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$X$ = $\mathbb{R}^n$ and

$f:X \times X \rightarrow X$

$($x$ , $y$) \rightarrow x+y$

This is for addition on $\mathbb{R}^n$, but why here $f$ can be continuous?

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    Try to follow the definition. – 2017-02-19
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    Because it is? Do you know what the definition of continuous is? Can you apply to definition? If not, why not? Where do you start to have trouble? – 2017-02-19
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    Are $X$ and $X \times X$ being considered as a metric spaces? If so, for what distances? – 2017-02-19

2 Answers 2

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Assuming you constructed two metric spaces $(X, d)$ and $(X*X, d')$ where $d'(u,v)=d'((u_1, u_2),(v_1,v_2))=\sqrt{d(u_1,v_1)^2+d(u_2,v_2)^2}$ for every $u,v \in X*X$.

For every sequence $\{ (x_n,y_n) \}$ that $(x_n,y_n)\rightarrow (x,y) $ in $X*X$ we have $x_n\rightarrow x$ and $y_n\rightarrow y$ in $X$, which implies that $f(x_n,y_n)=x_n+y_n\rightarrow x+y=f(x,y)$.

I hope my answer will help you well.

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    Sorry I forgot to conclude that $f$ is continuous on $X*X$ – 2017-02-19
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    you can use \times for $\times$ – 2017-03-03
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More generally, if $(E, \| \|)$ is a normed vector space, $f$ is Lipschitz continuous: in fact, if $(x_1, y_1),(x_2,y_2) \in E\times E$, where we consider the following distance on $E\times E$ $$ d( (x_1, y_1),(x_2,y_2))= \|x_1 -x_2 \| + \|y_1-y_2\| $$ then $$ \|f(x_1, y_1)-f(x_2,y_2)\| = \|x_1+y_1-(x_2+y_2)\| = \| (x_1-x_2) + (y_1 -y_2)\| \leq \| (x_1-x_2) \| + \| (y_1 -y_2)\|= d( (x_1, y_1),(x_2,y_2)).$$ Hence, $f$ is $1-$Lipschitz continuous which implies continuity.