$\mathbb R^n$ is a topological group.

Question

A group $(G,\cdot)$ is said to be topological if the binary operations $(a,b)\mapsto ab$ and $a\mapsto a^{-1}$ are continuous. Further it becomes a Lie groups if the corresponding binary operations are smooth. I want to show that $\mathbb{R}^n$ is a topological group. How can I show that the point wise addition is continuous?

I am trying to show that the inverse image of any open set in $\mathbb{R}^n$ is open $\mathbb{R}^n\times \mathbb{R}^n$ is open. Someone suggested me that it is a polynomial map so continuous. But I can not see this as a polynomial map.

Answer 1

The map in question is$$\begin{array}{rccc}a\colon&\Bbb R^n\times\Bbb R^n&\longrightarrow&\Bbb R^n\\&\bigl((a_1,\ldots,a_n),(b_1,\ldots,b_n)\bigr)&\mapsto&(a_1+b_1,\ldots,a_n+b_n).\end{array}$$It is continuous because, for each $k\in\{1,2,\ldots,n\}$, the map$$\begin{array}{rccc}a_k\colon&\Bbb R^n\times\Bbb R^n&\longrightarrow&\Bbb R\\&\bigl((a_1,\ldots,a_n),(b_1,\ldots,b_n)\bigr)&\mapsto&a_k+b_k\end{array}$$is continuous. And this is trua because $a_k$ is the sum of$$\begin{array}{rccc}a_{k,1}\colon&\Bbb R^n\times\Bbb R^n&\longrightarrow&\Bbb R\\&\bigl((a_1,\ldots,a_n),(b_1,\ldots,b_n)\bigr)&\mapsto&a_k\end{array}$$with$$\begin{array}{rccc}a_{k,2}\colon&\Bbb R^n\times\Bbb R^n&\longrightarrow&\Bbb R\\&\bigl((a_1,\ldots,a_n),(b_1,\ldots,b_n)\bigr)&\mapsto&b_k,\end{array}$$both of which are continuous.

Answer 2

Hint

To show that a map $f(x)=(f_1(x), \dots, f_n(x))$ taking values in $\mathbb R^n$ is continuous, it is enough to prove that each $f_i(x)$ is continuous. And here each $f_i$ is the usual addition.