A periodic function with no fundamental period and continous at one point is constant. Theorm : Let  $f:\mathbb{R} \to \mathbb{R}$ be a periodic function and suppose $f$ is continous at some $\zeta \in \mathbb{R}$ and that $f$ has no fundamental period then prove that $f$ is constant .
My trial proof using sequences
Let $\{p_n\}$ be a decreasing sequence of periods of $f$ converging to $0$.
If $f$ is not constant then  $\exists $  a point $a$  such that $f(a) \neq f(\zeta)$.
Let $ a\gt \zeta$.
There exist  $ m\in \mathbb{N}$ such that $0\lt p_n \lt a-\zeta, \forall n \gt m$
We choose $x_1, x_2 , ..., x_m$
as the same real number $a$
For $n\gt m$, we select  $x_n \in (\zeta, \zeta+p_n) $ such that $f(x_n)=f(a)$ which is possible by the periodicity of $f$
Clearly $x_n \to \zeta$ as $n\to \infty$ but the corresponding functional sequence $f(x_n)=f(a)\to f(a)\neq f(\zeta) $ as $n\to \infty$ thus contradicting that $f$ is continous at $\zeta$
Similar technique for $a\lt \zeta$
Thus there is no such $a$ and so the result follows.
I know there are several questions like this posted here but as far as I have seen none of them use sequences.
My proof looks too simple . Is everything correct or am I overlooking something?
Thanks for your time.
 A: Your proof seems correct, but there is a much simpler way to prove the statement.
Fix $x \in \mathbb{R}$, and let $\varepsilon > 0$.
Because $f$ is continuous at $\zeta$, there exists $\eta > 0$ such that for all $y \in [\zeta-\eta, \zeta + \eta]$, $|f(y)-f(\zeta)| \leq \varepsilon$. Let $T$ be a period of $f$ such that $0< T < 2\eta$. There exists $N \in \mathbb{Z}$ such that $x + NT \in [\zeta-\eta, \zeta + \eta]$, so you deduce that $|f(x)-f(\zeta)| = |f(x+NT)-f(\zeta)|  \leq \varepsilon$. Because this has to be true for all $\varepsilon > 0$, you deduce that $f(x)=f(\zeta)$.
Hence $f$ is constant.
A: You can further simplify  by showing that if $P$ is the set of periods of $f$ and if $a\in \Bbb R$ then the set $S(a)=\{a+mp: m\in \Bbb Z\land p\in P\}$ is dense in $\Bbb R.$ And if $x\in S(a)$ then (obviously) $f(x)=f(a).$ By the denseness of $S(a)$ there exists a sequence $(x_j)_{j\in \Bbb N}$ of members of $S(a)$ that converges to $\zeta.$ Hence $f(\zeta)=\lim_{j\to \infty}f(x_j)=\lim_{j\to \infty}f(a)=f(a).$
Addendum: To show that $S(a)$ is dense: Suppose $b<c.$ Take $p\in P\cap (0,c-b). $ There exist $n_1,n_2\in \Bbb N$ with $a-n_1p\le b$ and $a+n_2p\ge c.$
Let $m_0 =\max \{m\in \Bbb Z: -n_1\le  m<n_2 \land a+mp\le b\}.$
Then $a+m_0p\le b<a+(1+m_0)p<c$ (because $0<p<c-b) \,$). So $a+(1+m_0)p\in S(a)\cap (b,c).$
