A compact identity neighbourhood in the dual

$N$ is a neighbourhood of the identity character: the constant character $1$ satisfies $1[K]=\{1\}\subseteq D$ by [F1], so $1\in N$; and the open arc $D^{\circ }=\{z\in \mathbb{T}:\mathrm{\mid }z-1\mathrm{\mid }<1\mathrm{/}2\}$ contains $1$. Hence $S(K,D^{\circ })$ is a subbasic compact-open neighbourhood of the identity by [F4], and $S(K,D^{\circ })\subseteq N$ because $D^{\circ }\subseteq D$. Thus $N$ is a neighbourhood, although it need not be open.

$N$ is equicontinuous: fix $\epsilon >0$ and put $O:=B(1,\epsilon )\cap \mathbb{T}$, an open neighbourhood of $1$ in $\mathbb{T}$; for $k\ge 1$ put $F_k:=\{z\in \mathbb{T}:z^j\in D\text{ for }j=1,\dots ,k\}$. Each $F_k$ is closed, being a finite intersection of preimages of the closed set $D$ under the power maps $z\mapsto z^j$, which are continuous by induction on $j$ from the continuity of multiplication on $\mathbb{T}$ by [F1]; the sequence $F_1\supseteq F_2\supseteq \cdots$ has intersection $\{1\}$ by [F2], because any $z\mathrm{\ne }1$ satisfies $z^j\mathrm{\notin }D$ for some $j$. Hence $F_k\subseteq O$ for some $k$: otherwise the sets $F_k\setminus O$ would be nonempty closed sets with the finite intersection property, being decreasing, and [F3] would produce $z\in {\bigcap }_k(F_k\setminus O)\subseteq \{1\}\setminus O=\mathrm{\varnothing }$.

With $k$ as in step 1.2 put $U:=\{u\in G:ju\in K\text{ for }j=0,1,\dots ,k\}$, a neighbourhood of $0$: for $j\ge 1$ the set $\{u:ju\in K\}$ contains the open set $\{u:ju\in K^{\circ }\}$, which is the preimage of the open $K^{\circ }$ under the continuous map $u\mapsto ju$ and contains $0$, while $0\in K$ for $j=0$; a finite intersection of neighbourhoods of $0$ is a neighbourhood of $0$ by [F8]. For $u\in U$ and $\gamma \in N$ one has $\gamma (u{)}^j=\gamma (ju)\in D$ for $j=1,\dots ,k$, so $\gamma (u)\in F_k\subseteq O$, that is $\mathrm{\mid }\gamma (u)-1\mathrm{\mid }<\epsilon$. Thus $N$ is equicontinuous at $0$.

$N$ is equicontinuous at every point $x_0\in G$: for $x-x_0\in U$ and $\gamma \in N$, $\gamma (x)=\gamma (x_0)\gamma (x-x_0)$ and hence $\mathrm{\mid }\gamma (x)-\gamma (x_0)\mathrm{\mid }=\mathrm{\mid }\gamma (x_0)\mathrm{\mid }\mathrm{\mid }\gamma (x-x_0)-1\mathrm{\mid }=\mathrm{\mid }\gamma (x-x_0)-1\mathrm{\mid }<\epsilon$ by step 2.1 and [F1].

$N$ is closed in $C(G,\mathbb{T})$ for the compact-open topology: let $\gamma$ lie in the compact-open closure of $N$. The compact-open topology is finer than the pointwise topology by [F4], so $\gamma$ lies in the pointwise closure of $N$; by [F6] some net $({\gamma }_j)$ in $N$ converges to $\gamma$ pointwise. All ${\gamma }_j$ are characters and the family $N$ is equicontinuous by step 3.1, so $\gamma$ is a character by [F5]; and $\gamma [K]\subseteq D$ because each ${\gamma }_j[K]\subseteq D$ and $D$ is closed by [F1]. Hence $\gamma \in N$ and $N$ is compact-open closed.

$N$ is compact in the compact-open topology: $N\subseteq C(G,\mathbb{T})$ is equicontinuous by step 3.1 and $\mathbb{T}$ is a compact metric space by [F1], so the compact-open closure of $N$ is compact by [F7]; by step 4.1 that closure is $N$ itself.

By steps 1.1, 3.1 and 5.1 the set $N$ is equicontinuous, compact in the compact-open topology, and a neighbourhood of the identity character in $\widehat{G}$.