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There is the following characterisation:

Let $G$ be a locally compact Abelian group. Then $G$ is second countable if, and only if, the Pontryagin dual $\widehat{G}$ is second countable with respect to the compact-open topology on $\widehat{G}$.

I have difficulties to show that $\widehat{G}$ is second countable whenever $G$ is. The other direction is just an application of Pontryagin's duality theorem.

Any help or comment is highly appreciated.

  • With the notation of your previous question, take a countable $\mathcal{U}$. – Daniel Fischer Sep 04 '16 at 17:55
  • @DanielFischer. I think my main problem is that I am not sure how to show that the collection of sets $L(K, U)$ form a topological base for $\widehat{G}$. This was also more or less the problem in the previous question. –  Sep 04 '16 at 18:02
  • @jvnv: what is your definition of the topology on the dual group? According to your previous question, it seems to me that it was precisely defined as the topology generated by the sets $L(K,U)$. – Watson Sep 04 '16 at 18:04
  • They don't (generally - they do for trivial $G$ of course). They form a subbasis. The family of finite intersections of the $L(K,U)$ form a basis. – Daniel Fischer Sep 04 '16 at 18:05
  • @Watson. Yes, that is correct. That does not mean that the sets $L(K,U)$ form a basis, right? They form a subbasis. –  Sep 04 '16 at 18:06
  • If you let $K$ run through all compact subsets of $G$, then you only get a countable family in exceptionally simple situations. You choose a countable basis $\mathscr{B}$ of the topology of $G$ consisting of relatively compact sets, and a countable basis $\mathscr{B}'$ of the topology of $\mathbb{T}$, and then look at the finite intersections of members of the subbasis ${ L(\overline{U},V) : U \in \mathscr{B},, V \in \mathscr{B}'}$. – Daniel Fischer Sep 04 '16 at 18:20
  • @DanielFischer. Okay, the approach you suggest might be easier (@Watson just placed an answer with this approach). I tried to show that any open set $O \subset \widehat{G}$ can be expressed as a union of elements of the basis $\mathcal{F}$, but I did not succeed. A similar problem occurred in the previous question. –  Sep 04 '16 at 18:33

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Let $\scr B$ be a countable basis of $G$. From your previous question, it follows that $\mathscr S = \{L(\overline V,U) \mid V \in \mathscr B \text{ open with compact closure },U \subset S^1 \text{ open}\}$ is a subbasis of $\widehat G$.

We will show that $$\mathscr S' = \{L(\overline V,U(a,b)) \mid V \text{ open with compact closure },a,b \in \Bbb Q,a<b\}$$ is also a subbasis of the dual group of $G$, where $U(a,b) = \{e^{i\theta} \mid a<\theta<b\}$. The basis generated by $\scr S'$ is countable, so we are done.

Clearly, $\mathscr S' \subset \mathscr S$, so that $\mathcal T_{\mathscr S'} \subset \mathcal T_{\mathscr S}$. To show the reverse inclusion, let $O \in \mathcal T_{\scr S}$. Write it as the union of some $B_a$'s where each $B_a$ is a finite intersection of $L(\overline V,U)$.

Then, it is sufficient to show that each $L(\overline V,U)$ belongs to $T_{\mathscr S'}$. Pick any $\gamma \in L(\overline V,U)$.

Since $\gamma(\overline V)$ is a compact set contained in the open set $U = \bigcup_{i \in I} U(a_i,b_i) \subset S^1$, it is actually contained in a finite number of $ U(a_i,b_i) $:

$$\gamma(\overline V) \subset \bigcup_{k=1}^n U(a_{i_k},b_{i_k})$$

Then I let you show that $$\gamma \in \bigcap_{k=1}^n L(\overline V, U(a_{i_k},b_{i_k})) =: V_{\gamma} \subset L(\overline V,U).$$

Therefore $$L(\overline V,U) = \bigcup_{\gamma \in L(\overline V,U)} V_{\gamma} \;\in\; T_{\mathscr S'}.$$

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