= Solution
A real-valued function $f$ on an interval $I$ is <uniformly continuous> when
$$
\forall\varepsilon>0\ \exists\delta>0\ \forall x,y\in I:\quad |x-y|<\delta\ \Longrightarrow\ |f(x)-f(y)|<\varepsilon.
$$
The essential point in <uniform continuity> is that $\delta$ is independent of the locations of $x$ and $y$. \b[Uniform continuity on the whole real line does not imply boundedness]: $f(x)=x$ is <Lipschitz continuous> with constant $1$, hence <uniformly continuous>, but is unbounded.
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