb Z_3$$ has a unique representation as $$x=\sum_{n\geq 0} a_n 3^n$$ with balanced ternary digits $$a_n\in\{-1,0,1\}.$$ Define $$f(x)=\sum_{n\geq 0} |a_n|/(n+1)\in[0,\infty].$$ Define $$d(x,y)=f(x-y).$$ It's fairly easy to show this satisfies the triangle inequality; it suffices to consider finite integers, and each carry $$3^n+3^n=3^{n+1}-3^n$$ reduces the contribution to $$f,$$ from $$1/(n+1) + 1/(n+1)$$ to at most $$1/(n+2) + 1/(n+1).$$

Take $$G$$ to be the abstract subgroup of $$x\in\mathbb Z_3$$ such that $$f(x)<\infty,$$ equipped with the metric $$d.$$ This is an abelian Polish group. Fix $$x\in G\setminus\{0\}.$$ To show that subgroups containing $$x$$ are not locally compact, let's take an arbitrary $$\epsilon>0$$ and try to show that there is an infinite sequence of integers $$p_i$$ with $$f(p_ix)<\epsilon$$ but $$f((p_i-p_j)x)>\epsilon/2$$ for all $$i

Lemma. For any $$N$$ we can pick $$p\in\mathbb N$$ such that $$f(px)\in (\epsilon/2,\epsilon)$$ and $$p\equiv 0\pmod {3^N}.$$

Proof. First write $$x=3^ky$$ where $$y\not\equiv 0\pmod 3.$$ Since $$\sum 1/n$$ diverges, a sufficiently good finite $$3$$-adic approximation $$p'$$ to $$(3^N+3^{N+1}+3^{N+2}+\dots)/y$$ will give $$f(p'x)>\epsilon/2.$$ Let $$m$$ be the greatest integer such that $$f(3^mp'x)>\epsilon/2.$$ Then $$f(3^mp'x)<\epsilon,$$ because multiplying by $$3$$ reduces $$f$$ by at most a factor of $$2.$$ So $$p=3^mp'$$ will do. $$\square$$

We can therefore pick sequences of integers $$p_0,p_1,\dots$$ and $$N_0,N_1,\dots$$ satisfying:

• $$f(p_ix)<\epsilon$$ for all $$i$$
• $$\sum_{n=0}^{N_i-1} |a_{i,n}|/(n+1)>\epsilon/2$$ where $$(a_{i,n})_{n\geq 0}$$ are the balanced ternary digits of $$p_ix$$
• $$p_jx\equiv 0\pmod{3^{N_i}}$$ for $$i

The last two conditions ensure $$f((p_i-p_j)x)>\epsilon/2$$ for $$i