Characters being everywhere dense in the character group Let $k$ be the completion of an algebraic number field at a prime divisor $\mathfrak{p}$. We note that $k$ is locally compact. Let $k^{+}$ be the additive group of $k$ which is a locally compact commutative group.
Tate's Thesis Lemma 2.2.1 states that

If $\xi \rightarrow \chi(\xi)$ is one non-trivial character of $k^{+}$, then for each $\eta \in k^{+}$, $\xi \rightarrow \chi(\eta\xi)$ is also a character. The correspondence $\eta \leftrightarrow \chi(\eta\xi)$ is an isomorphism, both topological and algebraic, between $k^{+}$ and its character group.

The proof of this lemma is divided up into 6 steps, one step is to show that the characters $\chi(\eta\xi)$ are everywhere dense in the character group. Tate writes

$\chi(\eta\xi) = 1$, all $\eta \implies k^{+}\xi \neq k^{+} \implies \xi = 0$. Therefore the characters of the form $\chi(\eta\xi)$ are everywhere dense in the character group.

My question is: How does he get from showing that the $\xi = 0$ to the the result that the $\chi(\eta\xi)$ are everywhere dense?
 A: Denote the character $\xi\rightarrow\chi(\eta\xi)$ by $\chi_\eta$. We want to show that image of the map $f_\chi:\eta\rightarrow\chi_\eta$ is dense in $\hat k$. Take a closed subgroup $H$ of $k$ and set $N_H=\lbrace \xi\in k:\chi_\eta(\xi)=1\ {\rm for\ all}\ \eta\in H\rbrace$. This is also a closed subgroup. We have the short exact sequence
$$0\rightarrow N_H\rightarrow k\rightarrow k/N_H\rightarrow 0$$
and the functoriality of Pontryagin duality turns this into 
$$0\rightarrow \widehat{k/N_H}\rightarrow \widehat{k}\rightarrow \widehat{N_H}\rightarrow 0$$
We have an isomorphism $\widehat{k/N_H}\simeq f_\chi(H)$ (this is basically Theorem 4.39 in Folland's "A Course in Abstract Harmonic Analysis"). Now, setting $H=k$, we see that $N_k=\lbrace 0\rbrace$, so the short exact sequence becomes
$$0\rightarrow f_\chi(k)\rightarrow \widehat{k}\rightarrow 0\rightarrow 0$$
Hence $f_\chi(k)\simeq \widehat k$. This is mildly stronger than what Tate has done at this point, but I'm not worried, since we're incorporating the topology directly in the argument. 
