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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?

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    $\begingroup$ You have managed to spell "continuous" four different ways in one short post. Congratulations! More to the point, your conclusion is correct, but there's a gap in the supporting reasoning. To use the theorem about the sum of two continuous functions, the functions and their sum must have the same domain. So you must think of the first function as $g(x,y)=\sin x$, the second as $h(x,y)=\sin y$, and reason that these two functions are continuous, before you apply the theorem. $\endgroup$ – Gerry Myerson Jun 6 at 2:53
  • $\begingroup$ This is correct. Since limits of functions of several variables are defined using the so-called $\varepsilon-\delta$ definition of limits, the Limits Laws are valid for several variables. This implies that the sum, multiplication by a scalar, products, quotient (when defined) and composition of continuous functions are continuous. $\endgroup$ – Taladris Jun 6 at 2:53
  • $\begingroup$ @Tal, but see my reservations about the domains. $\endgroup$ – Gerry Myerson Jun 6 at 2:54
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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.

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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.

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For $a,b \in \mathbb R$ we have $|\sin (a)- \sin (b)| \le |a-b|$, by the mean value theorem. Hence

$|f(x,y)-f(u,v)| \le |x-u|+|y-v|=||(x,y)-(u,v)||_1.$

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