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Let $X$ be a topological space, and let $K \subset X$ be compact and $U \subset \mathbb{C}$ be open. Define $$ L (K,U) := \{ f : X \to \mathbb{C} \; | \; f(K) \subset U \}. $$

Then the topology generated by the sets $L(K,U)$, as $K$ and $U$ vary, is the compact-open topology on the space of continuous functions $C(X, \mathbb{C})$.

There is the following result:

Let $\mathcal{U}$ be a basis for the locally compact Abelian group $G$ and let $\mathcal{U}_c$ denote the set of all relatively compact sets $U \in \mathcal{U}$, i.e., the set $\overline{U}$ is compact. Then the sets $L(\overline{U}, V)$, where $U \in \mathcal{U}_c$ and $V \subset \mathbb{T}$ open, generate the compact-open topology on the Pontryagin dual group $\widehat{G}$.

I understand all the necessary concepts. My idea to prove this is to show that the collection $\mathcal{F}$ of all finite intersections of sets $L(\overline{U}, V)$ forms a basis for the compact-open topology on $\widehat{G}$. However, I am unable to do so.

Any help and/or comment is highly appreciated.

  • What are your problems? Is it a problem that in the first paragraph you describe the compact open topology on $C(X,\mathbb{C})$, and for the dual group you look at the CO topology on $\widehat{G}\subset C(G,\mathbb{T})$? – Daniel Fischer Sep 04 '16 at 14:25
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    I do not see why it suffices to take only the relatively compact sets in the basis for the topology on $G$ rather than taking all compact sets in $G$. –  Sep 04 '16 at 14:28
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    Okay. So it seems it's clear to you that the topology generated by the $L(\overline{U},V)$ with $U \in \mathcal{U}_c$ is coarser than the compact-open topology, and all that remains is to see that it's also finer? – Daniel Fischer Sep 04 '16 at 14:30
  • @DanielFischer. Exactly. It is clear to me that the topology generated by the sets $L(\overline{U}, V)$ is a subtopology of the compact-open topology on $\widehat{G}$. –  Sep 04 '16 at 14:37

1 Answers1

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Let $\mathscr S = \{L (K,U) \mid K \text{ compact }, U \subset S^1 \text{ open }\}$ and $\mathscr S' = \{L (\overline V,U) \mid V \text{ open with compact closure }, U \subset S^1 \text{ open }\}$.

Clearly, $\scr S' \subset \scr S$, so that $\mathcal T_{\scr S'} \subset \mathcal T_{\scr S}$. To prove the reverse inclusion, it is sufficient to prove that for all $S:=L(K,U) \in \scr S$, and for all $x \in S$, there is $S' \in \scr S'$ such that $x \in S' \subset S=L(K,U)$.

[Indeed, if $U$ is in the topology generated by $\scr B$, then it is the union of some $B_a$'s where each $B_a$ is a finite intersection of elements $S_{a,i}$ of $\scr S$. Then, for fixed $a$ and $i$, $$S_{a,i} = \bigcup\limits_{x \in S_{a,i}} S'_x \in \mathcal T_{\scr S'}$$ where $S'_x \subset S_{a,i}$ is the open set given by the condition given above. Therefore each $B_a$ and then $U$ itself belong to $\mathcal T_{\scr S'}$.]

Let $S :=L(K,U) \ni \gamma$. Notice that $K \subset W:=\gamma^{-1}(U)$. Since $\gamma$ is continuous, $W \subset G$ is open. We just have to find an open set $V \supset K$ whose closure is compact and contained in $W$: $$K \subset V \subset \overline V \subset W$$ This condition will give us $$\gamma \in S':= L(\overline V,U) \subset S$$ as desired.

Recall that $G$ is locally compact (and also Hausdorff, let's say). Then for every point $x$ of $W$, you get some neighborhood $W_x \ni x$ such that $\overline W_x$ is compact and contained in $W$. In particular, $K \subset \bigcup\limits_{x \in K} W_x$ and by compactness of $K$, we get a finite covering, say $K \subset \bigcup\limits_{j=1}^n W_{x_j}=:V$. Then $V$ has the desired properties (that is: $K \subset V \subset \overline V \subset W$).

Watson
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