Prove that a specific continuous function is bounded Let $f:$ $\mathbb{R}$ $\to$ $\mathbb{R}$ be a continuous function such that $f(x +1) = f(x)$ for all $x \in \mathbb{R}$. Show that $f$ is bounded.
My first thought is that $f(x)$ is a constant function, so it will be clearly limited. But, I can't link the definition of continuous
( $\forall \varepsilon>0\exists\delta>0(\forall y \in \mathbb{R}, || x - y||<\delta\rightarrow ||f(x) - f(y)||<\varepsilon$)   with it, because there isn't any relation that I can figure out between $x,y, \delta$. Could someone help me how to do it? Thanks
 A: Suppose the function is not bounded from above. Then for each $n>0$ there must be an $x_n\in \Bbb R$ such that $f(x_n)>n$. Since the function is periodic we can find one of such $x_n$'s in the interval $[0,1]$. In this way we can construct a sequence $(x_n)_{n\in\Bbb Z^+}$, that is bounded. By Completeness principle, therefore, $(x_n)$ must have a converging subsequence, say $(z_n)$, whose limit, say $\overline z$, must also lie in the closed interval $[0,1]$. But this contradicts continuity since $(z_n) \to \overline z$ with $(f(z_n)) \not\to f(\overline z)$. $\blacksquare$

EDIT Without Completeness the assertion becomes obviously false. Consider, for instance, in $\Bbb Q$ the function
$$g(x) = \begin{cases}\frac{1}{|2x^2-1|} & (0 \leq x <1) \\ 0 & (\mbox{otherwise})\end{cases}$$
and generate from it the function $f:\Bbb Q \to \Bbb Q$
$$f(x) = \sum_{k=-\infty}^{+\infty} g(x-k).$$
This function is continuous in $\Bbb Q$ and periodic with period $1$, and yet it is unbounded. 
A: Attempt:
1) Since $f$ is continuous it attains its max and min on the compact (closed and bounded) interval $[0,1]$.
Hence $f$ is bounded on $[0,1].$
2) Let $x >1$, real.
There is a $n \in \mathbb{Z^+}$ s.t.
$n \le x <n+1$, or 
$0 \le x-n <1$.
3) $f(x+1)=f(x)$, i.e. is periodic with period $1$.
$f(x)=f(x-1)$, successive applications give $f(x)=f(x-n)$.
4) Hence for $x >1$ , with $(x-n) \in [0,1]$
$f(x)=f(x-n)$, bounded, 
A similar argument for $x <0.$
