Problem: Let $ f: \Bbb R \to \Bbb R $ be continuous s.t. the set $ A = \{ T > 0 : f(x+T) = f(x) , \forall x \in \Bbb R \} $ isn't empty. Prove $ f $ is constant $ \iff $ $ inf A = 0 $.
Attempt:
$ (\rightarrow ) $ [ I managed to prove this direction ].
$ (\leftarrow ) $ Suppose $ inf A =0 $. We want to show that $ \forall x,y \in \Bbb R $ s.t. $ x\neq y $ that $ f(x) = f(y) $. Suppose for the sake of contradiction that $ \exists x',y' \in \Bbb R . x' \neq y'$ so that $ f(x') \neq f(y') $. Assume WLOG $ f(x') < f(y') $. Since $ inf A =0 $ then there exists a sequence $ ( x_n) \subset A $ s.t. $ \lim x_n = 0 $. [ Now I'm stuck and don't know how to proceed ].
Can you please help me as to how do I continue my proof? My intuition breaks down and I can't see what can I do next in order to help me in my proof - how would you also recommend getting an intuition for how to continue the proof?
If $f$ is not constant then there is some $x_0$ such that $f(x_0)\ne f(0)$. Let $|f(x_0)-f(0)|=\epsilon>0$. Since $f$ is continuous at $0$, we have:
$\exists\delta>0 \ \ (\forall x: |x|<\delta)\ \ [\ \ |f(x)-f(0)|<\epsilon]$
In paricular, if $x\in [0,\delta]$ then $f(x)\ne f(x_0)$. So now we will show that if $0<T<\delta$ then $T$ doesn't belong to the set $A$. (which will obviously imply that $\inf(A)>0$). So let $0<T<\delta$ and suppose that $T\in A$. Note that for $m=\lfloor\frac{x_0}{T}\rfloor$ we have $x_0-mT\in [0,T]\subseteq [0,\delta]$. As we have seen, this implies $f(x_0)\ne f(x_0-mT)$. But this is contradicting the assumption $T\in A$.