How to show that $f(x,y)=x+y$ and $g(x,y)=xy$ are continuous functions?

Question

A function $\phi:X\rightarrow Y$ between two topological space $(X,\tau)$ and $(Y,\sigma)$ is continuous in $x\in X$ if and only if for any open set $V$ in $Y$ such that $\phi(x)\in V$ there is a open set $U$ in $X$ such that $x\in U$ and $\phi(U)\subseteq V$.

Well with this definition we consider the euclidean space $\mathbb{R}^2$ and the functions \begin{align}f&:\mathbb{R}^2\owns(x,y)\rightarrow(x+y)\in\mathbb{R}\\g&:\mathbb{R}^2\owns(x,y)\rightarrow(xy)\in\mathbb{R}\end{align} and we prove to demonstrate that are continuous functions.

Therefore we consider the point $(\alpha,\beta)$ and its image under $f$ or $g$ and we chose a open range $(a,b)$ that contains $f(\alpha,\beta)$ and $g(\alpha,\beta)$: so it result that $a<\alpha+\beta<b$ and $a<\alpha\beta<b$.

Now let be $$ \varepsilon_f=\inf\Big\{d\big((\alpha,\beta);(x,a-x)\big),d\big((x,y);(x,b-x)\big)\Big\}\\ \varepsilon_g=\inf \biggl\{d\Big((\alpha,\beta);\Big(x,\frac{a}{x}\Big)\Big),d\biggl((x,y);\Big(x,\frac{b}{x}\Big)\Big)\biggl\}. $$ If we consider the balls $B_f:=B\big((\alpha,\beta),\varepsilon_f\big)$ and $B_g:=B\big((\alpha,\beta),\varepsilon_g\big)$ that contain $(\alpha,\beta)$ is true that $f(B_f)\subseteq(a,b)$ and $g(B_g)\subseteq(a,b)$ and that if $(x,y)\in B_f$ or $(x,y)\in B_g$ it result that $f(x,y)\in(a,b)$ or $g(x,y)\in(a,b)$? Then if this is true will we have proved the continuity of $f$ and $g$?

Could someone help me, please?

Answer 1

So it is a well know result that any normed space is a topologcial vector space so that the statement could follows immediately observig that the usual multiplication on $\Bbb R$ is an inner product and thus rememering that any inner product space is a normed space too. Anyway for sake of completeness we let prove to follow the continuity of $s$ and $p$ directely.

Let be $p:\mathbb{R}^2\owns(x,y)\rightarrow xy\in\mathbb{R}$ and we prove that it is continuous. So let be $(a,b)\in\mathbb{R}$ and $\epsilon>0$ and we define $$ \delta=\mathscr{min}\biggl\{\frac{\epsilon}{2(1+|b|)},1\biggl\} $$ Since the euclidean topology is the same of the rectangle topology we consider the open rectangle $$ V=\biggl\{(x,y)\in\mathbb{R}^2:|a-x|<\delta, |b-y|<\frac{\epsilon}{2(1+|a|)}\biggl\} $$ that contains the point $(a,b)$. Now previously we observe that if $|a-x|<1$ then it result that $$ -(|a|+1)\le a-1<x<a+1\le(|a|+1) $$ or rather $|x|<|a|+1$. So for some $\epsilon>0$ we consider the basic neighborhood $B(p(x,y),\epsilon)$ of $p(x,y)$ and we observe that for any $(x,y)$ in $V$ it result that $$ |p(x,y)-p(a,b)|=|xy-ab|=|x(y-b)+b(x-a)|\le|x||y-b|+|b|x-a|<(|a|+1)\frac{\epsilon}{2(|a|+1)}+(|b|+1)\frac{\epsilon}{2(|b|+1)}=\epsilon $$ and so $p$ is continuous.

Now let $s:\mathbb{R}^2\owns(x,y)\rightarrow(x+y)\in\mathbb{R}$ and we prove that it is continuous function. Remembering what we observed before, for some $(a,b)\in\mathbb{R}^2$ and $\epsilon>0$ we consider the ope n rectangle $$ V=\biggl\{(x,y)\in\mathbb{R}^2:|x-a|<\frac{\epsilon}2, |y-b|<\frac{\epsilon}2\biggl\} $$ that contains the point $(a,b)$. So for some $\epsilon>0$ we consider the basic neighborhood $B(s(a,b),\epsilon)$ of $s(a,b)$ and we observe that for any $(x,y)$ in $V$ it result that $$ |s(x,y)-s(a,b)|=|(x+y)-(a+b)|=|(x-a)+(y-b)|\le|x-a|+|y-b|<\frac{\epsilon}2 +\frac{\epsilon}2=\epsilon$$ and so the function $s$ is continuous.

Answer 2

Without going through your computation, I think in general for metric vector spaces over $\mathbb{R}$ or $\mathbb{C}$, the maps $(x,y)\mapsto$ and $(x,y)\mapsto xy$ are Lipschitz continuous, and as such continuous. The reason why I mention this is that Lipschitz continuity is a lot simpler to check then to go through the definition of continuity each time in my opinion.