Uniform Continuity on an unbounded domain The function $f(x) = sin(x)cos(x)$ is continuous on $(-\infty, \infty)$. Is it uniformly continuous? 
-No, since the domain is not closed or bounded, the domain is not compact, thus the function is not uniformly continuous on  $(-\infty, \infty)$. 
-I'm not over simplifying this correct? 
 A: From the definition of uniform continuity, $f$ is uniformly continuous on a domain $D$ if for any $\epsilon>0$, there exists a $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $x,y\in D$ satisfy the relation $|x-y|<\delta$.
Knowing this, we proceed to find such a $\delta$ for any given $\epsilon$. From the Mean Value Theorem, we know that if $f$ is continuously differentiable, then for any $x\ne y$, there exists $\xi$ between $x$ and $y$ such that
$$
f(y)-f(x)=f'(\xi)(y-x)
$$
which implies that
$$
|f(y)-f(x)|=|f'(\xi)||y-x|.
$$
Now, we should be almost finished if we can find a bound for $|f'(\xi)|$, but recall that 
$$
|f'(x)|=|\cos^2(x)-\sin^2(x)|\le|\cos^2(x)|+|\sin^2(x)|\le1+1=2\ ,
$$
combining the 2 previous inequalities, we have
$$
|f(x)-f(y)|\le 2|x-y|.
$$
Choose $\delta=\frac \epsilon2$ and we are done.
A: You're not correct.  Let $f\colon X\to\mathbb R$ be continuous.   Then it is true that

\begin{align}
X\textrm{ is compact  }&\Longrightarrow f\textrm{ is uniformly continuous}
\end{align}

However, the converse is not true:

\begin{align}
f\textrm{ is uniformly continuous  } &\mathrel{\rlap{\hskip .5em/}}\Longrightarrow X\textrm{ is compact}
\end{align}

In your case, you have a domain $X$ that is not compact, but the function $f$ is nevertheless uniformly continuous - for example, because it is differentiable and its derivative is bounded.
A: Cleary a constant function on $\mathbb R$ is uniformly continuous on $\mathbb R.$ So uniform continuity does not imply the domain is compact.
The answer to your specific problem follows from a general result: Every continuous periodic function on $\mathbb R$ is uniformly continuous on $\mathbb R.$ How could this not be true? If $|y-x|< p,$ where $p$ is the period, then then there exists an integer $n\in \mathbb Z$ such that $x-np,y-np \in [0,2p].$ Now $f$ is continuous on $[0,2p],$ hence is uniformly continuous there. The result follows.
