Boundedness of periodic functions Recently I was reading answers to a question of functional equations here,
and I saw a claim on a property of periodic functions which was strange to me: 

we then have $f$ to be periodic with period $f(0)$. This means $f(n)$ can take only finitely many values.   

I have never heard of this property and failed to find one reference to it! Also I have a counter example to this claim: $tan(x)$ is also periodic but it's not BOUNDED and can take INFINITELY many values as $x$ approaches $\frac{\pi}{2}$.
Is this claim true? Is there any proof to it or does it have any additional conditions such as continuity ?!
 A: Assume that $f(x)$ is defined in $[0,\infty)$ (to give evaluations for $f(n)$, $n=0,1,2,3,\ldots$) and $T$ periodic. Assume that $T=\left[T\right]+\left\{T\right\}$, where $\left[T\right]$ is the integer part of $T$ and $\left\{T\right\}$ is the  fractional part of $T$. Then if $T$ is rational, $0\leq\left\{T\right\}<1$ and exist positive integers $\mu,\nu$ such $0\leq\left\{T\right\}=\frac{\mu}{\nu}<1$. In this case set $k=[T]$. Then the sequence
$$
t=\left\{f(0),f(1),f(2),f(3),\ldots,f(k-1),f(k),f(k+1),f(k+2),\ldots\right\}
$$ 
will be
$$
\left\{f(0),f(1),f(2),f(3),\ldots,f([T]-1),f([T]),f([T]+1),f([T]+2),\ldots\right\}
$$
But 
$$
f(k)=f(T-\left\{T\right\})=f(-\left\{T\right\}).
$$
$$
f(k+1)=f([T]+\left\{T\right\}+1-\left\{T\right\})=f(T+1-\left\{T\right\})=f(1-\left\{T\right\}),
$$
$$
0\leq\left\{T\right\}<1.
$$ 
Hence the period lead us back to (near) start . In the same way 
$$
f(k+2)=f([T]+2)=f([T]+\left\{T\right\}+2-\left\{T\right\})=f(T+2-\left\{T\right\})=f(2-\left\{T\right\}).
$$
$$
f(2k+1)=f(2[T]+2\left\{T\right\}+1-2\left\{T\right\})=f(1-2\left\{T\right\}),
$$
$$
f(2k+2)=f(2-2\left\{T\right\}),
$$
$$
f(2k+3)=f(3-2\left\{T\right\}),
$$
$$
\ldots
$$
$$
f(3k)=f(3[T]),
$$
$$
f(3k+1)=f(1-3\left\{T\right\}),
$$
$$
f(3k+2)=f(2-3\left\{T\right\}),\ldots,
$$ 
Until we meet 
$$
f(\mu-\nu\left\{T\right\})=f(0).
$$
Hence when $T=rational$, the periodic function $f$ will have repeated "pattern" $t$. 
In case $n_1,n_2$ positive integers with $n_1<n_2$ and $f(n_1)=f(n_2)$, we have again repeated pattern 
$t_1=\left\{f(0),f(1),\ldots,f(n_1-1),f(n_1),f(n_1+1),\ldots,f(n_2-1)\right\}$, since, in this case there exist two positive integers $k_1<k_2$ such  $n_1=k_1T+\theta_1$, $n_2=k_2T+\theta_2$, with $0\leq\theta_1<T$ and $0\leq\theta_2<T$. Hence from $f(n_1)=f(n_2)$, we get $f(\theta_1)=f(\theta_2)$, where $\theta_{1,2}\in[0,T)$.
Hence
Theorem 1.
If $f$ is ''1-1'' in $[0,T)$ and periodic with period $T>0$, then:
$t$ have no repeated values and all $f(n)$ are different (hence infinite in number). In this case also $T=irrational$.
Examples.
1) $f(x)=x-\pi\left[\frac{x}{\pi}\right]$
2) $f(x)=\tan(x)$
