Is $f(x,y)=\sin{(x)}+\sin{(y)}$ continuous on $\mathbb{R}^2$? We know $\sin{(x)}$ is continuous for all $x\in{\mathbb{R}}$. Let's "define" $f(x,y)=\sin{(x)}+\sin{(y)}$ with $x,y \in \mathbb{R}$. Then as the sum of two continuous functions, we can assure $f$ is continuous on $\mathbb{R}^2$.
Is this correct? I don't see a problem a with it. If it's not correct, how can I know in what subsect of $\mathbb{R}^2$ this function is continuous?
 A: Yes, $f$ is a sum of two continuous functions, so it is indeed continuous on all of $\mathbb{R}^2$, but I think you've missed out one line of reasoning (which is usually obvious to many).
Define $p_1:\mathbb{R^2} \to \mathbb{R}$ by $p_1(\xi, \eta) = \xi$, and define
$p_2:\mathbb{R^2} \to \mathbb{R}$ by $p_2(\xi, \eta) = \eta$. These are the functions which project a tuple onto either the first or second entry. They are linear and hence continuous (it's also good to give a direct $\varepsilon$-$\delta$ proof if you know it) Then, $f$ is a sum of composite functions:
\begin{align}
f(x,y) = (\sin \circ p_1)(x,y) + (\sin \circ p_2)(x,y)
\end{align}
Or, as an equality of functions, we have
\begin{equation}
f = \sin \circ p_1 + \sin \circ p_2
\end{equation}
Since $p_1$ and $\sin$ are continuous, their composition is also continuous; likewise for $p_2$. Hence, their sum is continuous. 
A: Yes, $f$ is continuous. If you want to prove this carefully, you could first prove that the functions $f_1(x,y) = \sin(x)$ and $f_2(x,y) = \sin(y)$ are continuous. One way to prove this is to note that $f_1 = \sin \circ h$, where $h(x,y) = x$. Since $h$ is continuous, and the composition of continuous functions is continuous, we can conclude that $f_1$ is continuous. A similar argument shows that $f_2$ is continuous. Because $f = f_1 + f_2$ is a sum of continuous functions, it follows that $f$ is continuous.
A: For $a,b \in \mathbb R$ we have $|\sin (a)- \sin (b)| \le |a-b|$, by the mean value theorem. Hence
$|f(x,y)-f(u,v)| \le |x-u|+|y-v|=||(x,y)-(u,v)||_1.$
