If $\lbrace y_{k}\rbrace_{k\in K}$ is a net such that $e^{ixy_{k}}\to 1$ pointwise, then $y_{k}\to 0$ Okay, so my brain doesn't seem to be working right now. I'm trying to prove that, given a net $\lbrace y_{k}\rbrace_{k\in K}$ in $\mathbb{R}$ such that $e^{ixy_{k}}$ converges pointwise to $1$ then $y_{k}$ necessarily goes to $0$. I know that if $y_{k}$ is bounded then there's a sufficiently large $n$ such that the whole net is contained in $[-n,n]$. Since this set is compact there must be a convergent subnet, say $y_{k_{j}}$. If we call its limit $y$ then $e^{ixy_{k_{i}}}\to e^{ixy}$ pointwise, so $e^{ixy_{k_{j}}}\to 1$ pointwise. Thus $e^{ixy}=1$ for all $x$. Then it must be the case that $y=0$. Otherwise we could take $x=\dfrac{\pi}{y}$ in $e^{ixy}=e^{i\pi}=-1$, which is a contradiction.
The only thing left to show is that $\lbrace y_{k}\rbrace_{k\in K}$ is bounded. If it wasn't we should be able to construct a subnet $\lbrace y_{k_{j}}\rbrace_{j\in J}$ (I'm pretty sure I can take it to be a sequence) that blows up to either $\infty$ or $-\infty$. In particular, we can assume without loss of generality that it blows up to $\infty$ (since the other case should be analogous). It seems intuitively true that this will lead to a contradiction, since plotting the real part of $e^{iax}$, $\cos(ax)$ for some $a\in\mathbb
{R}$ and letting $a\to\infty$ the resulting set of points appears to coincide (or at least be dense) in the band $(-\infty,\infty)\times[-1,1]$. This seems to imply that the set $(-\infty,\infty)\times\lbrace 1\rbrace$ is dense in this band, which is obviously false.
So, how should I go about proving that the original net $\lbrace y_{k}\rbrace_{k\in K}$ is bounded? Or am I approaching this in the wrong way? For context, I am trying to prove that $\mathbb{R}$ is homeomorphic to $\widehat{\mathbb{R}}$ (its dual group with the corresponding topology) via the identification $y\longmapsto e^{ixy}$. I already know this is a bijection (moreover, I know it defines a group isomorphism) and I know that mapping $y$ to $e^{ixy}$ is continuous. But I've been trying a lot of different approaches to show continuity in the other direction and nothing I've tried so far has worked. The idea I outlined in the post seems the most promising so far, since I know that convergence in the topology of $\widehat{\mathbb{R}}$ implies uniform convergence on compact sets (in particular, it implies pointwise convergence). All of that led me to the approach I described at the beginning.
For reference, I am following Chapter VII of John B. Conway's "A Course in Functional Analysis" (Second Edition). According to the author the fact that $\mathbb{R}$ and $\widehat{\mathbb{R}}$  are homeomorphic is easy to prove, but I've been thinking about this for two days straight and I've honestly ran out of ideas. Maybe I'm just too dumb to see it, I don't know. But I would really appreciate your help.
 A: A net $(y_k)_{k \in K}$ such that $e^{ixy_k} \to 1$ pointwise need not converge to $0$.
Take an arbitrary finite set $\{ x_{\nu} : 1 \leqslant \nu \leqslant n\} \subset \mathbb{R}$ and consider the point
$$\xi = (e^{ix_1}, \dotsc, e^{ix_n}) \in T^n\,.$$
Since $T^n$ is compact, the sequence $(\xi^m)_{m \in \mathbb{N}}$ has a limit point $\eta$. Then for every $\varepsilon > 0$ there are infinitely many $m \in \mathbb{N}$ with $\lvert e^{ix_{\nu}m} - \eta_{\nu}\rvert < \varepsilon$ for all $\nu$. Taking two such $m$, we see that there are arbitrarily large integers $r = m_1 - m_2$ such that $\lvert e^{ix_{\nu}r} - 1\rvert < 2\varepsilon$ for $1 \leqslant \nu \leqslant n$.
Thus if we take for $K$ the directed set $\mathscr{F} \times (0,1) \times \mathbb{N}$, where $\mathscr{F}$ is the set of finite subsets of $\mathbb{R}$, and the preorder is given by
$$(F_1, \varepsilon_1, m_1) \leqslant (F_2, \varepsilon_2, m_2) \iff F_1 \subseteq F_2 \land \varepsilon_2 \leqslant \varepsilon_1 \land m_1 \leqslant m_2$$
we obtain a net $(y_k)_{k \in K}$ such that $y_k \to +\infty$ and $e^{ixy_k} \to 1$ for all $x \in \mathbb{R}$ by setting
$$y_{F, \varepsilon, m} = \min \{ \mu \in \mathbb{N} : \mu \geqslant m \land \lvert e^{ix\mu} - 1\rvert \leqslant \varepsilon \text{ for all } x \in F\}\,.$$
The topology of pointwise convergence is too coarse. You need to use the uniform convergence on compact sets, and the fact that $\mathbb{R}$ is locally compact. Thus convergence in $\widehat{\mathbb{R}}$ implies (is) locally uniform convergence, and for $a > 0$ the set of $y \in \mathbb{R}$ such that $\lvert e^{ixy} - 1\rvert \leqslant \varepsilon$ holds for all $x \in [-a,a]$ is (for small enough $\varepsilon$) easy to determine.
