Suppose $f:\mathbb{R} \to \mathbb{R}$ is uniformly continuous. Show that $f(x+1)-f(x)$ is bounded Suppose $f:\mathbb{R} \to \mathbb{R}$ is uniformly continuous. Show that $f(x+1)-f(x)$ is bounded.
I have considered the question, and my current approach is to show that there exists $\delta > 1$ such that there is some fixed $\epsilon > 0$ for which $|x-p| < \delta \implies |f(x)- f(p)| < \epsilon$, thus bounding $f(x+1)-f(x)$ by this $\epsilon$. Is my current approach correct, and are there any hints or alternative approaches to this question?
Thanks in advance.
 A: Pick an arbitrary $\epsilon>0$, say $1$. By definition of uniform continuity, there is a $\delta>0$ such that for any $d$ with $|d|<\delta$ we have $|f(x+d)-f(x)|<1$.

Side track: You cannot in any way exert control over the $\delta$ you get back. The only thing you know is that it will be positive, and it will be small enough to do what the definition requires from it. For any $\delta$ that does what the definition demands, any $\delta$ that is smaller also does what the definition demands.
Maybe the math gods have vowed to never return a $\delta$ larger than $1/10^{100}$ today. They can do that and still every $\delta$ they give you can do all that is required from it. Maybe they are even meaner. You have no control here. You will have to make do with what you get.

What can you now say about $|f(x+d)-f(x)|$ for any $d$ with $|d|<2\delta$? What about $|d|<3\delta$? Eventually (the exact number of steps depending on $\delta$, of course), we will be allowed to reach a conclusion about $d=1$.
You can turn this argument into the following alternative interpretation: Let's say we want some control over the $\delta$. Let's say we want to know the supremum of all possible values $\delta$ can have for our chosen $\epsilon$ (it exists, but is potentially infinite). Then the above paragraph shows that if we double $\epsilon$, this supremum must at least double as well. At some point this supremum becomes larger than $1$ and we're done.
A: By definition, there exists $N\in\mathbb{N}$ such that $y\in (x-N^{-1},x+N^{-1})$ implies
$$|f(y)-f(x)|<1$$
This same logic applies to $x_k=x+\frac{k}{2N}$:
$$|f(y)-f(x_k)|<1$$
Then
$$ |f(x+1)-f(x)|=\left|f\left(x+\frac{2N}{2N}\right)-f(x)\right|$$
$$=\left|\sum_{k=1}^{2N}\left[f\left(x+\frac{k}{2N}\right)-f\left(x+\frac{k-1}{2N}\right)\right]\right|$$
$$\leq \sum_{k=1}^{2N}\left|f\left(x+\frac{k}{2N}\right)-f\left(x+\frac{k-1}{2N}\right)\right|$$
Since
$$\frac{k}{2N}-\frac{k-1}{2N}=\frac{1}{2N}<\frac{1}{N}$$
we may conclude that the whole sum is bounded by $|f(x+1)-f(x)|<2N$.
