A random variable $X$ has a lattice distribution if the support of $X$ is $\{a+nb: n\in Z\}$. Let $X$ have characteristic function $\phi$.
I have proved earlier that $X$ has a lattice distribution $\iff$ $|\phi(t)| = 1$ for some $t$
The question now is if $|\phi(t)| = 1$ and $|\phi(t')| = 1$ for $t/t'=irrational$, prove that $P(X=c) = 1$.
My attempt:
$|\phi(t)| = 1\iff E[e^{iXt}]=1 \iff tX$ mod $2\pi = \theta \iff X = \frac{\theta}{t}+\frac{2\pi n}{t}$
$|\phi(t')| = 1\iff E[e^{iXt'}]=1 \iff t'X$ mod $2\pi = \theta' \iff X = \frac{\theta'}{t'}+\frac{2\pi n}{t'}$
When $X = \frac{\theta'}{t'}+\frac{2\pi n}{t'}, E[e^{iXt}] = E[e^{it(\frac{\theta'}{t'}+\frac{2\pi n}{t'})}] = 1\iff \frac{t}{t'}\in Z$.
By symmetry, $\frac{t'}{t}\in Z\implies t'=-t$ because $t\neq t'$
So why can't the support of $X$ be $\{\frac{\theta}{t}+\frac{2\pi n}{t}:n\in Z\}\cup \{-\frac{\theta'}{t}-\frac{2\pi n}{t}:n\in Z\}$ and why does $X$ have to be a single value a.s.? Thanks and appreciate a hint!