TITLE: prove that a non constant periodic, continuous function has a "smallest period"
QUESTION [6 upvotes]: Let $\ f:\mathbb{R}\to\mathbb{R} \ $ be a non constant, continuous and periodic function. Prove that $f$ has smallest/minimum period.
The definition of period that I work with is:
$p$ is a period of a function $f$ if $\ p\gt0 \ \land \ \forall x\in\mathbb{R}:f(x+p)=f(x) $.

REPLY [4 votes]: Outline of the proof: Assume by contradiction that there is no smallest period. Use first the fact that the difference between two periods is also a period to show that you can find a decreasing sequence of periods which converge to 0.
This means that for each $\delta >0$ you can find some $T$ period such that 
$$0 < T <\delta$$
Now, pick $x,y$ arbitrary. 
Fix $\epsilon >0$, then there exists a $\delta$ such that for all $z$ with 
$$|y-z| < \delta \Rightarrow |f(y)-f(z)|<\epsilon$$
Pick some $0< T < \delta$. 
Show now that there exists some $n \in \mathbb Z$ such that $|(x+nT)-y|<\delta$.
Then 
$$|f(x)-f(y)|=|f(x+nT)-f(y)| <\epsilon$$
Since this is true for all $\epsilon$ we get $f(x)=f(y)$. As those are arbitrary, you are done.