$f :\mathbb R\to \mathbb R$ is a continuous function of period $1$ , then $f$ is uniformly continuous on $\mathbb R$ Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?
 A: Take any interval $[x,x+1]$. Then f is a continuous function defined on a compact interval, so it is uniformly-continuous there. Then use the pair $\delta -\epsilon$ that works in $[0,1]$ anywhere in the Real line by periodicity of $f$. Another way of looking at this is that this is a function defined on the compact set $[0,1]$/~ , where $0$~$1$, which is homeomorphic to $S^1$, a compact space. Then $f$ is continuous, defined on a compact space, so uniformly-continuous.
A: First take $I = [0,2]$. Since $I$ is compact and $f$ is continuous on $I$ we know $f$ is uniformly continuous on $I$. Now let $x,y \in \mathbb{R}$. Let $\epsilon > 0$, since $f$ is uniformly continuous we know $\forall z,w \in I \; \exists \delta' > 0$ where
$$
|z - w| < \delta' \implies |f(x) - f(y)| < \epsilon
$$
Now put $\delta = \min \{ \delta', 1 \}$. For every $x,y \in \mathbb{R}$ so that
$$
| x - y | < \delta
$$
we know that $\exists n \in \mathbb{Z}$ so that $x + n,y+n \in I$. So, finally
$$
\begin{eqnarray}
| x - y | < \delta & \implies & | x + n - y - n | < \delta \\
& \implies & | x + n - y - n | < \delta' \text{ since } \delta \le \delta'\\
& \implies & | f(x+n) - f(y + n) | < \epsilon \\
& \implies & |f(x) - f(y)| < \epsilon
\end{eqnarray}
$$
